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the hole is small enough it acts like a point source of circular waves. This bending around the edges of the hole is called diﬁraction. To illustrate this behaviour we start by with Huygen’s principle. Huygen’s Principle Huygen’s principle states that each point on a wavefront acts like a point source or circular waves. The waves emitted from each point interfere to form another wavefront on which each point forms a point source. A long straight line of points emitting waves of the same frequency leads to a straight wave front moving away. To understand what this means lets think about a whole lot of peaks moving in the same direction. Each line represents a peak of a wave. If we choose three points on the next wave front in the direction of motion and make each of them emit waves isotropically (i.e. the same in all directions) we will get the sketch below: What we have drawn is the situation if those three points on the wave front were to emit waves of the same frequency as the moving wave fronts. Huygens principle says that every point on the wave front emits waves isotropically and that these waves interfere to form the next wave front. To see if this is possible we make more points emit waves isotropically to get the sketch below: 28 You can see that the lines from the circles (the peaks) start to overlap in straight lines. To make this clear we redraw the sketch with dashed lines showing the wavefronts which would form. Our wavefronts are not perfectly straight lines because we didn’t draw circles from every point. If we had it would be hard to see clearly what is going on. It Huygen’s principle is a method of analysis applied to problems of wave propagation. recognizes that each point of an advancing wave front is in fact the center of a fresh disturbance and the source of a new train of waves; and that the advancing wave as a whole may be regarded as the sum of all the secondary waves arising from points in the medium already traversed. This view of wave propagation helps better understand a variety of wave phenomena, such as diﬁraction. Wavefronts Moving Through an Opening Now if allow the wavefront to impinge on a barrier with a hole in it, then only the points on the wavefront that move into the hole can continue emitting forward moving waves - but because a lot of the wavefront have been removed the points on the edges of the
hole emit waves that bend round the edges. 29 The wave front that impinges (strikes) the wall cannot continue moving forward. Only the points moving into the gap can. If you employ Huygens’ principle you can see the eﬁect is that the wavefronts are no longer straight lines. For example, if two rooms are connected by an open doorway and a sound is produced in a remote corner of one of them, a person in the other room will hear the sound as if it originated at the doorway. As far as the second room is concerned, the vibrating air in the doorway is the source of the sound. The same is true of light passing the edge of an obstacle, but this is not as easily observed because of the short wavelength of visible light. This means that when waves move through small holes they appear to bend around the sides because there aren’t enough points on the wavefront to form another straight wavefront. This is bending round the sides we call diﬁraction. 2.3.7 Properties of Waves : Dispersion Dispersion is a property of waves where the speed of the wave through a medium depends on the frequency. So if two waves enter the same dispersive medium and have diﬁerent frequencies they will have diﬁerent speeds in that medium even if they both entered with the same speed. We will come back to this topic in optics. 2.4 Practical Applications of Waves: Sound Waves 2.4.1 Doppler Shift The Doppler shift is an eﬁect which becomes apparent when the source of sound waves or the person hearing the sound waves is moving. In this case the frequency of the sounds waves can 30 be diﬁerent. This might seem strange but you have probably experienced the doppler eﬁect in every day life. When would you notice it. The eﬁect depends on whether the source of the sound is moving away from the listener or if it is moving towards the listener. If you stand at the side fo the road or train tracks then a car or train driving by will at ﬂrst be moving towards you and then away. This would mean that we would experience the biggest change in the eﬁect. We said that it eﬁects the frequency of the sound so the sounds from the car or train would sound diﬁerent, have a diﬁerent
frequency, when the car is coming towards you and when it is moving away from you. Why does the frequency of the sound change when the car is moving towards or away from you? Lets convince ourselves that it must change! Imagine a source of sound waves with constant frequency and amplitude. Just like each of the points on the wave front from the Huygen’s principle section. Remember the sound waves are disturbances moving through the medium so if the source moves or stops after the sound has been emitted it can’t aﬁect the waves that have been emitted already. The Doppler shift happens when the source moves while emitting waves. So lets imagine we have the same source as above but now its moving to the right. It is emitting sound at a constant frequency and so the time between peaks of the sound waves will be constant but the position will have moved to the right. In the picture below we see that our sound source (the black circle) has emitted a peak which moves away at the same speed in all directions. The source is moving to the right so it catches up a little bit with the peak that is moving away to the right. 31 When the second peak is emitted it is from a point that has moved to the right. This means that the new or second peak is closer to the ﬂrst peak on the right but further away from the ﬂrst peak on the left. If the source continues moving at the same speed in the same direction (i.e. with the same velocity which you will learn more about later). then the distance between peaks on the right of the source is the constant. The distance between peaks on the left is also constant but they are diﬁerent on the left and right. This means that the time between peaks on the right is less so the frequency is higher. It is higher than on the left and higher than if the source were not moving at all. On the left hand side the peaks are further apart than on the right and further apart than if the source were at rest - this means the frequency is lower. So what happens when a car drives by on the freeway is that when it approaches you hear higher frequency sounds and then when it goes past you it is moving away so you hear a lower frequency. 2.4.2 Mach Cone Now we know that the waves move away from the source at the speed of sound. What happens if the source moves at the speed of sound? This means that the wave
peaks on the right never get away from the source so each wave is emitted on top of the previous on on the right hand side like in the picture below. 32 If the source moves faster than the speed of sound a cone of wave fronts is created. This is called a Mach cone. Sometimes we use the speed of sound as a reference to describe the speed of the object. So if the object moves at exactly the speed of sound we can say that it moves at 1 times the speed of sound. If it moves at double the speed of sound then we can say that it moves at 2 times the speed of sound. A convention for this is to use Mach numbers, so moving at the speed of sound is Mach one (one times the speed of sound) and moving at twice the speed of sound is called Mach two (twice the speed of sound). 2.4.3 Ultra-Sound Ultrasound is sound with a frequency greater than the upper limit of human hearing, approximately 20 kilohertz. Some animals, such as dogs, dolphins, and bats, have an upper limit that is greater than that of the human ear and can hear ultrasound. Ultrasound has industrial and medical applications. Medical ultrasonography can visualise muscle and soft tissue, making them useful for scanning the organs, and obstetric ultrasonography is commonly used during pregnancy. Typical diagnostic ultrasound scanners operate in the frequency range of 2 to 13 megahertz. More powerful ultrasound sources may be used to generate local heating in biological tissue, with applications in physical therapy and cancer treatment. Focused ultrasound sources may be used to break up kidney stones. Ultrasonic cleaners, sometimes called supersonic cleaners, are used at frequencies from 20-40 kHz for jewellery, lenses and other optical parts, watches, dental instruments, surgical instruments and industrial parts. These cleaners consist of containers with a uid in which the object to be cleaned is placed. Ultrasonic waves are then sent into the uid. The main mechanism for cleaning action in an ultrasonic cleaner is actually the energy released from the collapse of millions of microscopic cavitation events occurring in the liquid of the cleaner. 33 Interesting Fact: Ultrasound generator/speaker systems are sold with claims that they frighten away rodents and insects, but there is no scientiﬂc evidence that the devices work; controlled tests have shown that rodents quickly learn that the speakers are harmless. Medical Ultrasonography Medical ultrasonography makes uses the fact that waves are partially reected when the medium
in which they are moving changes density. If the density increases then the reected waves undergoes a phase shift exactly like the case where the waves in a string were reected from a ﬂxed end. If the density decreases then the reected waves has the same phase exactly like the case where the waves in a string were reected from a free end. Combining these properties of waves with modern computing technology has allowed medical professionals to develop an imaging technology to help with many aspects of diagnosis. Typical ultrasound units have a hand-held probe (often called a scan head) that is placed directly on and moved over the patient: a water-based gel ensures good contact between the patient and scan head. Ultrasonic waves are emitted from the scan head and sent into the body of the patient. The scan head also acts a receiver for reected waves. From detailed knowledge of interference and reection an image of the internal organs can be constructed on a screen by a computer programmed to process the reected signals. Interesting Fact: Medical ultrasonography was invented in 1953 at Lund University by cardiologist Inge Edler and Carl Hellmuth Hertz, the son of Gustav Ludwig Hertz, who was a graduate student at the department for nuclear physics. Uses Ultrasonography is widely utilized in medicine, primarily in gastroenterology, cardiology, gynaecology and obstetrics, urology and endocrinology. It is possible to perform diagnosis or therapeutic procedures with the guidance of ultrasonography (for instance biopsies or drainage of uid collections). Strengths of ultrasound imaging It images muscle and soft tissue very well and is particularly useful for ﬂnding the interfaces between solid and uid-ﬂlled spaces. It renders "live" images, where the operator can dynamically select the most useful section for diagnosing and documenting changes, often enabling rapid diagnoses. It shows the structure as well as some aspects of the function of organs. † † † 34 It has no known long-term side eﬁects and rarely causes any discomfort to the patient. Equipment is widely available and comparatively exible; examinations can be performed at the bedside. Small, easily carried scanners are available. † † † Much cheaper than many other medical imaging technology. † Weaknesses of ultrasound imaging † † † Ultrasound has trouble penetrating bone and performs very poorly when there is air between the scan head and the organ of interest. For
example, overlying gas in the gastrointestinal tract often makes ultrasound scanning of the pancreas di–cult. Even in the absence of bone or air, the depth penetration of ultrasound is limited, making it di–cult to image structures that are far removed from the body surface, especially in obese patients. The method is operator-dependent. A high level of skill and experience is needed to acquire good-quality images and make accurate diagnoses. Doppler ultrasonography Ultrasonography can be enhanced with Doppler measurements, which employ the Doppler eﬁect to assess whether structures (usually blood) are moving towards or away from the probe. By calculating the frequency shift of a particular sample volume, e.g. within a jet of blood ow over a heart valve, its speed and direction can be determined and visualised. This is particularily useful in cardiovascular studies (ultrasonography of the vasculature and heart) and essential in many areas such as determining reverse blood ow in the liver vasculature in portal hypertension. The Doppler information is displayed graphically using spectral Doppler, or as an image using colour Doppler or power Doppler. It is often presented audibly using stereo speakers: this produces a very distinctive, despite synthetic, sound. 2.5 Important Equations and Quantities Frequency: f = 1 T : (2.3) Quantity Amplitude Period Wavelength Frequency Speed Symbol A T ‚ f v S.I. Units m s m or m:s¡1 Hz s¡1 Direction \| \| \| \| \| Table 2.1: Units used in Waves 35 Speed: v = f ‚ ‚ T = 36 Chapter 3 Geometrical Optics In this chapter we will study geometrical optics. Optics is the branch of physics that describes the behavior and properties of light and the interaction of light with matter. By geometrical optics we mean that we will study all the optics that we can treat using geometrical analysis (geometry), i.e. lines, angles and circles. As we have seen in the previous chapters, light propagates as a wave. The waves travel in straight lines from the source. So we will consider light as a set of rays. The wavelike nature will become apparent when the waves go from one medium to another. Using rays and Snell’s law to describe what happens when the light ray moves from one medium to another we can solve all the geomet
rical optics problems in this chapter. 3.1 Refraction re-looked We have seen that waves refract as they move from shallower to deeper water or vise versa, thus light also refracts as it moves between two mediums of diﬁerent densities. We may consider the parallel beams of light in Fig 1(a) as a set of wheels connected by a straight rod placed through their centres as in Fig1(b). As the wheels roll onto the grass (representing higher density), they begin to slow down, while those still on the tarmac move at a relatively faster speed. This shifts the direction of movtion towards the normal of the grasstarmac barrier. Likewise light moving from a medium of low density to a medium of high density (Fig 2) moves towards the normal, hence the angle of incidence (i) is greater than the angle of Refraction (r): i > r for d1 < d2 (NOTE TO SELF: Diagram of ray moving into more dense medium) (NOTE TO SELF: Discussion of Snell’s Law) (NOTE TO SELF: Follow by discussion of how we can reverse rays) (NOTE TO SELF: Show its all the same) 37 where both angles are measured from the normal to the ray. I is the incident ray and R is the refracted ray. Rays are reversible 3.1.1 Apparent and Real Depth: If you submerge a straight stick in water and look at the stick where it passes through the surface of the water you will notice that it looks like the stick bends. If you remove you will see that the stick did not bend. The bending of the stick is a very common example of refraction. How can we explain this? We can start with a simple object under water. We can see things because light travels from the object into our eyes where it is absorbed and sent to the brain for processing. The human brain interprets the information it receives using the fact that light travels in straight lines. This is why the stick looks bent. The light did not travel in a straight line from the stick underwater to your eye. It was refracted at the surface. Your brain assumes the light travelled in a straight line and so it intrepets the information so that it thinks the stick is in a diﬁerent place. This phenomenon is easily explained using ray diagrams as in Fig??. The real light rays are represented with a solid line while dashed lines depict
the virtual rays. The real light rays undergo refraction at the surface of the water hence move away from the normal. However the eye assumes that light rays travel in straight lines, thus we extend the refracted rays until they converge to a point. These are virtual rays as in reality the light was refracted and did not originate from that point. 38 air water (NOTE TO SELF: Insert an example with water - can be worked example!) We note that the image of is seen slightly higher and ahead of the object. Where would we see the object if it was submerged in a uid denser than water? (NOTE TO SELF: Insert an example with denser medium show change - can be worked example!) 3.1.2 Splitting of White Light How is a rainbow formed? White is a combination of all other colours of light. Each colour has a diﬁerent wavelength and is thus diﬁracted through diﬁerent angles. red yellow blue The splitting of white light into its component colours may be demonstrated using a triangular prism (Fig4). White light is incident on the prism. As the white light enters the glass its component colours are diﬁracted through diﬁerent angles. This separation is further expanded as the light rays leave the prism. Why? What colour is diﬁracted the most? If red has the longest wavelength and violet the shortest, what is the relation between refraction and wavelength? As the sun appears after a rainstorm, droplets of water still remain in the atmosphere. These act as tiny prisms that split the suns light up into its component colours forming a rainbow. 39 white light red yellow blue 3.1.3 Total Internal Reection Another useful application of refraction is the periscope. We know that as light move from higher to lower density mediums, light rays tend to be diﬁracted towards the normal. We also know that the angle of refraction is greater than the angle of incidence and increases as we increase the angle of incidence (Fig 5(a),(b)). At a certain angle of incidence, c, the refracted angle equals 90o (Fig 5(c)), this angle is called the critical angle. For any angle of incidence greater than c the light will be refracted back into the incident medium, such refraction is called Total Internal reection (Fig 5(d)). A periscope uses two 90o triangular prisms as total internal reect
ors. Light enters the periscope and is reected by the ﬂrst prism down the chamber, where again the light is reected to the observer. This may be illustrated using a ray diagram as in (Fig 6). 3.2 Lenses Lenses are used in many aspects of technology ranging from contact lenses to projectors. We shall again use light rays to explain the properties on lenses. There are two types of lenses, namely: Converging: Lenses that cause parallel light rays to converge to a point (Fig 7). Diverging: Lenses that cause parallel light rays to diverge (Fig 8). Such deviation of light rays are caused by refraction. It is now a good time to introduce a few new deﬂnitions: Optical Centre: The centre of the lens Principal Axis: The line joining two centres of curvature of the surfaces of the lens. Focal Point: The point at which light rays, parallel to the principal axis, converge, or appear to converge, after passing through the lens. Focal Length: The distance between the optical centre and the focal point. 40 As seen in Fig 7 and 8, the focal point of a converging lens is real, while the focal point of a diverging lens is vitual. 3.2.1 Convex lenses Convex lenses are in general converging lenses. Hence they possess real focal points (with one exception, discussed later). Such focal points allow for real images to be formed. One may verify the above by placing an illuminated object on one side of a convex lens, ensuring that the distance between them is greater than the focal length (explained later). By placing a screen on the other side of the lens and varing the distance, one will acquire a sharp image of the object on the screen. The ray diagrams below illustrate the images formed when an object is placed a distance d from the optical centre of a convex lens with focal length F. d Image type Magniﬂcation Orientation Image Position (I) Figure >2F Real <1 Inverted F<I<2F 9(a) 2F Real 1 Inverted 2F 9(b) F<d<2F Real >1 Inverted >2F 9(c) F No Image - - - 9(d) <F Virtual >1 Original >d 9(e) 3.2.2 Concave Lenses Concave lenses disperse parallel rays of light and are hence
diverging lenses. All images formed by concave lenses are virtual and placed between the lens and the object. Furthermore, the image retains its original orientation while it is smaller in size (Fig 10). B0 A0 B A 41 B0 A0 B A 3.2.3 Magniﬂcation By Deﬂnition the magniﬂcation, m, is: m = (Height of Image) / (Height of Object) however, using similar triangles, we may prove that: m = (Distance of Image) / (Distance of Object) where both distances are measured from the optical centre. The above method allows us to accurately estimate the magniﬂcation of an image. This is commonly used in a compound microscope as discussed in the next section. Worked Example 2 A n object is placed 0.5m in front of a convex a lens. † † At what distance should a screen be placed in order to create an image with a magniﬂcation of 1.5? If the height of the image and object are equal, what is the focal length of the lens? Solutions: a)m = (Distance of Image) / (Distance of Object) therefore Distance of Image = (Distance of Object) x m Distance of Image = 0.5m x 1.5 = 0.75m b)This implies that the magniﬂcation m = 1. Therefore from the above table it is seen that d = I = 2F Therefore F = d/2 = 0.25m 42 3.2.4 Compound Microscope This type of microscope uses two convex lenses. The ﬂrst creates a real magniﬂed image of the object that is in turn used by the second lens to create the ﬂnal image. This image is virtual and again enlarged. The ﬂnal image, as seen in Fig 11, is virtual, enlarged and inverted. The lens, L1 that forms the real image is called the objective lens, while L2 is referred to as the eyepiece. B A A1 B1 3.2.5 The Human Eye The eye also contains a biconvex lens that is used to focus objects onto the retina of the eye. However, in some cases the lens maybe abnormal and cause defects in ones vision. Hyperopia (Long-Sightedness) This occurs when the image is focussed
beyond the retina. Hyperopia is due to the eyeball being too short or the lens not being convex enough. A convex lens is used to correct this defect (Fig12). Miopia (Short-Sightedness) Images are focussed before the retina. The lens being too convex or the eyeball being too long causes this. A concave lens corrects this defect (Fig 13). Astigmatism When this occurs, one is able to focus in one plane (e.g. vertical) but not another. This again is due to a distortion in the lens and may be cured by using relevant lenses. 43 3.3 Introduction Light is at ﬂrst, something we feel incredibly familiar with. It can make us feel warm, it allows us to see, allows mirrors and lenses to ‘work’, allows for... Under more careful study light exhibits many fascinating and wonderful properties. The study of light has led to many important and amazing scientiﬂc discoveries and much progress. For example ﬂbre optics, lasers and CCD’s play a huge role in modern technology. † † † † Light is a form of energy. This is demonstrated by Crookes Radiometer or Solar Cells Light travels in straight lines. This is demonstrated by an experiment involving a card with an hole looking at light source e.g. candle. Also the simple camera: candle black box one side pin hole other side grease proof paper. We see the candle upside down on the paper Light travels at a constant speed. The speed depends on the medium it is in. (substance like air or water)1. Nothing travels faster than the speed of light in a vacuum c = 2:99790 of the fundamental constants in physics. £ 108. This is one Probably the most important use of light by the majority of living things on earth is that it allows them to see. If the light from an object enters our optical detectors/sensors i.e. our eyes!, we can see that object. In some cases the light originates at the object itself. Objects which give out light are said to be luminous objects e.g. a lighted candle/torch/bulb, the Sun, stars. The moon is not a luminous object! Why? Most objects however are not luminous, objects which do not give out their own light. We can see them because they reect light into our eyes. 3
.4 Reection 3.4.1 Diﬁuse reection (NOTE TO SELF: diag of light rays hitting a rough surface) (NOTE TO SELF: diag of light rays hitting a polished surface) Most objects do not have perfectly smooth surfaces. Because of this diﬁerent parts of the surface reect light rays in diﬁerent directions (angles). 3.4.2 Regular reection Mirrors and highly polished surfaces give regular reection. A mirror is a piece of glass with a thin layer of silver (aluminium is commonly used) on the rear surface Light is reected according to the laws of reection. 3.4.3 Laws of reection (NOTE TO SELF: diag of i N r rays striking a mirror deﬂne i r N) Laws of reection: The angle of incidence is always equal to the angle of reection The incident ray, the normal and the reected ray are all in the same plane 1There are some substances where light moves so slowly you can walk faster than it moves - more on this later! 44 Note that the angle of incidence i is between the incident ray and the normal2 to the surface, not between the incident ray and the surface of the mirror. There are two forms of an image formed by reection: real and virtual. A real image is formed by the actual intersection of light rays. It is always inverted (upside down) and can be formed on a screen A virtual image is formed by the apparent intersection of light rays. It is always erect and cannot be formed on a screen. Images formed in plane mirrors are virtual images. Real images are formed by lenses (e.g. the image on a cinema screen) or by curved mirrors. 3.4.4 Lateral inversion Where some thing is back to front e.g. AMBULANCE (NOTE TO SELF: How a periscope works, why images are right!) 3.5 Curved Mirrors 3.5.1 Concave Mirrors (Converging Mirrors) (NOTE TO SELF: how to remember a con cave from a convex mirror) (NOTE TO SELF: diag of light rays striking a concave mirror base ray diag! deﬂne things) P = pole of mirror F = focal point (focus) C = centre of curvature \|CP\| = radius of curvature (r) \|FP\| = focal length
(f) r = 2f Rules for ray tracing: Rays of light that arrive parallel to principal axis leave the surface of the mirror through the focal point Rays of light that arrive through the focal point leave the surface of the mirror parallel to principal axis Rays of light that arrive through the centre of curvature leave the surface back through the centre of curvature (NOTE TO SELF: Maybe diag for each of these above!) In order to obtain the position, nature (real or virtual) and size of the image we need just apply the rules above. The details depend on the distance of the object from the mirror surface.. Object outside C diag Image - located between c and f - inverted (upside down) - diminished (reduced in size, smaller) - real Object at C diag Image - located at c - inverted (upside down) - same size - real Object between C and f diag Image - located outside c - inverted (upside down) - magniﬂed (increased in size, larger) - real Object at f diag 2The normal is a line that is perpendicular to the surface i.e. the angle between the line and the surface is 90o 45 Image - located at inﬂnity Object between f and P diag Image - located behind the mirror - erect (right side up) - magniﬂed (increased in size, larger) - virtual 3.5.2 Convex Mirrors diag base ray diag! deﬂne things Image - located behind the mirror - erect (right side up) - diminished (reduced in size, smaller) - virtual We can also arrive at the position and nature of the image by calculation, using the following formula 1/u + 1/v = 1/f AND Magniﬂcation (=M) m = v/u (We will deal with the ‘proﬁ’ for these later) N.B. distance between f and P = f (focal length negative for convex mirror) distance between o and P = u (object distance) distance between i and P = v (image distance negative for virtual image) concave mirror f is positive convex mirror f is negative real image v is positive virtual image v is negative Lots of examples with numbers e.g. u = 10cm v = 20cm m=2 things are bigger if m > 1 (and -1) things are smaller if
0 > m > 1 (and -1) Uses of Convex mirrors image is always erect wide range of view † † They often used in shops, double decker busses, dangerous bends in roads, wing mirrors of cars Disadvantage False sense of distance (objects seem closer than they actually are) uses of concave mirrors They are usually used as make-up mirrors or shaving mirrors and in reecting telescopes Advantages if object is inside f the image is magniﬂed † MAYBE MENTION ‘Virtual object’ in convex mirror for completeness 3.5.3 Refraction When light travels from one medium to another it changes direction, except when it is incident normally on the separating surface. The change of direction is caused by the change in the velocity of the light as it passes from one medium to the other. 46 4 3 2 1 0 -1 -2 -3 -4 D A C B -4 -5 -3 diag deﬂning angles i, r and N Light is refracted according to the laws of refraction: -1 -2 2 1 0 3 4 5 3.5.4 Laws of Refraction Laws of Refraction: sin i /sin r is a constant for two given media (Snell’s Law) The incident ray, the normal and the refracted ray are all in the same plane Sin i /sin r is known as the refractive index (n) EXP verify Snell’s law EXP coin in an empty cup, move head till it disappears. Then ﬂll with water. Can you see it? diag real apparent depth Due to refraction a body which is at O beneath the water appears to be at I when seen from above. As a result of refraction the pool appears to be 1.5m instead of 2m. The relationship between the real depth and the apparent depth is determined by the refractive index of water, which is 4/3 diag of water depth in pool n = sin i /sin r =\|AP\|/\|PI\| / \|AP\|/\|PO\| \|PO\|/\|PI\| However the above diagram is greatly exaggerated in size. In practise \|PO\| \|AO\| and \|PI\| \|AI\| So n =\|AO\|/\|AI\| == real/apparent depth! More than two media.g. air glass water The refractive index of glass is 3
/2 and the refractive index of water is 4/3. The refractive £ index from water to glass a n w= 4/3 => w n a =3/=4 £ 3=2 = 9=8 £ 47 3.5.5 Total Internal Reection This can only happen when light is travelling into a less dense medium Refrectionnumbers are n1 = 2 and n2 = 11 -2 -3 6 4 3 0 5 1 2 -2 -3 -7 -1 -4 -5 -8 -6 diag air glass critical angle i r When light travels from glass to air it is refracted away from the normal as in a above. As the angle of incidence is increased the angle of refraction eventually reaches 90o as in b If the angle of incidence is increased further beyond this value total internal reection occurs as in c The critical angle c is the angle of incidence corresponding to the angle of refraction of 90o From the above the refractive index from from air to glass n=sin90/sin c = 1/sin c Also sin c = 1 / n = 2/3 in the case of glass. Sin c = 0.6667 => critical angle for glass is 41o490 Total internal reection has some very useful properties eg diag prism for turning light 180o daig prism for turning light 90o The size of the ﬂnal image in a pair of binoculars depends on the distance travelled by the light within the binoculars. By using two prisms this distance can be increased without increasing the length of the binoculars. Fibre Optics In the same Water can be directed from one place to another by conﬂning it within a pipe. way light can be directed from one place to another by conﬂning it within a single glass ﬂbre. The light is kept within the ﬂbre by total internal reection. The amount of light which can be carried by a single ﬂbre is very small so it is usual to form a light tube tapping a few thousand ﬂbres together. On great advantage of such a light tube is exibility; it can be ties in knots 48 B 7 8 and still function. However since total internal reection only occurs when light is going from a medium to a less dense medium, it is necessary ti coat each ﬂbre with glass of a lower refractive index. Otherwise light would leak from one �
��bre at their points of contact. Light tube can be used to bring light from a lamp to an object, thus illuminating the object. A second light tube can then used to carry light from the illuminated object to an observer, thus enabling the object to be seen and photographed. The procedure has been used to photograph the digestive system the reproductive system and many other parts of the human body. In the case of the light tube carrying light from the object to the observer, it is vital that the individual ﬂbres in the tube do not cross each other, otherwise the image will become garbled. Like radio waves, light waves are electromagnetic. (cf section em) However, their shorter wavelength and higher frequency means that a single light beam can carry far more telephone conservations at one time compared with a radio wave. In the case of long ﬂbre cables it would be necessary to incorporate a device to boost the intensity of the light to make up for losses due to absorption. Nevertheless the system has great potential for the communication industry, including the possibility of transmitting pictures over long distances. As we said earlier, the reason why light bends when going from one medium to another is because of the change of velocity. This will be dealt with in more detail later. For the moment we consider only the implication for the refractive index. 1 n 2 = velocity medium 1 / velocity medium 2 3.5.6 Mirage Yet another eﬁect of refraction is the mirage. The most common mirage occurs in warm weather when motorists see what appears to be a pool of water on the road close to the horizon. The explanation is this: when air is heated it expands and becomes less dense and when it cools it contracts and becomes more dense. In summer the ground is hot and the layer of air nearest the ground is hot. The layer above that is cooler, etc etc. A of light coming from the blue sky passes down through the layers of air which are getting progressively less dense. As it does so it is progressively bent, as shown in the diagram. As a result an image of the blue sky is seen on the road and is taken to be a pool of water! 3.6 The Electromagnetic Spectrum They are transverse waves. They can travel through empty space. They travel at the speed of light. Short wavelength Long wavelength Dangers Microwaves kill living cells Ultraviolet light causes skin cancer and can kill living cells Uses Radio waves are used in TV’s and radios Mic
s in the equations imply we are calculating average quantities. Mention that we take the limit of a small time interval to give instantaneous quantities. Perhaps the example of a parabola with average gradient and gradient of 51 tangent can be used as an illustration. Else defer until chapter on Graphs and Equations of Motion. Instantaneous velocity: reading on the speedometer in a direction tangent to the path. Instantaneous speed is magnitude of instantaneous velocity but average speed is not equal to magnitude of average velocity. Average speed and average velocity are the total distance and resultant displacement over the time interval related to that part of the path. The example of the circular track uses these deﬂnitions and is an important illustration of the diﬁerences. Instantaneous calculated at a certain instant in time while average is calculated over an interval. Include relative velocity. Address PGCE comments above and comments in the text. † † 4.3 Introduction \A vector is ‘something’ that has both magnitude and direction." \‘Things’? What sorts of ‘things’?" Any piece of information which contains a magnitude and a related direction can be a vector. A vector should tell you how much and which way. Consider a man driving his car east along a highway at 100 km=h. What we have given here is a vector{ the car’s velocity. The car is moving at 100 km=h (this is the magnitude) and we know where it is going{ east (this is the direction). Thus, we know the speed and direction of the car. These two quantities, a magnitude and a direction, form a vector we call velocity. Deﬂnition: A vector is a measurement which has both magnitude and direction. In physics magnitudes often have directions associated with them. If you push something it is not very useful knowing just how hard you pushed. A direction is needed too. Directions are extremely important, especially when dealing with situations more complicated than simple pushes and pulls. Diﬁerent people like to write vectors in diﬁerent ways. Anyway of writing a vector so that it has both magnitude and direction is valid. Are vectors physics? No, vectors themselves are not physics. Physics is just a description of the world around us. To describe something we need to use a language. The most common language used to describe physics is mathematics. Vectors form a very important part of the mathematical description of physics, so much
so that it is absolutely essential to master the use of vectors. 4.3.1 Mathematical representation Numerous notations are commonly used to denote vectors. In this text, vectors will be denoted by symbols capped with an arrow. As an example, ¡!s, ¡!v and ¡!F are all vectors (they have both magnitude and direction). Sometimes just the magnitude of a vector is required. In this case, is another the arrow is ommitted. In other words, F denotes the magnitude of vector ¡!F. way of representing the size of a vector. ¡!F j j 52 4.3.2 Graphical representation Graphically vectors are drawn as arrows. An arrow has both a magnitude (how long it is) and a direction (the direction in which it points). For this reason, arrows are vectors. In order to draw a vector accurately we must specify a scale and include a reference direction in the diagram. A scale allows us to translate the length of the arrow into the vector’s magnitude. For instance if one chose a scale of 1cm = 2N (1cm represents 2N ), a force of magnitude 20N would be represented as an arrow 10cm long. A reference direction may be a line representing a horizontal surface or the points of a compass. Worked Example 3 Drawing vectors Question: Using a scale of 1cm = 2m:s¡1 represent the following velocities: a) 6m:s¡1 north b) 16m:s¡1 east Answer: scale 1cm = 2m:s¡1 m c 3 a) b) W N S E 8cm 4.4 Some Examples of Vectors 4.4.1 Displacement Imagine you walked from your house to the shops along a winding path through the veld. Your route is shown in blue in Figure 12.3. Your sister also walked from the house to the shops, but she decided to walk along the pavements. Her path is shown in red and consisted of two straight stretches, one after the other. Although you took very diﬁerent routes, both you and your sister walked from the house to the shops. The overall eﬁect was the same! Clearly the shortest path from your house to the shops is along the straight line between these two points. The length of this line and the direction from the start point (the house) to the end point (the shops)
forms a very special vector known as displacement. Displacement is assigned the symbol ¡!s. Deﬂnition: Displacement is deﬂned as the magnitude and direction of the straight line joining one’s starting point to one’s ﬂnal point. 53 Finish (Shop) D isplace m ent Start (House) Figure 4.1: Illustration of Displacement OR Deﬂnition: Displacement is a vector with direction pointing from some initial (starting) point to some ﬂnal (end) point and whose magnitude is the straight-line distance from the starting point to the end point. (NOTE TO SELF: choose one of the above) In this example both you and your sister had the same displacement. This is shown as the black arrow in Figure 12.3. Remember displacement is not concerned with the actual path taken. It is only concerned with your start and end points. It tells you the length of the straight-line path between your start and end points and the direction from start to ﬂnish. The distance travelled is the length of the path followed and is a scalar (just a number). Note that the magnitude of the displacement need not be the same as the distance travelled. In this case the magnitude of your displacement would be considerably less than the actual length of the path you followed through the veld! 4.4.2 Velocity Deﬂnition: Velocity is the rate of change of displacement with respect to time. The terms rate of change and with respect to are ones we will use often and it is important that you understand what they mean. Velocity describes how much displacement changes for a certain change in time. We usually denote a change in something with the symbol ¢ (the Greek letter Delta). You have probably seen this before in maths{ the gradient of a straight line is ¢y ¢x. The gradient is just how much y changes for a certain change in x. In other words it is just the rate of change of y with respect to x. This means that velocity must be ¡!v = ¢¡!s ¢t = ¡!s f inal ¡ ¡!s initial tinitial tf inal ¡ : (4.1) 54 (NOTE TO SELF: This is actually average velocity. For instantaneous ¢’s change to differentials. Explain that if ¢
is large then we have average velocity else for inﬂnitesimal time interval instantaneous!) What then is speed? Speed is how quickly something is moving. How is it diﬁerent from velocity? Speed is not a vector. It does not tell you which direction something is moving, only how fast. Speed is the magnitude of the velocity vector (NOTE TO SELF: instantaneous speed is the magnitude of the instantaneous velocity.... not true of averages!). Consider the following example to test your understanding of the diﬁerences between velocity and speed. Worked Example 4 Speed and Velocity Question: A man runs around a circular track of radius 100m. It takes him 120s to complete a revolution of the track. If he runs at constant speed calculate: 1. his speed, 2. his instantaneous velocity at point A, 3. his instantaneous velocity at point B, 4. his average velocity between points A and B, 5. his average velocity during a revolution. A 100m B W N S E Direction the man runs Answer: 1. To determine the man’s speed we need to know the distance he travels and how long it takes. We know it takes 120s to complete one revolution of the track. Step 1 : First we ﬂnd the distance the man travels What distance is one revolution of the track? We know the track is a circle and we know its radius, so we can determine the perimeter or distance around the circle. We start with the equation for the circumference of a circle C = 2…r = 2…(100m) = 628:3 m: So we know the distance the man covers in one revolution is 628:3 m. 55 Step 2 : Now we determine speed from the distance and time. We know that speed is distance covered per unit time. So if we divide the distance covered by the time it took we will know how much distance was covered for every unit of time. v = Distance travelled time taken = 628:3m 120s = 5:23 m:s¡1 2. Step 3 : Determine his instantaneous velocity at A Consider the point A in the diagram. A Direction the man runs We know which way the man is running around the track and we know his speed. His velocity at point A will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of his velocity). The instant that he arrives at A he is moving as indicated in the diagram below. A So
his velocity vector will be 5:23 m:s¡1 West. 3. Step 4 : Determine his instantaneous velocity at B Consider the point B in the diagram. B Direction the man runs 56 We know which way the man is running around the track and we know his speed. His velocity at point B will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of his velocity). The instant that he arrives at B he is moving as indicated in the diagram below. B So his velocity vector will be 5:23 m:s¡1 South. 4. Step 5 : Now we determine his average velocity between A and B (NOTE TO SELF: add this here to further stress the diﬁerence between average and instantaneous velocities, as well as the diﬁerence between magnitude of average velocity and average speed!) 5. Step 6 : Now we calculate his average velocity over a complete revolution. The deﬂnition of average velocity is given earlier and requires that you know the total displacement and the total time. The total displacement for a revolution is given by the vector from the initial point to the ﬂnal point. If the man runs in a circle then he ends where he started. This means the vector from his initial point to his ﬂnal point has zero length. So a calculation of his average velocity follows: ¡!v = ¢¡!s ¢t 0m 120s = 0 m:s¡1 = Remember displacement can be zero when even is distance not! 4.4.3 Acceleration Deﬂnition: Acceleration is the rate of change of velocity with respect to time. Acceleration is also a vector. Remember that velocity was the rate of change of displacement with respect to time so we expect the velocity and acceleration equations to look very similar. In fact, ¡!a = ¢¡!v ¢t = ¡!v f inal ¡ ¡!v initial tinitial tf inal ¡ (4.2) (NOTE TO SELF: average and instantaneous distiction again! expand further- what does it mean.) Acceleration will become very important later when we consider forces. 57 Using vectors is an important skill you MUST master! 4.5 Mathematical Properties of Vectors Vectors are mathematical objects and we will use them to describe physics in the language of mathematics. However, �
��rst we need to understand the mathematical properties of vectors (e.g. how they add and subtract). We will now use arrows representing displacements to illustrate the properties of vectors. Remember that displacement is just one example of a vector. We could just as well have decided to use forces to illustrate the properties of vectors. 4.5.1 Addition of Vectors If we deﬂne a displacement vector as 2 steps in the forward direction and another as 3 steps in the forward direction then adding them together would mean moving a total of 5 steps in the forward direction. Graphically this can be seen by ﬂrst following the ﬂrst vector two steps forward and then following the second one three steps forward: 2 steps + 3 steps = = 5 steps We add the second vector at the end of the ﬂrst vector, since this is where we now are after the ﬂrst vector has acted. The vector from the tail of the ﬂrst vector (the starting point) to the head of the last (the end point) is then the sum of the vectors. This is the tail-to-head method of vector addition. As you can convince yourself, the order in which you add vectors does not matter. In the example above, if you decided to ﬂrst go 3 steps forward and then another 2 steps forward, the end result would still be 5 steps forward. The ﬂnal answer when adding vectors is called the resultant. Deﬂnition: The resultant of a number of vectors is the single vector whose eﬁect is the same as the individual vectors acting together. In other words, the individual vectors can be replaced by the resultant{ the overall eﬁect is the same. If vectors ¡!a and ¡!b have a resultant ¡!R, this can be represented mathematically as, ¡!R = ¡!a + ¡!b : Let us consider some more examples of vector addition using displacements. The arrows tell you how far to move and in what direction. Arrows to the right correspond to steps forward, while arrows to the left correspond to steps backward. Look at all of the examples below and check them. 1 step 1 step 1 step 1 step + + = = 2 steps 2 steps = = 2 steps 2 steps 58 Let us test the ﬂrst one. It says one step
forward and then another step forward is the same as an arrow twice as long{ two steps forward. It is possible that you end up back where you started. In this case the net result of what you have done is that you have gone nowhere (your start and end points are at the same place). In this case, your resultant displacement is a vector with length zero units. We use the symbol ¡!0 to denote such a vector: 1 step 1 step 1 step 1 step + + = = 1 step 1 step 1 step 1 step = ¡!0 = ¡!0 Check the following examples in the same way. Arrows up the page can be seen as steps left and arrows down the page as steps right. Try a couple to convince yourself! + = = + = = + = = ¡!0 + = = ¡!0 It is important to realise that the directions aren’t special{ ‘forward and backwards’ or ‘left and right’ are treated in the same way. The same is true of any set of parallel directions: + = = + = = + = = ¡!0 + = = ¡!0 In the above examples the separate displacements were parallel to one another. However the same ‘tail-to-head’ technique of vector addition can be applied to vectors in any direction. + = = + = = 59 + = = Now you have discovered one use for vectors; describing resultant displacement{ how far and in what direction you have travelled after a series of movements. Although vector addition here has been demonstrated with displacements, all vectors behave in exactly the same way. Thus, if given a number of forces acting on a body you can use the same method to determine the resultant force acting on the body. We will return to vector addition in more detail later. 4.5.2 Subtraction of Vectors What does it mean to subtract a vector? Well this is really simple; if we have 5 apples and we subtract 3 apples, we have only 2 apples left. Now lets work in steps; if we take 5 steps forward and then subtract 3 steps forward we are left with only two steps forward: 5 steps - 3 steps 2 steps = What have we done? You originally took 5 steps forward but then you took 3 steps back. That backward displacement would be represented by an arrow pointing to the left (backwards) with length 3. The net result of adding these two vectors is 2 steps forward:
5 steps + 3 steps 2 steps = Thus, subtracting a vector from another is the same as adding a vector in the opposite direction (i.e. subtracting 3 steps forwards is the same as adding 3 steps backwards). This suggests that in this problem arrows to the right are positive and arrows to the left are negative. More generally, vectors in opposite directions diﬁer in sign (i.e. if we deﬂne up as positive, then vectors acting down are negative). Thus, changing the sign of a vector simply reverses its direction: - = - = - = - = - = - = 60 In mathematical form, subtracting ¡!a from ¡!b gives a new vector ¡!c : ¡!c = ¡!b ¡ ¡!a = ¡!b + ( ¡¡!a ) This clearly shows that subtracting vector ¡!a from ¡!b is the same as adding ( Look at the following examples of vector subtraction. ¡¡!a ) to ¡!b. - = + = ¡!0 - = + = 4.5.3 Scalar Multiplication What happens when you multiply a vector by a scalar (an ordinary number)? Going back to normal multiplication we know that 2 2 is just 2 groups of 2 added together to give 4. We can adopt a similar approach to understand how vector multiplication works. £ 2 x = + = 4.6 Techniques of Vector Addition Now that you have been acquainted with the mathematical properties of vectors, we return to vector addition in more detail. There are a number of techniques of vector addition. These techniques fall into two main categories- graphical and algebraic techniques. 4.6.1 Graphical Techniques Graphical techniques involve drawing accurate scale diagrams to denote individual vectors and their resultants. We next discuss the two primary graphical techniques, the tail-to-head technique and the paralelogram method. The Tail-to-head Method In describing the mathematical properties of vectors we used displacements and the tail-tohead graphical method of vector addition as an illustration. In the tail-to-head method of vector addition the following strategy is followed: † † Choose a scale and include a reference direction. Choose any of the vectors to be summed and draw it as an arrow in the correct direction and of the correct length{ remember to put an arrowhead on the end to denote
its direction. 61 † † † Take the next vector and draw it as an arrow starting from the arrowhead of the ﬂrst vector in the correct direction and of the correct length. Continue until you have drawn each vector{ each time starting from the head of the previous vector. In this way, the vectors to be added are drawn one after the other tail-to-head. The resultant is then the vector drawn from the tail of the ﬂrst vector to the head of the last. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram. Worked Example 5 Tail-to-Head Graphical Addition I Question: A ship leaves harbour H and sails 6km north to port A. From here the ship travels 12km east to port B, before sailing 5:5km south-west to port C. Determine the ship’s resultant displacement using the tail-to-head technique of vector addition. Answer: Now, we are faced with a practical issue: in this problem the displacements are too large to draw them their actual length! Drawing a 2km long arrow would require a very big book. Just like cartographers (people who draw maps), we have to choose a scale. The choice of scale depends on the actual question{ you should choose a scale such that your vector diagram ﬂts the page. Before choosing a scale one should always draw a rough sketch of the problem. In a rough sketch one is interested in the approximate shape of the vector diagram. Step 1 : Draw a rough sketch of the situation Its easy to understand the problem if we ﬂrst draw a quick sketch. N A 6km H W E S B 45o 5:5km 12km 62 C In a rough sketch one should include all of the information given in the problem. All of the magnitudes of the displacements are shown and a compass has been included as a reference direction. Step 2 : Next we choose a scale for our vector diagram It is clear from the rough sketch that choosing a scale where 1cm represents 1km (scale: 1cm = 1km) would be a good choice in this problem ){ the diagram will then take up a good fraction of an A4 page. We now start the accurate construction. Step 3 : Now we construct our scaled vector diagram Contruction Step 1: Starting at the harbour H we draw the ﬂrst vector 6
cm long in the direction north (remember in the diagram 1cm represents 1km): m c 6 A H W N S E 1cm = 1km Construction Step 2: Since the ship is now at port A we draw the second vector 12cm long starting from this point in the direction east: 63 A m c 6 H 12cm B W N S E 1cm = 1km 64 Construction Step 3: Since the ship is now at port B we draw the third vector 5:5cm long starting from this point in the direction south-west. A protractor is required to measure the angle of 45o. A m c 6 H 12cm B 45o C 5.5c m W N S 1cm = 1km E Construction Step 4: As a ﬂnal step we draw the resultant displacement from the starting point (the harbour H) to the end point (port C). We use a ruler to measure the length of this arrow and a protractor to determine its direction A m c 6 H 12cm B 45o C 5.5c m W N S 1cm = 1km E 75:4o 8. 3 8 c m 65 Step 4 : Apply the scale conversion We now use the scale to convert the length of the resultant in the scale diagram to the actual displacement in the problem. Since we have chosen a scale of 1cm = 1km in this problem the resultant has a magnitude of 8:38 km. The direction can be speciﬂed in terms of the angle measured either as 75:4o east of north or on a bearing of 75:4o. Step 5 : Quote the ﬂnal answer The resultant displacement of the ship is 8:38 km on a bearing of 75:4o! Worked Example 6 Tail-to-Head Graphical Addition II Question: A man walks 40 m East, then 30 m North. a) What was the total distance he walked? b) What is his resultant displacement? N S E W m 0 3 40m Answer: Step 1 : a) Determine the distance that the man traveled In the ﬂrst part of his journey he traveled 40 m and in the second part he traveled 30 m. This gives us a total distance traveled of 40 + 30 = 70 m. Step 2 : b) Determine his resultant displacement - start by drawing a rough sketch The man’s resultant displacement is the vector from where he started to where he ended. It is the
sum of his two separate displacements. We will use the tail-to-head method of accurate construction to ﬂnd this vector. Here is our rough sketch: 66 N S E W m 0 3 R es ulta n t 40m Step 3 : Choose a suitable scale A scale of 1cm represents 5m (1cm = 5m) is a good choice here. Now we can begin the process of construction. Step 4 : Draw the ﬂrst vector according to scale We draw the ﬂrst displacement as an arrow 8cm long (according to the scale 8cm = 8 5m = 40m) in the direction east: £ W N S E 1cm = 5m 8cm Step 5 : Draw the second vector according to scale Starting from the head of the ﬂrst vector we draw the second displacement as an 5m = 30m) in the direction north: arrow 6cm long (according to the scale 6cm = 6 £ 67 W N S E 1cm = 5m m c 6 8cm Step 6 : Determine the resultant vector Now we connect the starting point to the end point and measure the length and direction of this arrow (the resultant) W N S E 1cm = 5m 36:9o m c 6 1 0 c m 8cm Step 7 : Apply the scale conversion Finally we use the scale to convert the length of the resultant in the scale diagram to the actual magnitude of the resultant displacement. According to the chosen scale 1cm = 5m. Therefore 10cm represents 50m. The resultant displacement is then 50m 36:9o north of east. 68 The Parallelogram Method When needing to ﬂnd the resultant of two vectors another graphical technique can be applied- the parallelogram method. The following strategy is employed: † † † † † Choose a scale and a reference direction. Choose either of the vectors to be added and draw it as an arrow of the correct length in the correct direction. Draw the second vector as an arrow of the correct length in the correct direction from the tail of the ﬂrst vector. Complete the parallelogram formed by these two vectors. The resultant is then the diagonal of the parallelogram. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram. Worked Example 7 Parallelogram Method of Graphical Addition I Question: A force
of F1 = 5N is applied to a block in a horizontal direction. A second force F2 = 4N is applied to the object at an angle of 30o above the horizontal. N 4 = F 2 30o F1 = 5N Determine the resultant force acting on the block using the parallelogram method of accurate construction. Answer: Step 1 : Firstly make a rough sketch of the vector diagram N 4 30o 5N 69 Step 2 : Choose a suitable scale In this problem a scale of 1cm = 0:5N would be appropriate, since then the vector diagram would take up a reasonable fraction of the page. We can now begin the accurate scale diagram. Step 3 : Draw the ﬂrst scaled vector Let us draw F1 ﬂrst. According to the scale it has length 10cm: 10cm 1cm = 0:5N Step 4 : Draw the second scaled vector Next we draw F2. According to the scale it has length 8cm. We make use of a protractor to draw this vector at 30o to the horizontal: c m 8 30o 10cm Step 5 : Determine the resultant vector Next we complete the parallelogram and draw the diagonal: 1cm = 0:5N c m 8 1 7 : 4 c m 13:3o 10cm 1cm = 0:5N Step 6 : Apply the scale conversion Finally we use the scale to convert the measured length into the actual magnitude. Since 1cm = 0:5N, 17:4cm represents 8:7N. Therefore the resultant force is 8:7N at 13:3o above the horizontal. 70 The parallelogram method is restricted to the addition of just two vectors. However, it is arguably the most intuitive way of adding two forces acting at a point. 4.6.2 Algebraic Addition and Subtraction of Vectors Vectors in a Straight Line Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique: † † † Choose a positive direction. As an example, for situations involving displacements in the directions west and east, you might choose west as your positive direction. In that case, displacements east are negative. Next simply add (or subtract) the vectors with the appropriate signs. As a ﬂnal step the direction of the
resultant should be included in words (positive answers are in the positive direction, while negative resultants are in the negative direction). Let us consider a couple of examples. Worked Example 8 Adding vectors algebraically I Question: A tennis ball is rolled towards a wall which is 10m away to the right. If after striking the wall the ball rolls a further 2:5m along the ground to the left, calculate algebraically the ball’s resultant displacement. (NOTE TO SELF: PGCE suggest a ‘more real looking’ diagram, followed by a diagram one would draw to solve the problem (like our existing one with the positive direction shown as an arrow)) Answer: Step 1 : Draw a rough sketch of the situation 10 m 2.5 m Wall Start Step 2 : Decide which method to use to calculate the resultant We know that the resultant displacement of the ball (¡!s resultant) is equal to the sum of the ball’s separate displacements (¡!s 1 and ¡!s 2): ¡!s resultant = ¡!s 1 + ¡!s 2 Since the motion of the ball is in a straight line (i.e. the ball moves left and right), we can use the method of algebraic addition just explained. 71 Step 3 : Choose a positive direction Let’s make to the right the positive direction. This means that to the left becomes the negative direction. Step 4 : Now deﬂne our vectors algebraically With right positive: ¡!s 1 = +10:0m and ¡!s 2 = 2:5m ¡ Step 5 : Add the vectors Next we simply add the two displacements to give the resultant: ¡!s resultant = (+10m) + ( 2:5m) ¡ = (+7:5)m Step 6 : Quote the resultant Finally, in this case right means positive so: ¡!s resultant = 7:5m to the right Let us consider an example of vector subtraction. Worked Example 9 Subtracting vectors algebraically I Question: Suppose that a tennis ball is thrown horizontally towards a wall at 3m:s¡1 to the right. After striking the wall, the ball returns to the thrower at 2m:s¡1. Determine the change in velocity of the ball. Answer: Step 1 : Draw a sketch A quick sketch will help us understand the problem (NOTE TO SELF
: Maybe a sketch here?) Step 2 : Decide which method to use to calculate the resultant Remember that velocity is a vector. The change in the velocity of the ball is equal to the diﬁerence between the ball’s initial and ﬂnal velocities: ¢¡!v = ¡!v f inal ¡ ¡!v initial Since the ball moves along a straight line (i.e. left and right), we can use the algebraic technique of vector subtraction just discussed. Step 3 : Choose a positive direction Let’s make to the right the positive direction. This means that to the left becomes the negative direction. 72 Step 4 : Now deﬂne our vectors algebraically With right positive: ¡!v initial = +3m:s¡1 and ¡!v f inal = 2m:s¡1 ¡ Step 5 : Subtract the vectors Thus, the change in velocity of the ball is: ¢¡!v = ( = ( 2m:s¡1) 5)m:s¡1 ¡ ¡ (+3m:s¡1) ¡ Step 6 : Quote the resultant Remember that in this case right means positive so: ¢¡!v = 5m:s¡1 to the left Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line. A More General Algebraic technique In worked example 3 the tail to head method of accurate construction was used to determine the resultant displacement of a man who travelled ﬂrst east and then north. However, the man’s resultant can be calculated without drawing an accurate scale diagram. Let us revisit this example. Worked Example 10 An Algebraic solution to Worked Example 3 Question: A man walks 40 m East, then 30 m North. Calculate the man’s resultant displacement. Answer: Step 1 : Draw a rough sketch As before, the rough sketch looks as follows: 73 N S E W m 0 3 R es ulta n t ﬁ 40m Step 2 : Determine the length of the resultant Note that the triangle formed by his separate displacement vectors and his resultant displacement vector is a right-angle triangle. We can thus use Pythogoras’ theorem to determine the length of the resultant. If the length of the resultant vector is called s then:
s2 = (40m)2 + (30m)2 s2 = 2500m2 s = 50m Step 3 : Determine the direction of the resultant Now we have the length of the resultant displacement vector but not yet its direction. To determine its direction we calculate the angle ﬁ between the resultant displacement vector and East. We can do this using simple trigonometry: tan ﬁ = tan ﬁ = opposite adjacent 30 40 ﬁ = arctan(0:75) ﬁ = 36:9o Step 4 : Quote the resultant Our ﬂnal answer is then: Resultant Displacement: 50 m at 36:9o North of East This is exactly the same answer we arrived at after drawing a scale diagram! In the previous example we were able to use simple trigonometry to calculate a man’s resultant displacement. This was possible since the man’s directions of motion were perpendicular (north 74 and east). Algebraic techniques, however, are not limited to cases where the vectors to be combined are along the same straight line or at right angles to one another. The following example illustrates this. Worked Example 11 Further example of vector addition by calculation Question: A man walks from point A to point B which is 12km away on a bearing of 45o. From point B the man walks a further 8km east to point C. Calculate the man’s resultant displacement. Answer: Step 1 : Draw a rough sketch of the situation 8km C B 45o G 12k m F A 45o B ^AF = 45o since the man walks initially on a bearing of 45o. Then, A ^BG = B ^AF = 45o (alternate angles parallel lines). Both of these angles are included in the rough sketch. Step 2 : Calculate the length of the resultant The resultant is the vector AC. Since we know both the lengths of AB and BC and the included angle A ^BC, we can use the cosine rule: AC 2 = AB2 + BC 2 = (12)2 + (8)2 = 343:8 2 2 ¢ ¢ ¡ ¡ AC = 18:5 km BC cos(A ^BC) AB (12)(8) cos(135o) ¢ Step 3 : Determine the direction of the resultant Next we use the sine rule to determine the angle : 75 sin 8 = sin = sin 1350 18:5 8 £
sin 135o 18:5 = arcsin(0:3058) = 17:8o Thus, F ^AC = 62:8o. Step 4 : Quote the resultant Our ﬂnal answer is then: Resultant Displacement: 18:5km on a bearing of 62:8o 4.7 Components of Vectors In the discussion of vector addition we saw that a number of vectors acting together can be combined to give a single vector (the resultant). In much the same way a single vector can be broken down into a number of vectors which when added give that original vector. These vectors which sum to the original are called components of the original vector. The process of breaking a vector into its components is called resolving into components. While summing a given set of vectors gives just one answer (the resultant), a single vector can be resolved into inﬂnitely many sets of components. In the diagrams below the same black vector is resolved into diﬁerent pairs of components. These components are shown in red. When added together the red vectors give the original black vector (i.e. the original vector is the resultant of its components). In practice it is most useful to resolve a vector into components which are at right angles to one another. Worked Example 12 Resolving a vector into components 76 Question: A motorist undergoes a displacement of 250km in a direction 30o north of east. Resolve this displacement into components in the directions north (¡!s N ) and east (¡!s E). Answer: Step 1 : Draw a rough sketch of the original vector W N S 30o E k m 0 5 2 Step 2 : Determine the vector component Next we resolve the displacement into its components north and east. Since these directions are orthogonal to one another, the components form a right-angled triangle with the original displacement as its hypotenuse 30o ¡!s E N ¡!s Notice how the two components acting together give the orginal vector as their resultant. Step 3 : Determine the lengths of the component vectors Now we can use trigonometry to calculate the magnitudes of the components of the original displacement: sN = 250 sin 30o = 125 km 77 and sE = 250 cos 30o = 216:5 km Remember sN and sE are the magnitudes of the components{ they are in the directions north and east respectively. (NOTE TO SELF: SW: alternatively
these results can be arrived at by construction. Include?) 4.7.1 Block on an incline As a further example of components let us consider a block of mass m placed on a frictionless surface inclined at some angle to the horizontal. The block will obviously slide down the incline, but what causes this motion? The forces acting on the block are its weight mg and the normal force N exerted by the surface on the object. These two forces are shown in the diagram below. N W k? W mg Now the object’s weight can be resolved into components parallel and perpendicular to the inclined surface. These components are shown as red arrows in the diagram above and are at right angles to each other. The components have been drawn acting from the same point. Applying the parallelogram method, the two components of the block’s weight sum to the weight vector. To ﬂnd the components in terms of the weight we can use trigonometry: Wk = mg sin W? = mg cos The component of the weight perpendicular to the slope W? exactly balances the normal force N exerted by the surface. The parallel component, however, Wk is unbalanced and causes the block to slide down the slope. 78 Figure 4.2: An example of two vectors being added to give a resultant 4.7.2 Vector addition using components In Fig 4.3 two vectors are added in a slightly diﬁerent way to the methods discussed so far. It might look a little like we are making more work for ourselves, but in the long run things will be easier and we will be less likely to go wrong. In Fig 4.3 the primary vectors we are adding are represented by solid lines and are the same vectors as those added in Fig 4.2 using the less complicated looking method. Each vector can be broken down into a component in the x-direction and one in the ydirection. These components are two vectors which when added give you the original vector as the resultant. Look at the red vector in ﬂgure 4.3. If you add up the two red dotted ones in the x-direction and y-direction you get the same vector. For all three vectors we have shown their respective components as dotted lines in the same colour. But if we look carefully, addition of the x components of the two original vectors gives the x component of the resultant. The same applies to the y components. So if we just added all the components together we would get the
same answer! This is another important property of vectors. Worked Example 13 Adding Vectors Using Components Question: Lets work through the example shown in Fig. 4.3 to determine the resultant. Answer: Step 1 : Decide how to tackle the problem The ﬂrst thing we must realise is that the order that we add the vectors does not matter. Therefore, we can work through the vectors to be added in any order. 79 Figure 4.3: Components of vectors can be added as well as the vectors themselves Step 2 : Resolve the red vector into components Let us start with the bottom vector. If you are told that this vector has a length of 5:385 units and an angle of 21:8o to the horizontal then we can ﬂnd its components. We do this by using known trigonometric ratios. First we ﬂnd the vertical or y component: sin = y hypotenuse sin(21:8) = y 5:385 y = 5:385 sin(21:8 Secondly we ﬂnd the horizontal or x component: x hypotenuse cos = cos(21:8) = x 5:385 x = 5:385 cos(21:8) x = 5 80 We now know the lengths of the sides of the triangle for which our vector is the hypotenuse. If you look at these sides we can assign them directions given by the dotted arrows. Then our original red vector is just the sum of the two dotted vectors (its components). When we try to ﬂnd the ﬂnal answer we can just add all the dotted vectors because they would add up to the two vectors we want to add. Step 3 : Now resolve the second vector into components. The green vector has a length of 5 units and a direction of 53.13 degrees to the horizontal so we can ﬂnd its components. sin(53:13) = sin = y hypotenuse y 5 y = 5 sin(53:13) y = 4 5 4 3 cos(53:13) = cos = x hypotenuse x 5 x = 5 cos(53:13) x = 3 Step 4 : Determine the components of the resultant vector Now we have all the components. If we add all the x-components then we will have the x-component of the resultant vector. Similarly if we add all the y-components then we will have the y-component of the resultant vector. The
x-components of the two vectors are 5 units right and then 3 units right. This gives us a ﬂnal x-component of 8 units right. The y-components of the two vectors are 2 units up and then 4 units up. This gives us a ﬂnal y-component of 6 units up. Step 5 : Determine the magnitude and direction of the resultant vector Now that we have the components of the resultant, we can use Pythagoras’ theorem to determine the length of the resultant. Let us call the length of the hypotenuse l and we can calculate its value; 81 l2 = (6)2 + (8)2 l2 = 100 l = 10: 8 6 1 0 The resultant has length of 10 units so all we have to do is calculate its direction. We can specify the direction as the angle the vectors makes with a known direction. To do this you only need to visualise the vector as starting at the origin of a coordinate system. We have drawn this explicity below and the angle we will calculate is labeled ﬁ. 82 Using our known trigonometric ratios we can calculate the value of ﬁ; tan ﬁ = 6 8 ﬁ = arctan ﬁ = 36:8o: 6 8 Step 6 : Quote the ﬂnal answer Our ﬂnal answer is a resultant of 10 units at 36:8o to the positive x-axis. 4.8 Do I really need to learn about vectors? Are they really useful? Vectors are essential to do physics. Absolutely essential. This is an important warning. If something is essential we had better stop for a moment and make sure we understand it properly. 4.9 Summary of Important Quantities, Equations and Con- cepts Vector A vector is a measurement which has both magnitude and direction. Displacement Displacement is a vector with direction pointing from some initial (starting) point to some ﬂnal (end) point and whose magnitude is the straight-line distance from the starting point to the ﬂnal point. 83 Distance The distance travelled is the length of your actual path. Velocity Velocity is the rate of change of displacement with respect to time. Acceleration Acceleration is the rate of change of velocity with respect to time. Resultant The resultant of a number of vectors is the single vector whose eﬁect is the same as the individual
vectors acting together. Quantity Displacement Velocity Distance Speed Acceleration Symbol ¡!s ¡!u,¡!v d v ¡!a S.I. Units Direction m m:s¡1 m m:s¡1 m:s¡2 X X { { X Table 4.1: Summary of the symbols and units of the quantities used in Vectors 84 Chapter 5 Forces 5.1 ‘TO DO’ LIST introduce concept of a system for use in momentum- in NIII for instance, also concept of isolated system and external forces add incline plane examples generally ‘bulk-up’ the Newton’s Laws sections † † † 5.2 What is a force? The simplest answer is to say a ‘push’ or a ‘pull’. If the force is great enough to overcome friction the object being pushed or pulled will move. We could say a force is something that makes objects move. Actually forces give rise to accelerations! In fact, the acceleration of a body is directly proportional to the net force acting on it. The word net is important{ forces are vectors and what matters in any situation is the vector sum of all the forces acting on an object. The unit of force is the newton (symbol N ). It is named after Sir Isaac Newton, whose three laws you will learn about shortly. Interesting Fact: Force was ﬂrst described by Archimedes. Archimedes of Syracuse (circa 287 BC - 212 BC), was a Greek mathematician, astronomer, philosopher, physicist and engineer. He was killed by a Roman soldier during the sack of the city, despite orders from the Roman general, Marcellus, that he was not to be harmed. 5.3 Force diagrams The resultant force acting on an object is the vector sum of the set of forces acting on that one object. It is very important to remember that all the forces must be acting on the same object. 85 The easiest way to determine this resultant force is to construct what we call a force diagram. In a force diagram we represent the object by a point and draw all the force vectors connected to that point as arrows. Remember from Chapter?? that we use the length of the arrow to indicate the vector’s magnitude and the direction of the arrow to show which direction it acts in. The second step is to rearrange the force vectors so that it is easy to add them together and �
�nd the resultant force. Let us consider an example to get started: Two people push on a box from opposite sides with a force of 5N. 5N 5N When we draw the force diagram we represent the box by a dot. The two forces are represented by arrows, with their tails on the dot. Force Diagram: 5N 5N See how the arrows point in opposite directions and have the same magnitude (length). This means that they cancel out and there is no net force acting on the object. This result can be obtained algebraically too, since the two forces act along the same line. Firstly we choose a positive direction and then add the two vectors taking their directions into account. Consider to the right as the positive direction Fres = (+5N ) + ( = 0N ¡ 5N ) As you work with more complex force diagrams, in which the forces do not exactly balance, 2N ). What does this mean? you may notice that sometimes you get a negative answer (e.g. Does it mean that we have something the opposite of force? No, all it means is that the force acts in the opposite direction to the one that you chose to be positive. You can choose the positive direction to be any way you want, but once you have chosen it you must keep it. ¡ Once a force diagram has been drawn the techniques of vector addition introduced in the previous chapter can be implemented. Depending on the situation you might choose to use a graphical technique such as the tail-to-head method or the parallelogram method, or else an algebraic approach to determine the resultant. Since force is a vector all of these methods apply! Always remember to check your signs Worked Example 14 Single Force on a block Question: A block on a frictionless at surface weighs 100N. A 75N force is applied to the block towards the right. What is the net force (or resultant force) on the block? Answer: 86 Step 1 : Firstly, draw a force diagram for the block Fnormal = 100N Fapplied = 75N Fweight = 100N Be careful not to forget the two forces perpendicular to the surface. Every object with mass is attracted to the centre of the earth with a force (the object’s weight). However, if this were the only force acting on the block in the vertical direction then the block would fall through the table to the ground. This does not happen because the table exerts an upward force (the normal force) which exactly
balances the object’s weight. Step 2 : Answer Thus, the only unbalanced force is the applied force. This applied force is then the resultant force acting on the block. 5.4 Equilibrium of Forces At the beginning of this chapter it was mentioned that resultant forces cause objects to accelerate. If an object is stationary or moving at constant velocity then either, no forces are acting on the object, or † the forces acting on that object are exactly balanced. † A resultant force would cause a stationary object to start moving or an object moving at constant velocity to speed up or slow down. In other words, for stationary objects or objects moving with constant velocity, the resultant force acting on the object is zero. The object is said to be in equilibrium. If a resultant force acts on an object then that object can be brought into equilibrium by applying an additional force that exactly balances this resultant. Such a force is called the equilibrant and is equal in magnitude but opposite in direction to the original resultant force acting on the object. Deﬂnition: The equilibrant of any number of forces is the single force required to produce equilibrium. 87 Objects at rest or moving with constant velocity are in equilibrium and have a zero resultant force Resultant of ¡!F1 and ¡!F2 ¡!F1 ¡!F2 ¡!F3 Equilibrant of ¡!F1 and ¡!F2 In the ﬂgure the resultant of ¡!F1 and ¡!F2 is shown in red. The equilibrant of ¡!F1 and ¡!F2 is then the vector opposite in direction to this resultant with the same magnitude (i.e. ¡!F3). ¡!F1, ¡!F2 and ¡!F3 are in equilibrium ¡!F3 is the equilibrant of ¡!F1 and ¡!F2 † † ¡!F1 and ¡!F2 are kept in equilibrium by ¡!F3 † As an example of an object in equilibrium, consider an object held stationary by two ropes in the arrangement below: 50o 40o Rope 1 Rope 2 Let us draw a force diagram for the object. In the force diagram the object is drawn as a dot and all forces acting on the object are drawn in the correct directions starting from that dot. In this
case, three forces are acting on the object. 88 50o ¡!T1 40o ¡!T2 ¡!W Each rope exerts a force on the object in the direction of the rope away from the object. These tension forces are represented by ¡!T1 and ¡!T2. Since the object has mass, it is attracted towards the centre of the earth. This weight is represented in the force diagram as ¡!W. Since the object is stationary, the resultant force acting on the object is zero. In other words the three force vectors drawn tail-to-head form a closed triangle: 40o ¡!T2 50o ¡!T1 ¡!W In general, when drawn tail-to-head the forces acting on an object in equilibrium form a closed ﬂgure with the head of the last vector joining up with the tail of the ﬂrst vector. When only three forces act on an object this closed ﬂgure is a triangle. This leads to the triangle law for three forces in equilibrium: Triangle Law for Three Forces in Equilibrium: Three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order. Worked Example 15 89 Equilibrium Question: A car engine of weight 2000N is lifted by means of a chain and pulley system. In sketch A below, the engine is suspended by the chain, hanging stationary. In sketch B, the engine is pulled sideways by a mechanic, using a rope. The engine is held in such a position that the chain makes an angle of 30o with the vertical. In the questions that follow, the masses of the chain and the rope can be ignored. 30o CHAIN CHAIN ENGINE ENGINE ROPE Sketch A Sketch B i) Draw a force diagram representing the forces acting on the engine in sketch A. ii) Determine the tension in the chain in sketch A. iii) Draw a force diagram representing the forces acting on the engine in sketch B. ii) In sketch B determine the magnitude of the applied force and the tension in the chain. Answer: Step 1 : Force diagram for sketch A i) Just two forces are acting on the engine in sketch A: ¡!T chain ¡!W 90 Step 2 : Determine the tension in the chain ii) Since the engine in sketch A is stationary, the resultant force on the engine is zero. Thus the tension in the chain exactly
balances the weight of the engine, Tchain = W = 2000N Step 3 : Force diagram for sketch B iii) Three forces are acting on the engine in sketch B: ¡!F applied 30o ¡!W ¡!T chain 30o Since the engine is at equilibrium (it is held stationary) the three forces drawn tailto-head form a closed triangle. Step 4 : Calculate the magnitude of the forces in sketch B iv) Since no method was speciﬂed let us calculate the magnitudes algebraically. Since the triangle formed by the three forces is a right-angle triangle this is easily done: Fapplied W = tan 30o Fapplied = (2000) tan 30o and = 1155 N Tchain W Tchain = = 1 cos 30o 2000 cos 30o = 2309 N 5.5 Newton’s Laws of Motion Our current laws of motion were discovered by Sir Isaac Newton. It is said that Sir Isaac Newton started to think about the nature of motion and gravitation after being struck on the head by a falling apple. Newton discovered 3 laws describing motion: 91 5.5.1 First Law Deﬂnition: Every object will remain at rest or in uniform motion in a straight line unless it is made to change its state by the action of an external force. For example, a cement block isn’t going to move unless you push it. A rocket in space is not going to speed up, slow down nor change direction unless the engines are switched on. Newton’s First Law may seem rather surprising to you when you ﬂrst meet it. If you roll a ball along a surface it always stops, but Newton’s First Law says that an item will remain in uniform motion in a straight line unless a force acts upon it. It doesn’t seem like there is any force acting on the ball once you let it go, but there is! In real life (unlike many physics problems) there is usually friction acting. Friction is the external force acting on the ball causing it to stop. Notice that Newton’s First Law says that the force must be external. For example, you can’t grab your belt and pull yourself up to the ceiling. Of course, you could get someone else to pull you up, but then that person would be applying an external force. Worked Example 16 Newton’s First Law in action Question: Why do passengers get
thrown to the side when the car they are driving in goes around a corner? Answer: Newton’s First Law Step 1 : What happens before the car turns Before the car starts turning both you and the car are travelling at the same velocity. (picture A) Step 2 : What happens while the car turns The driver turns the wheels of the car, which then exert a force on the car and the car turns. This force acts on the car but not you, hence (by Newton’s First Law) you continue moving with the same velocity. (picture B) Step 3 : Why passengers get thrown to the side If the passenger is wearing a seatbelt it will exert a force on the passenger until the passenger’s velocity is the same as that of the car (picture C). Without a seatbelt the passenger may hit the side of the car or even the windscreen! 92 A: Both the car and the person travelling at the same velocity B: The cars turns but not the person C: Both the car and the person are travelling at the same velocity again 5.5.2 Second Law Deﬂnition: The resultant force acting on a body results in an acceleration which is in the same direction as the resultant force and is directly proportional to the magnitude of this force and inversely proportional to the mass of the object. In mathematical form this law states, ¡!F Res = m¡!a (5.1) So if a resultant force ¡!F Res acts on an object with mass m it will result in an acceleration ¡!a. It makes sense that the direction of the acceleration is in the direction of the resultant force. If you push something away from you it doesn’t move toward you unless of course there is another force acting on the object towards you! Worked Example 17 Newton’s Second Law Question: A block of mass 10kg is accelerating at 2m:s¡2. What is the magnitude of the resultant force acting on the block? Answer: Step 1 : Decide what information has been supplied We are given the block’s mass the block’s acceleration † † all in the correct units. 93 Step 2 : Determine the force on the block We are asked to ﬂnd the magnitude of the force applied to the block. Newton’s Second Law tells us the relationship between acceleration and force for an object. Since we are only asked for the magnitude we do not
need to worry about the directions of the vectors: FRes = ma 2m:s¡2 = 10kg = 20N £ Thus, there must be a resultant force of 20N acting on the box. Worked Example 18 Newton’s Second Law 2 Question: A 12N force is applied in the positive x-direction to a block of mass 100mg resting on a frictionless at surface. What is the resulting acceleration of the block? Answer: Step 1 : Decide what information has been supplied We are given the block’s mass the applied force † † but the mass is not in the correct units. Step 2 : Convert the mass into the correct units 100mg = 100 1000g = 1kg £ 10¡3g = 0:1g 1 1000g 1 = 1kg £ 1kg 1000g = 0:1g = 0:1g = 0:1g 1 1kg 1000g £ £ = 0:0001 kg Step 3 : Determine the direction of the acceleration We know that net force results in acceleration. Since there is no friction the applied force is the resultant or net force on the block (refer to the earlier example of the block pushed on the surface of the table). The block will then accelerate in the direction of this force according to Newton’s Second Law. 94 Step 4 : Determine the magnitude of the acceleration FRes = ma 12N = (0:0001kg)a a = = 120000 12N 0:0001kg N kg kg:m s2:kg m s2 = 120000 = 120000 = 1:2 £ 105 m:s¡2 From Newton’s Second Law the direction of the acceleration is the same as that of the 105 m:s¡2 resultant force. The ﬂnal result is then that the block accelerates at 1:2 in the positive x-direction. £ Weight and Mass You must have heard people saying \My weight is 60kg". This is actually incorrect because it is mass that is measured in kilograms. Weight is the force of gravity exerted by the earth on an object with mass: Fweight = mg (5.2) As such, weight is measured in newtons. If you compare this equation to Newton’s Second Law you will see that it looks exactly the same with the a replaced by g. Thus, when weight is the only force acting on an object (i.e. when Fweight is the resultant
force acting on the object) the object has an acceleration g. Such an object is said to be in free fall. The value for g is the same for all objects (i.e. it is independent of the objects mass): g = 9:8ms¡2 10ms¡2 (5.3) … You will learn how to calculate this value from the mass and radius of the earth in Chapter??. Actually the value of g varies slightly from place to place on the earth’s surface. The reason that we often get confused between weight and mass, is that scales measure your weight (in newtons) and then display your mass using the equation above. (NOTE TO SELF: include example of skydiver- calculate the acceleration before and after opening parachute{ free-fall and then negative acceleration) Worked Example 19 Calculating the resultant and then the acceleration 95 Question: A block (mass 20kg) on a frictionless at surface has a 45N force applied to it in the positive x-direction. In addition a 25N force is applied in the negative x-direction. What is the resultant force acting on the block and the acceleration of the block? Answer: Step 1 : Decide what information has been supplied We are given the block’s mass Force F1 = 45N in the positive x-direction Force F2 = 25N in the negative x-direction † † † all in the correct units. Step 2 : Figure out how to tackle the problem We are asked to determine what happens to the block. We know that net force results in an acceleration. We need to determine the net force acting on the block. Step 3 : Determine the magnitude and direction of the resultant force Since F1 and F2 act along the same straight line, we can apply the algebraic technique of vector addition discussed in the Vectors chapter to determine the resultant force. Choosing the positive x-direction as our positive direction: Positive x-direction is the positive direction: FRes = (+45N ) + ( = +20N = 20N in the positive x 25N ) ¡ direction ¡ where we remembered in the last step to include the direction of the resultant force in words. By Newton’s Second Law the block will accelerate in the direction of this resultant force. Step 4 : Determine the magnitude of the acceleration FRes = ma 20N = (20kg)a a = = 1 20N 20kg N kg kg
:m s2:kg = 1 m:s¡2 = 1 The ﬂnal result is then that the block accelerates at 1 m:s¡2 in the positive x-direction (the same direction as the resultant force). 96 Worked Example 20 Block on incline Question: insert question here maybe with friction! N? W mg W k 5.5.3 Third Law Deﬂnition: For every force or action there is an equal but opposite force or reaction. Newton’s Third Law is easy to understand but it can get quite di–cult to apply it. An important thing to realise is that the action and reaction forces can never act on the same object and hence cannot contribute to the same resultant. Worked Example 21 Identifying action-reaction pairs Question: Consider pushing a box on the surface of a rough table. 1. Draw a force diagram indicating all of the forces acting on the box. 2. Identify the reaction force for each of the forces acting on the box. Answer: 1. The following force diagram shows all of the forces acting on the box 97 Remember to draw the object in your force diagram as a dot Fnormal Fpush Ff riction Fweight 2. The following table lists each of forces acting on the box (the actions) together with their corresponding reactions: Action Fpush : Person pushing on the box Reaction Box pushing on the person Fweight : The earth attracting The box attracting the box (the weight) Fnormal : The box pushing on the table the earth The table pushing on the box Ff riction : The table acting on The box acting on the box the table Notice that to ﬂnd the reaction force you need to switch around the object supplying the force and the object receiving the force. Be careful not to think that the normal force is the reaction partner to the weight of the box. The normal force balances the weight force but they are both forces acting on the box so they cannot possibly form an action-reaction pair. There is an important thing to realise which is related to Newton’s Third Law. Think about dropping a stone oﬁ a cliﬁ. It falls because the earth exerts a force on it (see Chapter??) and it doesn’t seem like there are any other forces acting. So is Newton’s Third Law wrong? No, the reactionary force to the weight of the stone is the force exerted by the stone on the
earth. This is illustrated in detail in the next worked example. Worked Example 22 Newton’s Third Law Question: A stone of mass 0:5kg is accelerating at 10 m:s¡2 towards the earth. 98 1. What is the force exerted by the earth on the stone? 2. What is the force exerted by the stone on the earth? 3. What is the acceleration of the earth, given that its mass is 5:97 Answer: 1. Step 1 : Decide what information has been supplied We are given 1027kg? £ the stone’s mass the stone’s acceleration (g) † † all in the correct units. Step 2 : The force applied by the earth on the stone This force is simply the weight of the stone. Step 3 : The magnitude of the force of the earth on the stone Applying Newton’s Second Law, we can ﬂnd the magnitude of this force, FRes = ma = 0:5kg = 5N £ 10 m:s¡2 Therefore the earth applies a force of 5N towards the earth on the stone. 2. By Newton’s Third Law the stone must exert an equal but opposite force on the earth. Hence the stone exerts a force of 5N towards the stone on the earth. 3. We have the force acting on the earth the earth’s mass † † in the correct units Step 4 : To ﬂnd the earth’s acceleration we must apply Newton’s Second Law to ﬂnd the magnitude FRes = ma 5N = (5:97 a = 5:97 £ = 8:37521 1027kg)a £ 5N 1027kg 10¡28 m:s¡2 £ The earth’s acceleration is directed towards the stone (i.e. in the same direction as the force on the earth). The earth’s acceleration is really tiny. This is why you don’t notice the earth moving towards the stone, even though it does. 99 Interesting Fact: Newton ﬂrst published these laws in Philosophiae Naturalis Principia Mathematica (1687) and used them to prove many results concerning the motion of physical objects. Only in 1916 were Newton’s Laws superceded by Einstein’s theory of relativity. The next two worked examples are quite long and involved but it is very important that you understand
the discussion as they illustrate the importance of Newton’s Laws. Worked Example 23 Rockets Question: How do rockets accelerate in space? Answer: † † † † Gas explodes inside the rocket. This exploding gas exerts a force on each side of the rocket (as shown in the picture below of the explosion chamber inside the rocket). Note that the forces shown in this picture are representative. With an explosion there will be forces in all directions. Due to the symmetry of the situation, all the forces exerted on the rocket are balanced by forces on the opposite side, except for the force opposite the open side. This force on the upper surface is unbalanced. This is therefore the resultant force acting on the rocket and it makes the rocket accelerate forwards. Systems and External Forces The concepts of a system and an external forces are very important in physics. A system is any collection of objects. If one draws an imaginary box around such a system then an external force is one that is applied by an object or person outside the box. Imagine for example a car pulling a trailer. 100 5.6 Examples of Forces Studied Later Most of physics revolves around forces. Although there are many diﬁerent forces we deal with them all in the same way. The methods to ﬂnd resultants and acceleration do not depend on the type of force we are considering. At ﬂrst glance, the number of diﬁerent forces may seem overwhelming - gravity, drag, electrical forces, friction and many others. However, physicists have found that all these forces can be classiﬂed into four groups. These are gravitational forces, electromagnetic forces, strong nuclear force and weak nuclear force. Even better, all the forces that you will come across at school are either gravitational or electromagnetic. Doesn’t that make life easy? 5.6.1 Newtonian Gravity Gravity is the attractive force between two objects due to the mass of the objects. When you throw a ball in the air, its mass and the earth’s mass attract each other, which leads to a force between them. The ball falls back towards the earth, and the earth accelerates towards the ball. The movement of the earth toward the ball is, however, so small that you couldn’t possibly measure it. 5.6.2 Electromagnetic Forces Almost all of the forces that we experience in everyday life are electromagnetic in origin. They have this unusual name because long ago people thought that electric forces
and magnetic forces were diﬁerent things. After much work and experimentation, it has been realised that they are actually diﬁerent manifestations of the same underlying theory. The Electric Force If we have objects carrying electrical charge, which are not moving, then we are dealing with electrostatic forces (Coulomb’s Law). This force is actually much stronger than gravity. This may seem strange, since gravity is obviously very powerful, and holding a balloon to the wall seems to be the most impressive thing electrostatic forces have done, but think about it: for gravity to be detectable, we need to have a very large mass nearby. But a balloon rubbed in someone’s hair can stick to a wall with a force so strong that it overcomes the force of gravity\|with just the charges in the balloon and the wall! Magnetic force The magnetic force is a diﬁerent manifestation of the electromagnetic force. It stems from the interaction between moving charges as opposed to the ﬂxed charges involved in Coulomb’s Law. Examples of the magnetic force in action include magnets, compasses, car engines, computer data storage and your hair standing on end. Magnets are also used in the wrecking industry to pick up cars and move them around sites. 101 Friction We all know that Newton’s First Law states that an object moving without a force acting on it will keep moving. Then why does a box sliding on a table stop? The answer is friction. Friction arises from the interaction between the molecules on the bottom of a box with the molecules on a table. This interaction is electromagnetic in origin, hence friction is just another view of the electromagnetic force. The great part about school physics is that most of the time we are told to neglect friction but it is good to be aware that there is friction in the real world. Friction is also useful sometimes. If there was no friction and you tried to prop a ladder up against a wall, it would simply slide to the ground. Rock climbers use friction to maintain their grip on cliﬁs. Drag Force This is the force an object experiences while travelling through a medium. When something travels through the air it needs to displace air as it travels and because of this the air exerts a force on the object. This becomes an important force when you move fast and a lot of thought is taken to try and reduce the amount of drag force a sports car experiences. The drag force is very useful for
parachutists. They jump from high altitudes and if there was no drag force, then they would continue accelerating all the way to the ground. Parachutes are wide because the more surface area you show, the greater the drag force and hence the slower you hit the ground. 5.7 Summary of Important Quantities, Equations and Con- cepts Equilibrium Objects at rest or moving with constant velocity are in equilibrium and have a zero resultant force. Equilibrant The equilibrant of any number of forces is the single force required to produce equilibrium. Triangle Law for Forces in Equilibrium Three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order. Newton’s First Law Every object will remain at rest or in uniform motion in a straight line unless it is made to change its state by the action of an external force. Newton’s Second Law The resultant force acting on a body results in an acceleration which is in the same direction as the resultant force and is directly proportional to the magnitude of this force and inversely proportional to the mass of the object. Quantity Mass Acceleration Force Symbol m ¡!a ¡!F S.I. Units kg m:s¡2 kg:m:s¡2 or N Direction \| X X Table 5.1: Summary of the symbols and units of the quantities used in Force 102 Newton’s Third Law For every force or action there is an equal but opposite force or reaction. 103 Chapter 6 Rectilinear Motion 6.1 What is rectilinear motion? Rectilinear motion means motion along a straight line. This is a useful topic to study for learning how to describe the movement of cars along a straight road or of trains along straight railway In this section you have only 2 directions to worry about: (1) along the direction of tracks. motion, and (2) opposite to the direction of motion. To illustrate this imagine a train heading east. Train W P N S E If it is accelerating away from the station platform (P), the direction of acceleration is the same as the direction of the train’s velocity - east. If it is braking the direction of acceleration is opposite to the direction of its motion, i.e. west. 6.2 Speed and Velocity Let’s take a moment to review our deﬂnitions of velocity and speed by looking at the worked example below: 104 Worked Example 24 Speed
and Velocity C B m 0 4 A 30m Question: A cyclist moves from A through B to C in 10 seconds. Calculate both his speed and his velocity. Answer: Step 1 : Decide what information has been supplied The question explicitly gives the distance between A and B the distance between B and C the total time for the cyclist to go from A through B to C † † † all in the correct units! Step 2 : Determine the cyclist’s speed His speed - a scalar - will be v = = s t 30m + 40m 10s = 7 m s Step 3 : First determine the cyclist’s resultant displacement Since velocity is a vector we will ﬂrst need to ﬂnd the resultant displacement of the cyclist. His velocity will be ¡!v = ¡!s t The total displacement is the vector from A to C, and this is just the resultant of the two displacement vectors, ie. ¡!s = ¡!AC = ¡¡!AB + ¡¡!BC 105 Remember to check the units! Using the rule of Pythagoras: ¡!s = (30m)2 + (40m)2 = 50m in the direction f rom A to C q Step 4 : Now we can determine the average velocity from the displacement and the time ) ¡!v = 50m 10s m s = 5 in the direction f rom A to C For this cyclist, his velocity is not the same as his speed because there has been a change in the direction of his motion. If the cyclist traveled directly from A to C without passing through B his speed would be v = 50m 10s m s = 5 and his velocity would be ¡!v = 50m 10s m s = 5 in the direction f rom A to C In this case where the cyclist is not undergoing any change of direction (ie. he is traveling in a straight line) the magnitudes of the speed and the velocity are the same. This is the deﬂning principle of rectilinear motion. Important: For motion along a straight line the magnitudes of speed and velocity are the same, and the magnitudes of the distance and displacement are the same. 6.3 Graphs In physics we often use graphs as important tools for picturing certain concepts. Below are some graphs that help us picture the concepts of displacement, velocity and acceleration. 6.3.1 Dis
placement-Time Graphs Below is a graph showing the displacement of the cyclist from A to C: 106 50 ) ¢¡!s ¢t time (s) 10 This graphs shows us how, in 10 seconds time, the cyclist has moved from A to C. We know the gradient (slope) of a graph is deﬂned as the change in y divided by the change in x, i.e ¢y ¢x. In this graph the gradient of the graph is just ¢¡!s - and this is just the expression for velocity. ¢t Important: The slope of a displacement-time graph gives the velocity. The slope is the same all the way from A to C, so the cyclist’s velocity is constant over the entire displacement he travels. In ﬂgure 6.1 are examples of the displacement-time graphs you will encounter) time ) time ) time Figure 6.1: Some common displacement-time graphs: a) shows the graph for an object stationary over a period of time. The gradient is zero, so the object has zero velocity. b) shows the graph for an object moving at a constant velocity. You can see that the displacement is increasing as time goes on. The gradient, however, stays constant (remember: its the slope of a straight line), so the velocity is constant. Here the gradient is positive, so the object is moving in the direction we have deﬂned as positive. c) shows the graph for an object moving at a constant acceleration. You can see that both the displacement and the velocity (gradient of the graph) increase with time. The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating. 107 6.3.2 Velocity-Time Graphs Look at the velocity-time graph below ¢¡!v ¢t time This is the velocity-time graph of a cyclist traveling from A to B at a constant acceleration, i.e. with steadily increasing velocity. The gradient of this graph is just ¢¡!v - and this is just the expression for acceleration. Because the slope is the same at all points on this graph, the acceleration of the cyclist is constant. ¢t Important: The slope of a velocity-time graph gives the acceleration. Not only can we get the acceleration of an object from its velocity-time graph, but we can also get some idea of the displacement traveled.
Look at the graph below: 10 ) time (s) 5 This graph shows an object moving at a constant velocity of 10m=s for a duration of 5s. The area between the graph and the time axis (the (NOTE TO SELF: SHADED) area) of the above plot will give us the displacement of the object during this time. In this case we just need to calculate the area of a rectangle with width 5s and height 10m/s area of rectangle = height width £ = ¡!v = 10 t £ m s £ 5s = 50m = ¡!s = displacement 108 So, here we’ve shown that an object traveling at 10m/s for 5s has undergone a displacement of 50m. Important: The area between a velocity-time graph and the ‘time’ axis gives the displacement of the object. Here are a couple more velocity-time graphs to get used to) time y t i c o l e v b) time Figure 6.2: Some common velocity-time graphs: In ﬂgure 6.2 are examples of the displacement-time graphs you may encounter. a) shows the graph for an object moving at a constant velocity over a period of time. The gradient is zero, so the object is not accelerating. b) shows the graph for an object which is decelerating. You can see that the velocity is decreasing with time. The gradient, however, stays constant (remember: its the slope of a straight line), so the acceleration is constant. Here the gradient is negative, so the object is accelerating in the opposite direction to its motion, hence it is decelerating. 6.3.3 Acceleration-Time Graphs In this chapter on rectilinear motion we will only deal with objects moving at a constant acceleration, thus all acceleration-time graphs will look like these two) time ) time Here is a description of the graphs below: a) shows the graph for an object which is either stationary or traveling at a constant velocity. Either way, the acceleration is zero over time. 109 b) shows the graph for an object moving at a constant acceleration. In this case the acceler- ation is positive - remember that it can also be negative. We can obtain the velocity of a particle at some given time from an acceleration time graph it is just given by the area between the graph and the time-axis. In the graph below, showing an object at a
constant positive acceleration, the increase in velocity of the object after 2 seconds corresponds to the (NOTE TO SELF: shaded) portion time (s) t area of rectangle = ¡!a £ m = 5 s2 £ m s = 10 2s = ¡!v Its useful to remember the set of graphs below when working on problems. Figure 6.3 shows how displacement, velocity and time relate to each other. Given a displacement-time graph like the one on the left, we can plot the corresponding velocity-time graph by remembering that the slope of a displacement-time graph gives the velocity. Similarly, we can plot an acceleration-time graph from the gradient of the velocity-time graph. 110 Displacement Velocity Acceleration Figure 6.3: A Relationship Between Displacement, Velocity and Acceleration 6.3.4 Worked Examples Worked Example 25 Relating displacement-, velocity-, and acceleration-time graphs Question: Given the displacement-time graph below, draw the corresponding velocitytime and acceleration-time graphs, and then describe the motion of the object time (s) 4 6 Answer: Step 1 : Decide what information is supplied The question explicitly gives a displacement-time graph. Step 2 : Decide what is asked? 3 things are required: 1. Draw a velocity-time graph 2. Draw an acceleration-time graph 111 3. Describe the behaviour of the object Step 3 : Velocity-time graph - 0-2 seconds For the ﬂrst 2 seconds we can see that the displacement remains constant - so the object is not moving, thus it has zero velocity during this time. We can reach this conclusion by another path too: remember that the gradient of a displacement-time graph is the velocity. For the ﬂrst 2 seconds we can see that the displacement-time graph is a horizontal line, ie. it has a gradient of zero. Thus the velocity during this time is zero and the object is stationary. Step 4 : Velocity-time graph - 2-4 seconds For the next 2 seconds, displacement is increasing with time so the object is moving. Looking at the gradient of the displacement graph we can see that it is not constant. In fact, the slope is getting steeper (the gradient is increasing) as time goes on. Thus, remembering that the gradient of a displacement-time graph is the velocity, the velocity must be increasing with time during this phase. Step 5 : Velocity-time graph - 4-6 seconds For the �
��nal 2 seconds we see that displacement is still increasing with time, but this time the gradient is constant, so we know that the object is now travelling at a constant velocity, thus the velocity-time graph will be a horizontal line during this stage. So our velocity-time graph looks like this one below. Because we haven’t been given any values on the vertical axis of the displacement-time graph, we cannot ﬂgure out what the exact gradients are and hence what the values of the velocity are. In this type of question it is just important to show whether velocities are positive or negative, increasing, decreasing or constant time (s) 4 6 Once we have the velocity-time graph its much easier to get the acceleration-time graph as we know that the gradient of a velocity-time graph is the just the acceleration. Step 6 : Acceleration-time graph - 0-2 seconds For the ﬂrst 2 seconds the velocity-time graph is horizontal at zero, thus it has a gradient of zero and there is no acceleration during this time. (This makes sense because we know from the displacement time graph that the object is stationary during this time, so it can’t be accelerating). Step 7 : Acceleration-time graph - 2-4 seconds For the next 2 seconds the velocity-time graph has a positive gradient. This gradient its constant) throughout these 2 seconds so there must be a is not changing (i.e. 112 constant positive acceleration. Step 8 : Accleration-time graph - 4-6 seconds For the ﬂnal 2 seconds the object is traveling with a constant velocity. During this time the gradient of the velocity-time graph is once again zero, and thus the object is not accelerating. The acceleration-time graph looks like this time (s) 4 6 Step 9 : A description of the object’s motion A brief description of the motion of the object could read something like this: At t = 0s and object is stationary at some position and remains stationary until t = 2s when it begins accelerating. It accelerates in a positive direction for 2 seconds until t = 4s and then travels at a constant velocity for a further 2 seconds. Worked Example 26 Calculating distance from a velocity-time graph Question: The velocity-time graph of a car is plotted below. Calculate the displacement of the car has after 15 seconds. 113 2 5 time (s) 12 15 Answer: Step
1 : Decide how to tackle the problem We are asked to calculate the displacement of the car. All we need to remember here is that the area between the velocity-time graph and the time axis gives us the displacement. Step 2 : Determine the area under the velocity-time graph For t = 0s to t = 5s this is the triangle on the left: Area 4 h £ b = 1 2 1 2 = 10m 5s = 4m=s £ For t = 5s to t = 12s the displacement is equal to the area of the rectangle h Area⁄ = w = 7s £ = 28m £ 4m=s For t = 12s to t = 14s the displacement is equal to the area of the triangle above the time axis on the right Area 4 = = 1 2 1 2 h b £ 2s £ 4m=s = 4m For t = 14s to t = 15s the displacement is equal to the area of the triangle below the 114 time axis Area 4 = = 1 2 1 2 h b £ 1s £ 2m=s = 1m Step 3 : Determine the total displacement of the car Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the last second (from t = 14s to t = 15s) the velocity of the car is negative, it means that the car was going in the opposite direction, i.e. back where it came from! So, to get the total displacement, we have to add the ﬂrst 3 areas (those with positive displacements) and subtract the last one (because it signiﬂes a displacement in the opposite direction). ¡!s = 10 + 28 + 4 1 ¡ = 41m in the positive direction Worked Example 27 Velocity from a displacement-time graph Question: Given the diplacement-time graph below, 1. what is the velocity of the object during the ﬂrst 4 seconds? 2. what is the velocity of the object from t = 4s to t = 7s? 2 ) m ( s 4 t (s) 7 Answer: Step 1 : The velocity during the ﬂrst 4 seconds The velocity is given by the slope of a displacement-time graph. During the ﬂrst 4 seconds, this is ¡!v = ¢s ¢t 2m 4s = 0:5m
=s = 115 Step 2 : The velocity during the last 3 seconds For the last 3 seconds we can see that the displacement stays constant, and that the gradient is zero. Thus ¡!v = 0m=s Worked Example 28 From an acceleration-time graph to a velocity-time graph Question: Given the acceleration-time graph below, assume that the object starts from rest and draw its velocity-time graph2 2 4 t (s) 6 Answer: Once again attempt to draw the graph in time sections, i.e. ﬂrst draw the velocity-time graph for the ﬂrst 2 seconds, then for the next 2 seconds and so ons) 6 116 6.4 Equations of Motion This section is about solving problems relating to uniformly accelerated motion. We’ll ﬂrst introduce the variables and the equations, then we’ll show you how to derive them, and after that we’ll do a couple of examples. u = starting velocity (m/s) at t = 0 v = ﬂnal velocity (m/s) at time t s = displacement (m) t = time (s) a = acceleration (m/s2) v = u + at s = (u + v) 2 t s = ut + at2 1 2 v2 = u2 + 2as (6.1) (6.2) (6.3) (6.4) Make sure you can rhyme these oﬁ, they are very important! There are so many diﬁerent types of questions for these equations. Basically when you are answering a question like this: 1. Find out what values you have and write them down. 2. Figure out which equation you need. 3. Write it down!!! 4. Fill in all the values you have and get the answer. Interesting Fact: Galileo Galilei of Pisa, Italy, was the ﬂrst to determined the correct mathematical law for acceleration: the total distance covered, starting from rest, is proportional to the square of the time. He also concluded that objects retain their velocity unless a force { often friction { acts upon them, refuting the accepted Aristotelian hypothesis that objects "naturally" slow down and stop unless a force acts upon them. This principle was incorporated into Newton’s laws of motion (1st law). Equation 6.1 By the deﬂnition of
acceleration a = ¢v t 117 where ¢v is the change in velocity, i.e. ¢v = v a = ¡ v u. Thus we have u ¡ t v = u + at Equation 6.2 In the previous section we saw that displacement can be calculated from the area between a velocity-time graph and the time-axis. For uniformly accelerated motion the most complicated velocity-time graph we can have is a straight line. Look at the graph below - it represents an object with a starting velocity of u, accelerating to a ﬂnal velocity v over a total time t time (s) To calculate the ﬂnal displacement we must calculate the area under the graph - this is just the area of the rectangle added to the area of the triangle. (NOTE TO SELF: SHADING) Area £ £ vt ¡ (v u) ¡ 1 2 ut h Area⁄ = w = t £ = ut £ u Displacement = Area⁄ + Area 4 1 2 ¡ ut s = ut + vt 1 2 (u + v) 2 t = Equation 6.3 This equation is simply derived by eliminating the ﬂnal velocity v in equation 6.2. Remembering from equation 6.1 that v = u + at 118 then equation 6.2 becomes s = = t u + u + at 2 2ut + at2 2 1 2 at2 = ut + Equation 6.4 This equation is just derived by eliminating the time variable in the above equation. From Equation 6.1 we know Substituting this into Equation 6.3 gives t = v u ¡ a s = u( u ) + v ¡ a u )2 ¡ a v2 v a( 1 2 1 2 v2 2a ¡ a( + u2 uv a ¡ a u2 uv a ¡ a 2u2 + v2 + u2 + = = 2as = ¡ v2 = u2 + 2as ¡ 2uv + u2 a2 ) uv a + u2 2a (6.5) This gives us the ﬂnal velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable. Worked Example 29 Question: A racing car has an initial velocity of 100m=s and it covers a displacement of 725m in 10s.
Find its acceleration. Answer: Step 1 : Decide what information has been supplied We are given the quantities u, s and t - all in the correct units. We need to ﬂnd a. Step 2 : Find an equation of motion relating the given information to the acceleration We can use equation 6.3 s = ut + at2 1 2 Step 3 : Rearrange the equation if needed We want to determine the acceleration so we rearrane equation 6.3 to put acceleration on the left of the equals sign: 2(s a = ut) ¡ t2 119 Step 4 : Do the calculation Substituting in the values of the known quantities this becomes 2(725m a = 100 m 10s) ¡ 102s2 s ¢ = = 2( 275m) ¡ 100s2 m 5:5 s2 ¡ Step 5 : Quote the ﬂnal answer The racing car is accelerating at -5.5 m s2, or we could say it is decelerating at 5.5 m s2. Worked Example 30 Question: An object starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64m in 4s. Calculate its acceleration its ﬂnal velocity at what time the object had covered half the total distance what distance the object had covered in half the total time. † † † † Answer: Step 1 : Decide what information is supplied We are given the quantities u, s and t in the correct units. Step 2 : Acceleration: Find an equation to calculate the acceleration and rearrange To calculate the acceleration we can use equation 6.3. Step 3 : Rearrange to make a subject of the formula 2(s a = ut) ¡ t2 Step 4 : Do the calculation Substituting in the values of the known quantities this becomes 2(64m 0 m s 4s ¡ 42s2 a = = 128m 16s2 m s2 = 8 Step 5 : Final velocity: Find an equation to calculate the ﬂnal velocity We can use equation 6.1 - remember we now also know the acceleration of the object. v = u + at 120 Step 6 : Do the calculation v = 0 + (8 m s2 )(4s) m s m s = 32 Step 7 : Time at half the distance: Find an equation to relate the unknown and known quantities Here we have the quantities s, u and a
so we do this in 2 parts, ﬂrst using equation 6.4 to calculate the velocity at half the distance, i.e. 32m: v2 = u2 + 2as = (0m)2 + 2(8m=s2)(32m) = 512m2=s2 v = 22:6m=s Now we can use equation 6.2 to calculate the time at this distance: t = 2s u + v = (2)(32m) 0m=s + 22:6m=s = 2:8s Step 8 : Distance at half the time: Find an equation to relate the distance and time To calculate the distance the object has covered in half the time. Half the time is 2s. Thus we have u, a and t - all in the correct units. We can use equation 6.3 to get the distance: s = ut + at2 1 2 = (0m=s)(2s) + = 16m 1 2 (8 m s2 )(2s)2 Worked Example 31 Question: A ball is thrown vertically upwards with a velocity of 10m=s from the balcony of a tall building. The balcony is 15m above the ground and gravitational accleration is 10m=s2. Find a) the time required for the ball to hit the ground, and b) the velocity with which it hits the ground. Answer: Step 1 : Draw a rough sketch of the problem In most cases helps to make the problem easier to understand if we draw ourselves a picture like the one below: 121 balcony s1 v1 u1 u2 a1; a2 s2 v2 ground where the subscript 1 refers to the upward part of the ball’s motion and the subscript 2 refers to the downward part of the ball’s motion. Step 2 : Decide how to tackle the problem First the ball goes upwards with gravitational acceleration slowing it until it reaches its highest point - here its speed is 0m=s - then it begins descending with gravitational acceleration causing it to increase its speed on the way down. We can separate the motion into 2 stages: Stage 1 - the upward motion of the ball Stage 2 - the downward motion of the ball. We’ll choose the upward direction as positive - this means that gravitation acceleraton is negative. We have used this and we’ll begin by solving for all the variables of Stage 1. Step 3 : For Stage 1
, decide what information is given We have these quantities: u1 = 10m=s v1 = 0m=s a1 = ¡ t1 =? s1 =? 10m=s2 Step 4 : Find the time for stage 1 Using equation 6.1 to ﬂnd t1: v1 = u1 + a1t1 v1 u1 t1 = ¡ a1 0m=s = ¡ = 1s 10m=s ¡ 10m=s2 Step 5 : Find the distance travelled during stage 1 122 We can ﬂnd s1 by using equation 6.4 s1 = 1 = u2 v2 1 + 2a1s1 u2 v2 1 ¡ 1 2a (0m=s)2 2( = = 5m ¡ (10m=s)2 ¡ 10m=s2) Step 6 : For Stage 2, decide what information is supplied For Stage 2 we have the following quantities: u2 = 0m=s v2 =? a2 = ¡ t2 =? s2 = 15m ¡ 10m=s2 5m = 20m ¡ Step 7 : Determine the velocity at the end of stage 2 We can determine the ﬂnal velocity v2 using equation 6.4: 2 = u2 v2 2 + 2a2s2 = (0m=s)2 + 2( = 400(m=s)2 ¡ 10m=s2)( 20m) ¡ v2 = 20m=s downwards Step 8 : Determind the time for stage 2 Now we can determine the time for Stage 2, t2, from equation 6.1: v2 = u2 + a2t2 v2 u2 t2 = ¡ a2 20m=s 0m=s = ¡ = 2s ¡ 10m=s2 ¡ Step 9 : Quote the answers to the problem Finally, a) the time required for the stone to hit the ground is t = t1 + t2 = 1s + 2s = 3s b) the velocity with which it hits the ground is just v2 = 20m=s ¡ These questions do not have the working out in them, but they are all done in the manner described on the previous page. Question: A car starts oﬁ at 10 m/s and accelerates at 1 m
/s2 for 10 seconds. What is it’s ﬂnal velocity? Answer: 20 m/s 123 Question: A car starts from rest, and accelerates at 1 m/s2 for 10 seconds. How far does it move? Answer: 50 m Question: A car is going 30 m/s and stops in 2 seconds. What is it’s stopping distance for this speed? Answer: 30 m Question: A car going at 20 m/s stops in a distance of 20 m/s. 1. What is it’s deceleration? 2. If the car is 1 Tonne (1000 Kg, or 1 Mg) how much force do the brakes exert? 124 6.5 Important Equations and Quantities Quantity Displacement Velocity Distance Speed Acceleration Units Symbol Unit ¡!s ¡!u,¡!v s v ¡!a - Base S.I. Units m + direction m:s¡1 + direction m m:s¡1 m:s¡1 + direction Table 6.1: Units used in Rectilinear Motion 125 Chapter 7 Momentum 7.1 What is Momentum? Momentum is a physical quantity which is closely related to forces. We will learn about this connection a little later. Remarkably momentum is a conserved quantity. This makes momentum extremely useful in solving a great variety of real-world problems. Firstly we must consider the deﬂnition of momentum. Deﬂnition: The momentum of an object is deﬂned as its mass multiplied by its velocity. Mathematically, ¡!p = m¡!v : momentum (kg:m:s¡1 + direction) ¡!p m : mass (kg) ¡!v : velocity (m:s¡1 + direction) Thus, momentum is a property of a moving object and is determined by its velocity and mass. A large truck travelling slowly can have the same momentum as a much smaller car travelling relatively fast. Note the arrows in the equation deﬂning momentum{ momentum is a vector with the same direction as the velocity of the object. Since the direction of an object’s momentum is given by the direction of its motion, one can calculate an object’s momentum in two steps: Momentum is a vector with the same direction as the velocity. calculate the magnitude of the object’s
momentum using, † p = mv : magnitude of momentum (kg:m:s¡1) p m : mass (kg) v : magnitude of velocity (m:s¡1) 126 include in the ﬂnal answer the direction of the object’s motion † Worked Example 32 Calculating Momentum 1 Question: A ball of mass 3kg moves at 2m:s¡1 to the right. Calculate the ball’s momentum. Answer: Step 1 : Decide what information has been supplied The question explicitly gives the ball’s mass, and the ball’s velocity † † in the correct units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the ball’s momentum. From the deﬂnition of momentum, ¡!p = m¡!v ; we see that we need the mass and velocity of the ball, which we are given. Step 3 : Do the calculation We calculate the magnitude of the ball’s momentum, p = mv = (3kg)(2m:s¡1) = 6 kg:m:s¡1: Step 4 : Quote the ﬂnal answer We quote the answer with the direction of the ball’s motion included, ¡!p = 6 kg:m:s¡1 to the right Worked Example 33 Calculating Momentum 2 Question: A ball of mass 500g is thrown at 2m:s¡1. Calculate the ball’s momentum. Answer: Step 1 : Decide what information is supplied The question explicitly gives the ball’s mass, and the magnitude of the ball’s velocity † † but with the ball’s mass in the incorrect units! 127 Remember to check the units! Remember to check the units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the momentum which is deﬂned as ¡!p = m¡!v : Thus, we need the mass and velocity of the ball but we have only its mass and the magnitude of its velocity. In order to determine the velocity of the ball we need the direction of the ball’s motion. If the problem does not give an explicit direction we are forced to be general. In a case like this we could say that the direction
of the velocity is in the direction of motion of the ball. This might sound silly but the lack of information in the question has forced us to be, and we are certainly not wrong! The ball’s velocity is then 2m:s¡1 in the direction of motion. Step 3 : Convert the mass to the correct units 1000g = 1kg 1 = 1kg 1000g 500g £ 1 = 500g £ = 0:500kg 1kg 1000g Step 4 : Do the calculation Now, let us ﬂnd the magnitude of the ball’s momentum, p = mv = (0:500kg)(2m:s¡1) = 1 kg:m:s¡1 Step 5 : Quote the ﬂnal answer Remember to include the direction of the momentum: ¡!p = 1 kg:m:s¡1 in the direction of motion of the ball Worked Example 34 Calculating the Momentum of the Moon Question: The moon is 384 400km away from the earth and orbits the earth in 1022kg1 what is the magnitude of its 27.3 days. If the moon has a mass of 7:35 momentum if we assume a circular orbit? Answer: Step 1 : Decide what information has been supplied The question explicitly gives £ 1This is 1 81 of the mass of the earth 128 with mass in the correct units but all other quantities in the incorrect units. The units we require are Remember to check the units! the moon’s mass, the distance to the moon, and the time for one orbit of the moon seconds (s) for time, and metres (m) for distance † † † † † Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate only the magnitude of the moon’s momentum (i.e. we do not need to specify a direction). In order to do this we require the moon’s mass and the magnitude of its velocity, since Step 3 : Find the speed or magnitude of the moon’s velocity Speed is deﬂned as, p = mv: speed = Distance time We are given the time the moon takes for one orbit but not how far it travels in that time. However, we can work this out from the distance to the moon and the fact that the moon’s orbit is circular. Firstly let us convert the distance to the moon to the
correct units, 1km = 1000m 1000m 1km 1 = 384 400km £ 1000m 1km 1 = 384 400km £ = 384 400 000m 108 m = 3:844 £ Using the equation for the circumference, C, of a circle in terms of its radius, we can determine the distance travelled by the moon in one orbit: C = 2…r = 2…(3:844 = 2:42 £ 109 m: 108 m) £ Next we must convert the orbit time, T, into the correct units. Using the fact that a day contains 24 hours, an hour consists of 60 minutes, and a minute is 60 seconds long, 1day = (24)(60)(60)seconds 1 = (24)(60)(60)s 1day 27:3days £ 1 = 27:3days (24)(60)(60)s 1day = 2:36 106s £ 129 Therefore, T = 2:36 106s: £ Combining the distance travelled by the moon in an orbit and the time taken by the moon to complete one orbit, we can determine the magnitude of the moon’s velocity or speed, v = Distance time = C T = 1:02 103 m:s¡1: £ Step 4 : Finally calculate the momentum and quote the answer The magnitude of the moon’s momentum is: p = mv = (7:35 = 7:50 £ £ 1022kg)(1:02 £ 1025 kg:m:s¡1: 103 m:s¡1) 7.2 The Momentum of a System In Chapter?? the concept of a system was introduced. The bodies that make up a system can have diﬁerent masses and can be moving with diﬁerent velocities. In other words they can have diﬁerent momenta. Deﬂnition: The total momentum of a system is the sum of the momenta of each of the objects in the system. Since momentum is a vector, the techniques of vector addition discussed in Chapter?? must be used to calculate the total momentum of a system. Let us consider an example. Worked Example 35 Calculating the Total Momentum of a System Question: Two billiard balls roll towards each other. They each have a mass of 0:3kg. Ball 1 is moving at v1 = 1 m:s¡1 to the right, while ball 2 is moving at v2 = 0
:8 m:s¡1 to the left. Calculate the total momentum of the system. Answer: Step 1 : Decide what information is supplied The question explicitly gives the mass of each ball, the velocity of ball 1, ¡!v1, and † † 130 the velocity of ball 2, ¡!v2, † all in the correct units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the total momentum of the system. In this example our system consists of two balls. To ﬂnd the total momentum we must sum the momenta of the balls, Remember to check the units! ¡!p total = ¡!p1 + ¡!p2 Since ball 1 is moving to the right, its momentum is in this direction, while the second ball’s momentum is directed towards the left. System m1 ¡!p 1 m2 ¡!p 2 Thus, we are required to ﬂnd the sum of two vectors acting along the same straight line. The algebraic method of vector addition introduced in Chapter?? can thus be used. Step 3 : Choose a positive direction Let us choose right as the positive direction, then obviously left is negative. Step 4 : Calculate the momentum The total momentum of the system is then the sum of the two momenta taking the directions of the velocities into account. Ball 1 is travelling at 1 m:s¡1 to the right or +1 m:s¡1. Ball 2 is travelling at 0:8 m:s¡1 to the lef t or 0:8 m:s¡1. Thus, ¡ Right is the positive direction ¡!p total = m1¡!v1 + m2¡!v2 = (0:3kg)(+1 m:s¡1) + (0:3kg)( = (+0:3 kg:m:s¡1) + ( = +0:06 kg:m:s¡1 = 0:06 kg:m:s¡1 to the right ¡ ¡ 0:24 kg:m:s¡1) 0:8 m:s¡1) In the last step the direction was added in words. Since the result in the second last line is positive, the total momentum of the system is in the positive direction (i.e
. to the right). 7.3 Change in Momentum If either an object’s mass or velocity changes then its momentum too will change. If an object has an initial velocity ¡!u and a ﬂnal velocity ¡!v, then its change in momentum, ¢¡!p, is 131 ¢¡!p = ¡!p f inal ¡ ¡!p initial = m¡!v ¡ m¡!u Worked Example 36 Change in Momemtum Question: A rubber ball of mass 0:8kg is dropped and strikes the oor at a velocity of 6 m:s¡1. It bounces back with an initial velocity of 4 m:s¡1. Calculate the change in momentum of the rubber ball caused by the oor. Answer: Step 1 : Decide what information has been supplied The question explicitly gives the ball’s mass, the ball’s initial velocity, and the ball’s ﬂnal velocity † † † all in the correct units. Do not be confused by the question referring to the ball bouncing back with an \initial velocity of 4 m:s¡1". The word \initial" is included here since the ball will obviously slow down with time and 4 m:s¡1 is the speed immediately after bouncing from the oor. Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the change in momentum of the ball, ¢¡!p = m¡!v m¡!u : ¡ We have everything we need to ﬂnd ¢¡!p. Since the initial momentum is directed downwards and the ﬂnal momentum is in the upward direction, we can use the algebraic method of subtraction discussed in the vectors chapter. Step 3 : Choose a positive direction Let us choose down as the positive direction. Then substituting, Down is the positive direction Step 4 : Do the calculation and quote the answer Remember to check the units! ¢¡!p = m¡!v m¡!u ¡ = (0:8kg)( = (0:8kg)( = = 8 kg:m:s¡1 up 8 kg:m:s¡1 ¡ ¡ ¡ 4 m:s¡1) ¡ 10 m:s�
�1) (0:8kg)(+6 m:s¡1) where we remembered in the last step to include the direction of the change in momentum in words. 132 ¡!u 1 m1 ¡!u 2 m2 Figure 7.1: Before the collision. 7.4 What properties does momentum have? You may at this stage be wondering why there is a need for introducing momentum. Remarkably momentum is a conserved quantity. Within an isolated system the total momentum is constant. No matter what happens to the individual bodies within an isolated system, the total momentum of the system never changes! Since momentum is a vector, its conservation implies that both its magnitude and its direction remains the same. This Principle of Conservation of Linear Momentum is one of the most fundamental principles of physics and it alone justiﬂes the deﬂnition of momentum. Since momentum is related to the motion of objects, we can use its conservation to make predictions about what happens in collisions and explosions. If we bang two objects together, by conservation of momentum, the total momentum of the objects before the collision is equal to their total momentum after the collision. Momentum is conserved in isolated systems! Principle of Conservation of Linear Momentum: The total linear momentum of an isolated system is constant. or In an isolated system the total momentum before a collision (or explosion) is equal to the total momentum after the collision (or explosion). Let us consider a simple collision of two pool or billiard balls. Consider the ﬂrst ball (mass m1) to have an initial velocity (¡!u1). The second ball (mass m2) moves towards the ﬂrst ball with an initial velocity ¡!u2. This situation is shown in Figure 7.1. If we add the momenta of each ball we get a total momentum for the system. This total momentum is then ¡!p total bef ore = m1¡!u1 + m2¡!u2; After the two balls collide and move away they each have a diﬁerent momentum. If we call the ﬂnal velocity of ball 1 ¡!v1 and the ﬂnal velocity of ball 2 ¡!v2 (see Figure 7.2), then the total momentum of the system after the collision is ¡!p total af ter = m1¡!v1 + m2�
�!v2; This system of two balls is isolated since there are no external forces acting on the balls. Therefore, by the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision. This gives the equation for the conservation of momentum in a collision of two objects, 133 ¡!v 1 m1 m2 ¡!v 2 Figure 7.2: After the collision. ¡!p total bef ore = ¡!p total af ter m1¡!u1 + m2¡!u2 = m1¡!v1 + m2¡!v2 m1 m2 : mass of object 1 (kg) : mass of object 2 (kg) ¡!u1 ¡!u2 ¡!v1 ¡!v2 : initial velocity of object 1 (m:s¡1 + direction) : initial velocity of object 2 (m:s¡1 + direction) : ﬂnal velocity of object 1 (m:s¡1 + direction) : ﬂnal velocity of object 2 (m:s¡1 + direction) This equation is always true- momentum is always conserved in collisions. The chapter ‘Collisions and Explosions’ (Chapter??) deals with applications of momentum conservation. 7.5 Impulse At the beginning of this chapter it was mentioned that momentum is closely related to force. We will now explain the nature of this connection. Consider an object of mass m moving with constant acceleration ¡!a. During a time ¢t the object’s velocity changes from an initial velocity ¡!u to a ﬂnal velocity ¡!v (refer to Figure 7.3). We know from Newton’s First Law that there must be a resultant force ¡!F Res acting on the object. Starting from Newton’s Second Law, Momentum is always conserved in collisions! ¡!F Res = m¡!a = m( ¡!v ) ¡ ¡!u ¢t m¡!u m¡!v = ¡ ¢t = ¡!p f inal ¡ ¡!p initial ¢t since ¡!a = ¡!v ¡ ¡!u ¢t = ¢¡!p ¢t
This alternative form of Newton’s Second Law is called the Law of Momentum. 134 ¡!F Res m ¡!u t = 0 t = ¢t ¡!v ¡!F Res m Figure 7.3: An object under the action of a resultant force. Law of Momentum: The applied resultant force acting on an object is equal to the rate of change of the object’s momentum and this force is in the direction of the change in momentum. Mathematically, ¡!F Res = ¢¡!p ¢t ¡!F Res ¢¡!p ¢t : resultant force (N + direction) : change in momentum (kg:m:s¡1 + direction) : time over which ¡!F Res acts (s) Rearranging the Law of Momentum, The product ¡!F Res¢t is called impulse, ¡!F Res¢t = ¢¡!p : Impulse · ¡!F Res¢t = ¢¡!p From this equation we see, that for a given change in momentum, ¡!F Res¢t is ﬂxed. Thus, if FRes is reduced, ¢t must be increased (i.e. the resultant force must be applied for longer). Alternatively if ¢t is reduced (i.e. the resultant force is applied for a shorter period) then the resultant force must be increased to bring about the same change in momentum. Worked Example 37 Impulse and Change in momentum 135 Question: A 150 N resultant force acts on a 300 kg object. Calculate how long it takes this force to change the object’s velocity from 2 m:s¡1 to the right to 6 m:s¡1 to the right. Answer: Step 1 : Decide what information is supplied The question explicitly gives the object’s mass, the object’s initial velocity, the object’s ﬂnal velocity, and the resultant force acting on the object † † † † all in the correct units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the time taken ¢t to accelerate the object from the given initial velocity to ﬂnal velocity. From the Law of Momentum, ¡!F Res¢t = ¢¡!p m
¡!u = m¡!v ¡ ¡ ¡!u ): = m(¡!v Thus we have everything we need to ﬂnd ¢t! Step 3 : Choose a positive direction Although not explicitly stated, the resultant force acts to the right. This follows from the fact that the object’s velocity increases in this direction. Let us then choose right as the positive direction. Step 4 : Do the calculation and quote the ﬂnal answer Right is the positive direction Remember to check the units! ¡!F Res¢t = m(¡!v ¡ ¡!u ) (+150N )¢t = (300kg)((+6 (+150N )¢t = (300kg)(+4 m s m s (300kg)(+4 m s ) +150N ) ¢t = (+2 ) ¡ m s )) ¢t = 8s Worked Example 38 Calculating Impulse Question: A cricket ball weighing 156g is moving at 54 km:hr¡1 towards a batsman. It is hit by the batsman back towards the bowler at 36 km:hr¡1. Calculate i) the 136 Remember to check the units! ball’s impulse, and ii) the average force exerted by the bat if the ball is in contact with the bat for 0:13s. Answer: Step 1 : Decide what information is supplied The question explicitly gives the ball’s mass, the ball’s initial velocity, the ball’s ﬂnal velocity, and the time of contact between bat and ball † † † † all except the time in the wrong units! Answer to (i): Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the impulse Impulse = ¢¡!p = ¡!F Res¢t: Since we do not have the force exerted by the bat on the ball (¡!F Res), we have to calculate the impulse from the change in momentum of the ball. Now, since ¢¡!p = ¡!p f inal ¡ ¡!p initial = m¡!v m¡!u ; ¡ we need the ball’s mass, initial velocity and ﬂnal velocity, which we are given. Step 3 : Convert to S.I
. units Firstly let us change units for the mass 1000g = 1kg 1 = 1kg 1000g 156g £ 1 = 156g £ = 0:156kg 1kg 1000g Next we change units for the velocity 1km = 1000m 1000m 1km 1 = 3600s = 1hr 1 = 1hr 3600s 54 km hr £ 1 £ 1 = 54 = 15 km hr £ m s 1000m 1km £ 1hr 3600s 137 36 km hr £ 1 £ 1 = 36 = 10 km hr £ m s 1000m 1km £ 1hr 3600s Step 4 : Choose your convention Next we must choose a positive direction. Let us choose the direction from the batsman to the bowler as the postive direction. Then the initial velocity of the ball 15 m:s¡1, while the ﬂnal velocity of the ball is ¡!v = +10 m:s¡1 is ¡!u = Step 5 : Calculate the momentum Now we calculate the change in momentum, ¡ Direction from batsman to bowler is the positive direction ¢¡!p = ¡!p f inal ¡ ¡!p initial = m¡!v m¡!u ¡ ¡ ¡!u ) = m(¡!v = (0:156kg)((+10 m:s¡1) = +3:9 kg:m:s¡1 = 3:9 kg:m:s¡1 in the direction from batsman to bowler 15 m:s¡1)) ( ¡ ¡ where we remembered in the last step to include the direction of the change in momentum in words. Step 6 : Determine the impulse Finally since impulse is just the change in momentum of the ball, Impulse = ¢¡!p = 3:9 kg:m:s¡1 in the direction from batsman to bowler Answer to (ii): Step 7 : Determine what is being asked What is being asked? We are asked to calculate the average force exerted by the bat on the ball, ¡!F Res. Now, Impulse = ¡!F Res¢t = ¢¡!p : We are given ¢t and we have calculated the change in momentum or impulse of the ball in part (i)! Step 8 : Choose a convention Next we choose a positive direction. Let us choose
the direction from the batsman to the bowler as the postive direction. Step 9 : Calculate the force Then substituting, Direction from batsman to bowler is the positive direction 138 ¡!F Res¢t = Impulse ¡!F Res(0:13s) = +3:9 kg:m s +3:9 kg:m s 0:13s ¡!F Res = = +30 kg:m s2 = 30N in the direction from batsman to bowler where we remembered in the ﬂnal step to include the direction of the force in words. 7.6 Summary of Important Quantities, Equations and Con- cepts Quantity Momentum Mass Velocity Change in momentum Force Impulse Units Symbol Unit ¡!p m ¡!u,¡!v ¢¡!p ¡!F J N - S.I. Units Direction kg:m:s¡1 kg m:s¡1 kg:m:s¡1 kg:m:s¡2 kg:m:s¡1 X \| X X X X Table 7.1: Summary of the symbols and units of the quantities used in Momentum Momentum The momentum of an object is deﬂned as its mass multiplied by its velocity. Momentum of a System The total momentum of a system is the sum of the momenta of each of the objects in the system. Principle of Conservation of Linear Momentum: ‘The total linear momentum of an isolated system is constant’ or ‘In an isolated system the total momentum before a collision (or explosion) is equal to the total momentum after the collision (or explosion)’. Law of Momentum: The applied resultant force acting on an object is equal to the rate of change of the object’s momentum and this force is in the direction of the change in momentum. 139 Chapter 8 Work and Energy 8.1 What are Work and Energy? During this chapter you will discover that work and energy are very closely related: We consider the energy of an object as its capacity to do work and doing work as the process of transferring energy from one object or form to another. In other words, † † an object with lots of energy can do lots of work. when work is done, energy is lost by the object doing work and gained by the object on which the work is done. Lifting objects or throwing them requires that
you do work on them. Even making electricity ow requires that something do work. Something must have energy and transfer it through doing work to make things happen. 8.2 Work To do work on an object, one must move the object by applying a force with at least a component in the direction of motion. The work done is given by W = Fks W : work done (N:m or J) Fk s : component of applied force parallel to motion (N ) : displacement of the object (m) It is very important to note that for work to be done there must be a component of the applied force in the direction of motion. Forces perpendicular to the direction of motion do no work. As with all physical quantities, work must have units. As follows from the deﬂnition, work is measured in N:m. The name given to this combination of S.I. units is the joule (J). Deﬂnition: 1 joule is the work done when an object is moved 1m under the application of a force of 1N in the direction of motion. 140 The work done by an object can be positive or negative. Since force (Fk) and displacement (s) are both vectors, the result of the above equation depends on their directions: † † If Fk acts in the same direction as the motion then positive work is being done. In this case the object on which the force is applied gains energy. If the direction of motion and Fk are opposite, then negative work is being done. This means that energy is transferred in the opposite direction. For example, if you try to push a car uphill by applying a force up the slope and instead the car rolls down the hill you are doing negative work on the car. Alternatively, the car is doing positive work on you! Worked Example 39 Calculating Work Done I Question: If you push a box 20m forward by applying a force of 15N in the forward direction, what is the work you have done on the box? Answer: Step 1 : Analyse the question to determine what information is provided † † † The force applied is F = 15N. The distance moved is s = 20m. The applied force and distance moved are in the same direction. Therefore, Fk = 15N. These quantities are all in the correct units, so no unit conversions are required. Step 2 : Analyse the question to determine what is being asked �
� We are asked to ﬂnd the work done on the box. We know from the deﬂnition that work done is W = Fks Step 3 : Next we substitute the values and calculate the work done W = Fks = (15N )(20m) = 300 N = 300 J m ¢ Remember that the answer must be positive as the applied force and the motion are in the same direction (forwards). In this case, you (the pusher) lose energy, while the box gains energy. 141 Worked Example 40 Calculating Work Done II Question: What is the work done by you on a car, if you try to push the car up a hill by applying a force of 40N directed up the slope, but it slides downhill 30cm? Answer: Step 1 : Analyse the question to determine what information is provided † † † The force applied is F = 40N The distance moved is s = 30cm. This is expressed in the wrong units so we must convert to the proper S.I. units (meters): s = 30cm = 30cm 1m 100cm ¢ = 0:3m The applied force and distance moved are in opposite directions. Therefore, if we take s = 0:3m, then Fk = 40N. ¡ Step 2 : Analyse the question to determine what is being asked † We are asked to ﬂnd the work done on the car by you. We know that work done is W = Fks Step 3 : Substitute the values and calculate the work done Again we have the applied force and the distance moved so we can proceed with calculating the work done: W = Fks = ( 40N )(0:3m) = = ¡ 12N 12 J ¡ ¡ m ¢ Note that the answer must be negative as the applied force and the motion are in opposite directions. In this case the car does work on the person trying to push. What happens when the applied force and the motion are not parallel? If there is an angle between the direction of motion and the applied force then to determine the work done we have to calculate the component of the applied force parallel to the direction of motion. Note that this means a force perpendicular to the direction of motion can do no work. Worked Example 41 Calculating Work Done III Question: Calculate the work done on a box, if it is pulled 5m along the ground by applying a
force of F = 10N at an angle of 60o to the horizontal. 142 F 60o Answer: Step 1 : Analyse the question to determine what information is provided The force applied is F = 10N The distance moved is s = 5m along the ground The angle between the applied force and the motion is 60o † † † These quantities are in the correct units so we do not need to perform any unit conversions. Step 2 : Analyse the question to determine what is being asked We are asked to ﬂnd the work done on the box. † Step 3 : Calculate the component of the applied force in the direction of motion Since the force and the motion are not in the same direction, we must ﬂrst calculate the component of the force in the direction of the motion. Fk F 60o Fjj From the force diagram we see that the component of the applied force parallel to the ground is Fjj = F cos(60o) ¢ = 10N = 5 N cos(60o) ¢ Step 4 : Substitute and calculate the work done Now we can calculate the work done on the box: W = Fks = (5N )(5m) = 25 J Note that the answer is positive as the component of the force Fk is in the same direction as the motion. We will now discuss energy in greater detail. 143 8.3 Energy As we mentioned earlier, energy is the capacity to do work. When positive work is done on an object, the system doing the work loses energy. In fact, the energy lost by a system is exactly equal to the work done by the system. Like work (W ) the unit of energy (E) is the joule (J). This follows as work is just the transfer of energy. A very important property of our universe which was discovered around 1890 is that energy is conserved. Energy is never created nor destroyed, but merely transformed from one form to another. Energy conservation and the conservation of matter are the principles on which classical me- chanics is built. Energy is conserved! IN THE ABSENCE OF FRICTION When work is done on an object by a system: -the object gains energy equal to the work done by the system Work Done = Energy Transferred IN THE PRESENCE OF FRICTION When work is done by a system: -only some of the energy lost by the system is transferred into useful energy -the rest of the energy transferred is
lost to friction Total Work Done = Useful Work Done + Work Done Against Friction 8.3.1 Types of Energy So what diﬁerent types of energy exist? Kinetic, mechanical, thermal, chemical, electrical, radiant, and atomic energy are just some of the types that exist. By the principle of conservation of energy, when work is done energy is merely transferred from one object to another and from one type of energy to another. 144 Kinetic Energy Kinetic energy is the energy of motion that an object has. Objects moving in straight lines possess translational kinetic energy, which we often abbreviate as Ek. The translational kinetic energy of an object is given by Ek = 1 2 mv2 : kinetic energy (J) Ek m : mass of object (kg) v : speed of the object (m:s¡1) Note the dependence of the kinetic energy on the speed of the object{ kinetic energy is related to motion. The faster an object is moving the greater its kinetic energy. Worked Example 42 Calculation of Kinetic Energy Question: If a rock has a mass of 1kg and is thrown at 5m=s, what is its kinetic energy? Answer: Step 1 : Analyse the question to determine what information is provided The mass of the rock m = 1kg The speed of the rock v = 5m=s † † These are both in the correct units so we do not have to worry about unit conversions. Step 2 : Analyse the question to determine what is being asked † We are asked to ﬂnd the kinetic energy. From the deﬂnition we know that to work out Ek, we need to know the mass and the velocity of the object and we are given both of these values. Step 3 : Substitute and calculate the kinetic energy mv2 Ek = = 1 2 1 2 (1kg)(5 )2 m s m2 ¢ s2 kg = 12:5 = 12:5 J 145 To check that the units in the above example are in fact correct: kg m2 ¢ s2 m kg ¢ s2 ¢ ¶ = = J m = N m ¢ The units are indeed correct! Study hint: Checking units is an important cross-check and you should get into a habit of doing this. If you, for example, ﬂnish an exam early then checking the units in your calculations is a very good idea. Worked Example 43 Mixing Units
and Kinetic Energy Calculations 1 Question: If a car has a mass of 900kg and is driving at 60km=hr, what is its kinetic energy? Answer: Step 1 : Analyse the question to determine what information is provided † † The mass of the car m = 900kg The speed of the car v = 60km=hr. These are not the units we want so before we continue we must convert to m=s. We do this by multiplying by one: 60 km hr £ 1 = 60 1000m 1km km hr £ m hr = 60000 Now we need to change from hours to seconds so we repeat our procedure: 60000 m hr £ 1 = 60000 = 16:67 1hr 3600s m hr £ m s and so the speed in the units we want is v = 16:67m=s. Step 2 : Analyse the question to determine what is being asked We are asked to ﬂnd the kinetic energy. † Step 3 : Substitute and calculate 146 We know we need the mass and the speed to work out Ek and we are given both of these quantities. We thus simply substitute them into the equation for Ek: Ek = = 1 2 1 2 mv2 (900kg)(16:67 m s )2 = 125 000 kgm2 s2 = 125 000 J Worked Example 44 Mixing Units and Kinetic Energy Calculations 2 Question: If a bullet has a mass of 150g and is shot at a muzzle velocity of 960m=s, what is its kinetic energy? Answer: Step 1 : Analyse the question to determine what information is provided † We are given the mass of the bullet m = 150g. This is not the unit we want mass to be in. We need to convert to kg. Again, we multiply by one: 150g ¢ 1 = 150g ¢ = 0:15kg 1kg 1000g † We are given the muzzle velocity which is just how fast the bullet leaves the barrel and it is v = 960m=s. Step 2 : Analyse the question to determine what is being asked We are asked to ﬂnd the kinetic energy. † Step 3 : Substitute and calculate We just substitute the mass and velocity (which are known) into the equation for Ek: Ek = = 1 2 1 2 mv2 (150kg)(960 m s )2 = 69 120 kgm2 s2 = 69 120 J 147 Potential Energy If
you lift an object you have to do work on it. This means that energy is transferred to the object. But where is this energy? This energy is stored in the object and is called potential energy. The reason it is called potential energy is because if we let go of the object it would move. Deﬂnition: Potential energy is the energy an object has due to its position or state. As an object raised above the ground falls, its potential energy is released and transformed into kinetic energy. The further it falls the faster it moves as more of the stored potential energy is transferred into kinetic energy. Remember, energy is never created nor destroyed, but merely transformed from one type to another. In this case potential energy is lost but an equal amount of kinetic energy is gained. In the example of a falling mass the potential energy is known as gravitational potential energy as it is the gravitational force exerted by the earth which causes the mass to accelerate towards the ground. The gravitational ﬂeld of the earth is what does the work in this case. Another example is a rubber-band. In order to stretch a rubber-band we have to do work on it. This means we transfer energy to the rubber-band and it gains potential energy. This potential energy is called elastic potential energy. Once released, the rubber-band begins to move and elastic potential energy is transferred into kinetic energy. Gravitational Potential Energy As we have mentioned, when lifting an object it gains gravitational potential energy. One is free to deﬂne any level as corresponding to zero gravitational potential energy. Objects above this level then possess positive potential energy, while those below it have negative potential energy. To avoid negative numbers in a problem, always choose the lowest level as the zero potential mark. The change in gravitational potential energy of an object is given by: ¢EP = mg¢h ¢EP : Change in gravitational potential energy (J) m : mass of object (kg) g ¢h : acceleration due to gravity (m:s¡2) : change in height (m) When an object is lifted it gains gravitational potential energy, while it loses gravitational potential energy as it falls. Worked Example 45 Gravitational potential energy 148 Question: How much potential energy does a brick with a mass of 1kg gain if it is lifted 4m. Answer: Step 1 : Analyse the question to determine what information is provided The mass of the brick is m = 1kg The height lifted is ¢h = 4
m † † These are in the correct units so we do not have to worry about unit conversions. Step 2 : Analyse the question to determine what is being asked We are asked to ﬂnd the gain in potential energy of the object. † Step 3 : Identify the type of potential energy involved Since the block is being lifted we are dealing with gravitational potential energy. To work out ¢EP, we need to know the mass of the object and the height lifted. As both of these are given, we just substitute them into the equation for ¢EP. Step 4 : Substitute and calculate ¢EP = mg¢h = (1kg) 10 = 40 kg ‡ m2 ¢ s2 = 40 J (4m) m s2 · 8.4 Mechanical Energy and Energy Conservation Kinetic energy and potential energy are together referred to as mechanical energy. The total mechanical energy (U ) of an object is then the sum of its kinetic and potential energies: U = EP + EK 1 2 U = mgh + mv2 (8.1) Now, IN THE ABSENCE OF FRICTION Mechanical energy is conserved Ubef ore = Uaf ter This principle of conservation of mechanical energy can be a very powerful tool for solving physics problems. However, in the presence of friction some of the mechanical energy is lost: 149 IN THE PRESENCE OF FRICTION Mechanical energy is not conserved (The mechanical energy lost is equal to the work done against friction) ¢U = Ubef ore ¡ Uaf ter = Work Done Against Friction Worked Example 46 Using Mechanical Energy Conservation Question: A 2kg metal ball is suspended from a rope. If it is released from point A and swings down to the point B (the bottom of its arc) what is its velocity at point B? A 0.5m B Answer: Step 1 : Analyse the question to determine what information is provided The mass of the metal ball is m = 2kg The change in height going from point A to point B is h = 0:5m The ball is released from point A so the velocity at point A is zero (vA = 0m=s). † † † These are in the correct units so we do not have to worry about unit conversions. Step 2 : Analyse the question to determine what is being asked Find the velocity of the metal ball at point B. † Step 3 : Determine the Mechanical Energy at A and
B To solve this problem we use conservation of mechanical energy as there is no friction. Since mechanical energy is conserved, UA = UB Therefore we need to know the mechanical energy of the ball at point A (UA) and at point B (UB). The mechanical energy at point A is UA = mghA + 1 2 m(vA)2 150 We already know m, g and vA, but what is hA? Note that if we let hB = 0 then hA = 0:5m as A is 0:5m above B. In problems you are always free to choose a line corresponding to h = 0. In this example the most obvious choice is to make point B correspond to h = 0. Now we have, UA = (2kg) 10 = 10 J ‡ m s2 · (0:5m) + (2kg)(0)2 1 2 As already stated UB = UA. Therefore UB = 10J, but using the deﬂnition of mechanical energy UB = mghB + 1 2 m(vB)2 = 1 2 m(vB)2 because hB = 0. This means that (2kg)(vB)2 10J = 1 2 (vB)2 = 10 J kg vB = p10 m s 8.5 Summary of Important Quantities, Equations and Con- cepts Quantity Work Kinetic Energy Potential Energy Mechanical Energy Units Symbol Unit W EK EP U J J J J S.I. Units N:m or kg:m2:s¡2 N:m or kg:m2:s¡2 N:m or kg:m2:s¡2 N:m or kg:m2:s¡2 Direction \| \| \| \| Table 8.1: Summary of the symbols and units of the quantities used in Energy Principle of Conservation of Energy: Energy is never created nor destroyed, but merely transformed from one form to another. Conservation of Mechanical Energy: In the absence of friction, the total mechanical energy of an object is conserved. 151 Essay 1 : Energy Author: Asogan Moodaly Asogan Moodaly received his Bachelor of Science degree (with honours) in Mechanical Engineering from the University of Natal, Durban in South Africa. For his ﬂnal year design project he worked on a 3-axis ﬂlament winding machine for composite (Glass re-en
going to last forever. 3. Nuclear power is cleaner in terms of emissions but theres no proven way of disposing of the nuclear waste. Oh, and it wont last forever either! Renewable Energy As the name suggests renewable energy lasts forever. Solar (sun), wind, geothermal, wave, hydro and biomass (organic) are all sources of energy that will last until the sun eventually explodes many millions of years from now. Hopefully the human race will have moved from the earth by then! Generally the principal of renewable electricity generation is similar to fossil fuel electricity generation in that electricity is generated by moving a magnet relative to a conducting coil. What is diﬁerent is the way energy is supplied to cause that motion. The below are a few diﬁerent types of available renewable energy technologies. Solar There are diﬁerent types of solar electricity technologies, the main ones being solar thermal and photovoltaic. 153 Sun Heated oil stored in insulated tank for night use Transmission cables & pylon Steam pipes Steam Turbine Generator Mirrored trough reflector, reflects sunlight onto black pipes Black pipes carrying oil Heat exchanger transfers heat from hot oil to water Water becomes steam Water pipes Solar thermal uses the heat of the sun to produce electricity. Sun is concentrated using mirrors. This heat either creates steam which drives a turbine which in turn drives a generator (as per fossil fuel generation), or drives an air engine (engine that uses expanding air to obtain motion) that drives a generator. Photovoltaic panels convert sunlight directly into electricity. The beneﬂt of photovoltaic panels is that there are no moving parts, and is therefore relatively maintenance free. The downside is that its very expensive at this stage (17/06/2004). Solar Water Heaters could save up to 30% of the total electricity used in a house. Heater on roof Hot Water out Insulated geyser to store water Cold water in Wooden box with glass top provides insulation. Silicon gel should be used to make the box airtight. If air escapes, so does heat. copper Water (in pipes painted black) is heated by sunlight. Cold water sinks. Hot water rises into the geyser. Wind Wind turbines catch wind that spins the blades. The blades are connected to a shaft that spins because of the wind. This spinning shaft spins another shaft that turns a permanent magnet relative to conducting coils. Note that gears are used to convert the slow spinning of the 1st shaft to a faster spin on the
2nd shaft. The generator shaft needs to spin at the correct speed to produce the right amount and quality of electricity. Some generators are now being modiﬂed to run at slower speeds. This saves money as gears are not needed. 154 Vessel Nacelle containing all moving parts Blades Detail of Nacelle Vessel (top view) to blades Shaft 1 (slow spin) Shaft 2 (Fast spin) Gears Generator Biomass Biomass is anything organic i.e. plant or animal matter. It can be used in the place of coal as per a normal coal ﬂred plant and is renewable as long as the biomass e.g. wood; is handled in a sustainable manner. By sustainable I mean that suitable farming practices are used so that the land is not over farmed which will result in the soil becoming barren and nothing growing there again. Pipe carrying biogas for heat or power Processed biomass fertilizer Biogas bubbles Upside down floating container to catch biogas Input biomass Water forms a seal to keep air out Input biomass being digested Biomass can also be processed using anaerobic digestion to produce a gas that can be burned for heat or electricity. This biogas is made up of a number of other gases that are similar to those found in fossil fuel natural gas Except the amount of the gases are diﬁerent. E.g. Natural gas has about 94 Anaerobic digestion: Anaerobic means No air. Therefore anaerobic digestion means to digest in the absence of air. Bacteria that naturally exist in organic matter will convert organic matter to biogas and fertilizer when all the air is removed. Thousands of anaerobic digesters have been installed in rural India, Nepal and China in rural areas where cow dung, human waste and chicken litter (faeces) are all processed using anaerobic digestion to produce gas that can be burned in the home for cooking and heating. The leftover is used as fertilizer. Geothermal Energy In some places on earth, the earths crust is thinner than others. As a result the heat from the earths core escapes. The heat can be captured by converting water to steam, and using the 155 steam to drive a steam generator as discussed above. Hydroelectric power Water from a river is diverted to turn a water turbine to create electricity similar to the principles of steam generation. The water is returned to the river after driving the turbine. Pipe diverting water Transmission cables Generator house River Wave Energy
Some wave energy generators work similarly to wind turbines except that underwater ocean currents turns the blades instead of wind; and of course most of the structure is under water! Ocean current blades Ocean floor Underwater pipe to shore carrying transmission cable Another concept uses the rising and falling of the tides to suck air in using a one way valve. As a result air becomes compressed in a chamber and the compressed air is let out to drive a turbine which in turn drives a generator One way valve allows air to be sucked in then shuts when air tries to get out. Generator house with turbine and generator Chamber with compressed air Cliff face Ocean floor As the water level falls and rises, air is sucked in and compressed 156 These are relatively new technologies. Liquid Fuels Liquid fuels are used mainly for transportation. Petrol and diesel are the most common liquid fuels and are obtained from oil. Sasol is the only company in the world that makes liquid fuels from coal; and will be one of the leading companies in the world to make liquid fuels from natural gas! The Sasol petro-chemical plants are based in Sasolburg on the border of the Free State and in Secunda in Mpumalanga. However, as discussed above coal, gas and oil are fossil fuels and are not renewable. Petrol and diesel are obtained from fossil fuels and therefore pollute and contribute to the green house eﬁect (global warming). Alternatives Biodiesel Oil can be extracted from plants such as the soya bean, sunower and rapeseed by pressing it through a ﬂlter. This oil if mixed correctly with either methanol or dry ethanol and Sodium Hydroxide will separate the plant oil into biodiesel, glycerol and fertilizer. The biodiesel can be used as produced in a conventional diesel engine with little or no mod- iﬂcations required. The glycerol can be reﬂned a bit further for pharmaceutical companies to use, or can be used to make soap. Ethanol Corn, maize and sugar cane can be used to make ethanol as a fuel substitute for petrol. Its made by the same fermentation process used to make alcohol. Enzymes are often used to speed up the process. In ethanol from sugar cane production, the leftover bagasse (the ﬂbre part of the sugar cane) can be burned in a biomass power station to produce electricity. Hydrogen Through the process of electrolysis electricity (hopefully clean, renewable electricity!) can split water into hydrogen and oxygen.
The stored hydrogen can be used in a fuel cell to create electricity in a process that is opposite to electrolysis; to drive electric motors in a car. The hydrogen can also be burned directly in a modiﬂed internal combustion engine. In both cases the waste product is water. 157 Essay 2 : Tiny, Violent Collisions Author: Thomas D. Gutierrez Tom Gutierrez received his Bachelor of Science and Master degrees in physics from San Jose State University in his home town of San Jose, California. As a Master’s student he helped work on a laser spectrometer at NASA Ames Research Centre. The instrument measured the ratio of diﬁerent isotopes of carbon in CO2 gas and could be used for such diverse applications as medical diagnostics and space exploration. Later, he received his Ph.D. in physics from the University of California, Davis where he performed calculations for various reactions in high energy physics collisions. He currently lives in Berkeley, California where he studies proton-proton collisions seen at the STAR experiment at Brookhaven National Laboratory on Long Island, New York. High Energy Collisions Take an orange and expanded it to the size of the earth. The atoms of the earth-sized orange would themselves be about the size of regular oranges and would ﬂll the entire \earth-orange". Now, take an atom and expand it to the size of a football ﬂeld. The nucleus of that atom would be about the size of a tiny seed in the middle of the ﬂeld. From this analogy, you can see that atomic nuclei are very small objects by human standards. They are roughly 10¡15 meters in diameter { one-hundred thousand times smaller than a typical atom. These nuclei cannot be seen or studied via any conventional means such as the naked eye or microscopes. So how do scientists study the structure of very small objects like atomic nuclei? The simplest nucleus, that of hydrogen, is called the proton. Faced with the inability to isolate a single proton, open it up, and directly examine what is inside, scientists must resort to a brute-force and somewhat indirect means of exploration: high energy collisions. By colliding protons with other particles (such as other protons or electrons) at very high energies, one hopes to learn about what they are made of and how they work. The American physicist Richard Feynman once compared this process to slamming delecate watches together and ﬂguring
out how they work by only examining the broken debris. While this analogy may seem pessimistic, with su–cent mathematical models and experimental precision, considerable information can be extracted from the debris of such high energy subatomic collisions. One can learn about both the nature of the forces at work and also about the sub-structure of such systems. The experiments are in the category of \high energy physics" (also known as \subatomic" physics). The primary tool of scientiﬂc exploration in these experiments is an extremely violent collision between two very, very small subatomic objects such as nuclei. As a general rule, the higher the energy of the collisions, the more detail of the original system you are able to resolve. These experiments are operated at laboratories such as CERN, SLAC, BNL, and Fermilab, just to name a few. The giant machines that perform the collisions are roughly the size of towns. For example, the RHIC collider at BNL is a ring about 1 km in diameter and can be seen from space. The newest machine currently being built, the LHC at CERN, is a ring 9 km in diameter! Let’s examine the kinematics of such a collisions in some detail... 158 Chapter 9 Collisions and Explosions In most physics courses questions about collisions and explosions occur and to solve these we must use the ideas of momentum and energy; with a bit of mathematics of course! This section allows you to pull the momentum and energy ideas together easily with some speciﬂc problems. 9.1 Types of Collisions We will consider two types of collisions in this section Elastic collisions Inelastic collisions † † In both types of collision, total energy and total momentum is always conserved. Kinetic energy is conserved for elastic collisions, but not for inelastic collisions. 9.1.1 Elastic Collisions Deﬂnition: An elastic collision is a collision where total momentum and total kinetic energy are both conserved. (NOTE TO SELF: this should be in an environment for deﬂnitions!!) This means that the total momentum and the total kinetic energy before an elastic collision is the same as after the collision. For these kinds of collisions, the kinetic energy is not changed into another type of energy. Before the Collision In the following diagram, two balls are rolling toward each other, about to collide ¡!p 1, K1 ¡!p 2, K
2 159 Before the balls collide, the total momentum of the system is equal to all the individual momenta added together. The ball on the left has a momentum which we call ¡!p 1 and the ball on the right has a momentum which we call ¡!p 2, it means the total momentum before the collision is ¡!p Before = ¡!p 1 + ¡!p 2 (9.1) We calculate the total kinetic energy of the system in the same way. The ball on the left has a kinetic energy which we call K1 and the ball on the right has a kinetic energy which we call K2, it means that the total kinetic energy before the collision is KBefore = K1 + K2 (9.2) After the Collision The following diagram shows the balls after they collide ¡!p 3, K3 ¡!p 4, K4 After the balls collide and bounce oﬁ each other, they have new momenta and new kinetic energies. Like before, the total momentum of the system is equal to all the individual momenta added together. The ball on the left now has a momentum which we call ¡!p 3 and the ball on the right now has a momentum which we call ¡!p 4, it means the total momentum after the collision is ¡!p After = ¡!p 3 + ¡!p 4 The ball on the left now has a kinetic energy which we call K3 and the ball on the right now has a kinetic energy which we call K4, it means that the total kinetic energy after the collision is (9.3) KAfter = K3 + K4 (9.4) Since this is an elastic collision, the total momentum before the collision equals the total momentum after the collision and the total kinetic energy before the collision equals the total kinetic energy after the collision Before ¡!p Before ¡!p 1 + ¡!p 2 After = ¡!p After = ¡!p 3 + ¡!p 4 and KBefore K1 + K2 = KAfter = K3 + K4 (9.5) (9.6) Worked Example 47 An Elastic Collision 160 We will have a look at the collision between two pool balls. Ball 1 is at rest and ball 2 is moving towards it with a speed of 2 [m:s¡1]. The mass of each ball is 0.3 [Kg
]. After the balls collide elastically, ball 2 comes to a stop and ball 1 moves oﬁ. What is the ﬂnal velocity of ball 1? Step 1 : Draw the \before" diagram Before the collision, ball 2 is moving; we will call it’s momentum P2 and it’s kinetic energy K2. Ball 1 is at rest, so it has zero kinetic energy and momentum. 2 1 ¡!p 2, K2 ¡!p 1 = 0, K1 = 0 Step 2 : Draw the \after" diagram After the collision, ball 2 is at rest but ball 1 has a momentum which we call P3 and a kinetic energy which we call K3. ¡!p 4 = 0, K4 = 0 ¡!p 3, K3 2 1 Because the collision is elastic, we can solve the problem using momentum conservation or kinetic energy conservation. We will do it both ways to show that the answer is the same, whichever method you use. Step 3 : Show the conservation of momentum We start by writing down that the momentum before the collision ¡!p Before is equal to the momentum after the collision ¡!p After Before After ¡!p Before = ¡!p After ¡!p 1 + ¡!p 2 = ¡!p 3 + ¡!p 4 0 + ¡!p 2 = ¡!p 3 + 0 ¡!p 2 = ¡!p 3 We know that momentum is just P = mv, and we know the masses of the balls, so we can rewrite the conservation of momentum in terms of the velocities of the balls (9.7) ¡!p 2 = ¡!p 3 m2v2 = m3v3 0:3v2 = 0:3v3 v2 = v3 So ball 1 exits with the velocity that ball 2 started with! v3 = 2[m:s¡1] 161 (9.8) (9.9) Step 4 : Show the conservation of kinetic energy We start by writing down that the kinetic energy before the collision KBefore is equal to the kinetic energy after the collision KAfter Before After KBefore = KAfter K1 + K2 = K3 + K4 0 + K2 = K3 + 0 K2 = K3 (9.10) We know that kinetic energy is just K = mv 2, and we
know the masses of the balls, so we can rewrite the conservation of kinetic energy in terms of the velocities of the balls 2 K2 = K3 (9.11) = m3v2 3 2 m2v2 2 2 0:15v2 2 = 0:15v2 3 2 = v2 v2 3 v2 = v3 So ball 1 exits with the velocity that ball 2 started with, which agrees with the answer we got when we used the conservation of momentum. v3 = 2[m:s¡1] (9.12) Worked Example 48 Elastic Collision 2 Question: Now for a slightly more di–cult example. We have 2 marbles. Marble 1 has mass 50 g and marble 2 has mass 100 g. I roll marble 2 along the ground towards marble 1 in the positive x-direction. Marble 1 is initially at rest and marble 2 has a velocity of 3 m:s¡1 in the positive x-direction. After they collide elastically, both marbles are moving. What is the ﬂnal velocity of each marble? Answer: Step 1 : Put all the quantities into S.I. units So: m1 = 0:05kg and m2 = 0:1kg Step 2 : Draw a rough sketch of the situation Before the collision: 2 1 ¡!p2, Ek2 ¡!p1 = 0, Ek1 = 0 162 After the collision: 2 1 ¡!p3, Ek3 ¡!p4, Ek4 Step 3 : Decide which equations to use in the problem Since the collision is elastic, both momentum and kinetic energy are conserved in the collision. So: EkBef ore = EkAf ter and ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter There are two unknowns (¡!v1 and ¡!v2) so we will need two equations to solve for them. We need to use both kinetic energy conservation and momentum conservation in this problem. Step 4 : Solve the ﬂrst equation Let’s start with energy conservation. Then: 1 2 2 m1¡!u1 + 2 m1¡!u1 2 EkBef ore = EkAf ter 1 2 + m2¡!u2 1 2 = m1¡!v1 m1¡!v1 2 m
2¡!u2 = 2 2 + 1 2 + m2¡!v2 2 m2¡!v2 2 But ¡!u1=0, and solving for ¡!v2 2 : 2 2 ¡!v2 ¡!v2 2 ¡!v2 2 = ¡!u2 ¡ = (3)2 = 9 ¡ ¡ 1 2 ¡!v1 2 m1 m2 ¡!v1 (0:05) (0:10) ¡!v1 2 2 (A) Step 5 : Solve the second equation Now we have simpliﬂed as far as we can, we move onto momentum conservation: ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter m1¡!u1 + m2¡!u2 = m1¡!v1 + m2¡!v2 But ¡!u1=0, and solving for ¡!v1: m2¡!u2 = m1¡!v1 + m2¡!v2 m2¡!v2 m1¡!v1 = m2¡!u2 m2 ¡!v2 ¡!u2 m1 ¡!v1 = ¡ m2 m1 ¡!v1 = 2(3) ¡!v1 = 6 ¡ ¡ 2¡!v2 ¡ 2¡!v2 (B) 163 Step 6 : Substitute one equation into the other Now we can substitute (B) into (A) to solve for ¡!v2: 2 2 2 2 2 2 ¡!v2 ¡!v2 ¡!v2 ¡!v2 3¡!v2 ¡!v2 = 9 = 9 = 9 ¡ ¡ ¡ 2¡!v2)2 2 1 2 ¡!v1 1 (6 2 1 2 ¡ 18 + 12¡!v2 (36 ¡ = 9 = ¡ = 4¡!v2 ¡ 9 + 12¡!v2 3 ¡ 24¡!v2 + 4¡!v2 2 ) 2 2¡!v2 ¡ 2 ¡!v2 (¡!
v2 ¡ ¡ 3)(¡!v2 4¡!v2 + 3 = 0 1) = 0 ¡!v2 = 3 or ¡!v2 = 1 ¡ We were lucky in this question because we could factorise. If you can’t factorise, then you can always solve using the formula for solving quadratic equations. Remember: b x = ¡ § 4ac pb2 2a ¡ So, just to check: ¡!v2 = 4 § 42 ¡ 2(1) 4(1)(3) 4 ¡ § 12 ¡!v2 = p p16 2 p4 § 2 1 ¡!v2 = 3 or ¡!v2 = 1 same as before ¡!v2 = ¡!v2 = 2 § 4 Step 7 : Solve for and quote the ﬂnal answers So ﬂnally, substituting into equation (B) to get ¡!v1: ¡!v1 = 6 2¡!v2 ¡ If ¡!v2 = 3 m:s¡1 then ¡!v1 = 6 ¡ 2(3) = 0 m:s¡1 But, according to the question, marble 1 is moving after the collision. So ¡!v1 ¡!v2 = 3. Therefore: = 0 and ¡!v2 = 1 m:s¡1 in the positive x and ¡!v1 = 4 m:s¡1 in the positive x direction direction ¡ ¡ 164 6 6 9.1.2 Inelastic Collisions Deﬂnition: An inelastic collision is a collision in which total momentum is conserved but total kinetic energy is not conserved; the kinetic energy is transformed into other kinds of energy. So the total momentum before an inelastic collisions is the same as after the collision. But the total kinetic energy before and after the inelastic collision is diﬁerent. Of course this does not mean that total energy has not been conserved, rather the energy has been transformed into another type of energy. As a rule of thumb, inelastic collisions happen when the colliding objects are distorted in some way. Usually they change their shape. To modify the shape of an object requires energy and this is where the \missing" kinetic energy goes. A classic
example of an inelastic collision is a car crash. The cars change shape and there is a noticeable change in the kinetic energy of the cars before and after the collision. This energy was used to bend the metal and deform the cars. Another example of an inelastic collision is shown in the following picture. Here an asteroid (the small circle) is moving through space towards the moon (big circle). Before the moon and the asteroid collide, the total momentum of the system is: ¡¡¡¡!pBef ore = ¡!pm + ¡!pa (¡!pm stands for ¡¡¡!pmoon and ¡!pa stands for ¡¡¡¡¡! pasteroid) and the total kinetic energy of the system is: EBef ore = Ekm + Eka ¡!pm, Ekm ¡!pa, Eka When the asteroid collides inelastically with the moon, its kinetic energy is transformed mostly into heat energy. If this heat energy is large enough, it can cause the asteroid and the area of the moon’s surface that it hit, to melt into liquid rock! From the force of impact of the asteroid, the molten rock ows outwards to form a moon crater. After the collision, the total momentum of the system will be the same as before. But since this collision is inelastic, (and you can see that a change in the shape of objects has taken place!), 165 total kinetic energy is not the same as before the collision. ¡¡¡¡!pAf ter, EkAf ter ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter ¡!pm + ¡!pa = ¡¡¡¡!pAf ter but = EkAf ter = EkAf ter EkBef ore Ekm + Eka So: Worked Example 49 Inelastic Collision Question: Let’s consider the collision of two cars. Car 1 is at rest and Car 2 is moving at a speed of 2m:s¡1 in the negative x-direction. Both cars each have a mass of 500kg. The cars collide inelastically and stick together. What is the resulting velocity of the resulting mass of metal? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: Car 1 ¡
!p1 = 0 Car 2 ¡!p2 After the collision: ¡¡¡¡!pAf ter 166 6 6 Step 2 : Decide which equations to use in the problem We know the collision is inelastic and there was a deﬂnite change in shape of the objects involved in the collision - there were two objects to start and after the collision there was one big mass of metal! Therefore, we know that kinetic energy is not conserved in the collision but total momentum is conserved. So: EkBef ore = EkAf ter but pT Bef ore = ¡¡¡¡!pAf ter ¡¡¡¡¡¡! Step 3 : Solve for and quote the ﬂnal velocity So we must use conservation of momentum to solve this problem. Take the negative x-direction to have a negative sign: pT Bef ore = ¡¡¡¡!pAf ter ¡¡¡¡¡¡! ¡!p1 + ¡!p2 = ¡¡¡¡!pAf ter m1¡!u1 + m2¡!u2 = (m1 + m2)¡!v 2) = (500 + 500)¡!v 0 + 500( ¡ ¡ 1000 = 1000¡!v ¡!v = 1 m:s¡1 ¡ Therefore, ¡!v = 1 m:s¡1 in the negative x direction: ¡ 9.2 Explosions When an object explodes, it breaks up into more than one piece and it therefore changes its shape. Explosions occur when energy is transformed from one kind e.g. chemical potential energy to another e.g. heat energy or kinetic energy extremely quickly. So, like in inelastic collisions, total kinetic energy is not conserved in explosions. But total momentum is always conserved. Thus if the momenta of some of the parts of the exploding object are measured, we can use momentum conservation to solve the problem! Interesting Fact: The Tunguska event was an aerial explosion that occurred near the Podkamennaya (Stony) Tunguska River in what is now Evenkia, Siberia, at 7:17 AM on June 30, 1908. The size of the blast was later estimated to be equivalent
to between 10 and 15 million tons of regular explosive. It felled an estimated 60 million trees over 2,150 square kilometers. At around 7:15 AM, Tungus natives and Russian settlers in the hills northwest of Lake Baikal observed a huge ﬂreball moving across the sky, nearly as bright as the Sun. A few minutes later, there was a ash that lit up half of the sky, followed by a shock wave that 167 6 knocked people oﬁ their feet and broke windows up to 650 km away (the same as the distance from Bloemfontein to Durban!). The explosion registered on seismic stations across Europe and Asia, and produced uctuations in atmospheric pressure strong enough to be detected in Britain. Over the next few weeks, night skies over Europe and western Russia glowed brightly enough for people to read by. Had the object responsible for the explosion hit the Earth a few hours later, it would have exploded over Europe instead of the sparsely-populated Tunguska region, producing massive loss of human life. In the following picture, a closed can of baked beans is put on a stove or ﬂre: BAKED BEANS ¡¡!pcan; Ekcan Before the can heats up and explodes, the total momentum of the system is: ¡¡¡¡!pBef ore = ¡¡!pcan = 0 and the total kinetic energy of the system is: EkBef ore = Ekcan = 0 since the can isn’t moving. Once the mixture of beans, juice and air inside the can reach a certain temperature, the pressure inside the can becomes so great that the can explodes! Beans and sharp pieces of metal can y out in all directions. Energy in the system has been transformed from heat energy into kinetic energy. ¡!p4; Ek4 ¡!p3; Ek3 ¡!p1; Ek1 ¡!p2; Ek2 168 After the explosion, the can is completely destroyed. But momentum is always conserved, so: ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter ¡¡¡¡!pBef ore = ¡!p1 + ¡!p2 + ¡!p3 + ¡!p4 0 = ¡!p1 + ¡!p2 + �
�!p3 + ¡!p4 However, the kinetic energy of the system is not conserved. The can’s shape was changed in the explosion. Before the explosion the can was not moving, but after the explosion, the pieces of metal and baked beans were moving when they were ying out in all directions! So: EkB 6 = EkA Safety tip: Never heat a closed can on a stove or ﬂre! Always open the can or make a hole in the lid to allow the pressure inside and outside the can to remain equal. This will prevent the can from exploding! Worked Example 50 Explosions 1 Question: An object with mass mT ot = 10 kg is sitting at rest. Suddenly it explodes into two pieces. One piece has a mass of m1 = 5 kg and moves oﬁ in the negative x-direction at ¡!v1 = 3ms¡1. What is the velocity of the other piece? Answer: Step 1 : Draw a rough sketch of the situation Before the explosion, the object is at rest: ¡¡!pT ot = 0, EkT ot = 0 After the explosion, the two pieces move oﬁ: 1 2 ¡!p1, Ek1 ¡!p2, Ek2 Step 2 : Decide which equations to use in the problem Now we know that in an explosion, total kinetic energy is not conserved. There is a deﬂnite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So: ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter ¡¡¡¡!pBef ore = ¡!p1 + ¡!p2 169 But the object was initially at rest so: 0 = ¡!p1 + ¡!p2 0 = m1¡!v1 + m2¡!v2 (A) Step 3 : Find the mass of the second piece Now we know that m1= 5 kg but we do not know what the mass of m2 is. However, we do know that: mT ot = m1 + m2 m2 = mT ot ¡ = 10 kg ¡ m1 5 kg = 5 kg Step 4 : Solve for and quote the velocity of the other piece Now we can substitute all the
values we know into equation (A) and solve for ¡!v2. Let’s choose the positive x-direction to have a postive sign and the negative x-direction to have a negative sign: (A) 0 = m1¡!v1 + m2¡!v2 0 = 5( 3) + 5¡!v2 0 = ¡ 5¡!v2 = 15 ¡!v2 = +3 m:s¡1 ¡ 15 + 5¡!v2 Therefore, ¡!v2 = 3 m:s¡1 in the positive x direction: ¡ Worked Example 51 Explosions 2 Question: An object with mass mT ot =15 kg is sitting at rest. Suddenly it explodes into two pieces. One piece has a mass of m1 = 5000 g and moves oﬁ in the positive x-direction at v1 = 7ms¡1. What is the ﬂnal velocity of the other piece? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: ¡¡!pT ot = 0, EkT ot = 0 170 After the collision: 2 1 ¡!p2, Ek2 ¡!p1, Ek1 Step 2 : Convert all units into S.I. units m1 = 5000 g m1 = 5 kg Step 3 : Decide which equations to use in the problem Now we know that in an explosion, total kinetic energy is not conserved. There is a deﬂnite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So: ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter ¡¡¡¡!pBef ore = ¡!p1 + ¡!p2 But the object was initially at rest so: 0 = ¡!p1 + ¡!p2 0 = m1¡!v1 + m2¡!v2 (A) Step 4 : Determine the mass of the other piece Now we know that m1= 5 kg but we do not know what the mass of m2 is. However, we do know that: mT ot = m1 + m2 m2 = mT ot ¡ = 15 kg ¡ m1 5 kg =
10 kg Step 5 : Solve for and quote the ﬂnal velocity of the other piece Now we can substitute all the values we know into equation (A) and solve for ¡!v2. Let’s choose the positive x-direction to have a postive sign and the negative x-direction to have a negative sign: (A) 0 = m1¡!v1 + m2¡!v2 0 = 5(7) + 10(¡!v2) 0 = 35 + 10(¡!v2) 35 3:5 m:s¡1 ¡ ¡ 10(¡!v2) = ¡!v2 = Therefore, ¡!v2 = 3:5 m:s¡1 in the negative x direction: ¡ 171 9.3 Explosions: Energy and Heat In explosions, you have seen that kinetic energy is not conserved. But remember that total energy is always conserved. Let’s look at what happens to the energy in some more detail. If a given amount of energy is released in an explosion it is not necessarily all transformed into kinetic energy. Due to the deformation of the exploding object, often a large amount of the energy is used to break chemical bonds and heat up the pieces. Energy is conserved but some of it is transferred through non-conservative processes like It will be radiated into the heating. This just means that we cannot get the energy back. environment as heat energy but it is all still accounted for. Now we can start to mix the ideas of momentum conservation with energy transfer to make longer problems. These problems are not more complicated just longer. We will start oﬁ short and them combine the diﬁerent ideas later on. Long problems should be treated like a number of smaller problems. Focus on them one at a time. Worked Example 52 Energy Accounting 1 Question: An object with a mass of mt = 17 kg explodes into two pieces of mass m1 = 7 kg and m2 = 10 kg. m1 has a velocity of 9ms¡1 in the negative x-direction and m2 has a velocity of 6:3ms¡1 in the positive x-direction. If the explosion released a total energy of 2000 J, how much was used in a non-conservative way? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: After the collision: ¡¡
!pT ot = 0, EkT ot = 0 E = 2000 J 2 1 ¡!p2, Ek2 ¡!p1, Ek1 Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Determine what is being asked We are asked how much energy was used in a nonconservative fashion. This is the diﬁerence between how much energy was used in a conservative fashion and how much was used in total. We are lucky because we have everything we need to determine the kinetic energy of both pieces. The kinetic energy of the pieces is energy that was used in a conservative way. 172 Step 3 : Determine the total kinetic energy The sum of the kinetic energy for the two blocks is the total kinetic energy of the pieces. So: EkT ot = Ek1 + Ek2 2 + = m1¡!v1 1 2 1 2 2 1 m2¡!v2 2 1 2 = (7)(9)2 + (10)(6:3)2 = 283:5 + 198:45 EkT ot = 481:95 J Step 4 : Solve for and quote the ﬂnal answer The total energy that was transformed into kinetic energy is 481:95 J. We know that 2000 J of energy were released in total. the question makes no statements about other types of energy so we can assume that the diﬁerence was lost in a non-conservative way. Thus the total energy lost in non-conservative work is: EkT ot = 2000 481:95 ¡ = 1518:05 J E ¡ Worked Example 53 Energy Accounting 2 Question: An object at rest, with mass mT ot = 4 kg, explodes into two pieces (m1, m2) with m1 = 2:3 kg. m1 has a velocity of 17ms¡1 in the negative x-direction. If the explosion released a total energy of 800 J, 1. What is the velocity of m2? 2. How much energy does it carry? 3. And how much energy was used in a non-conservative way? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: After the collision: ¡¡!pT ot = 0, EkT ot = 0 173 E = 800 J 1 ¡!p1; Ek1 2 ¡!p2; Ek2 Step 2 : N ow we know that in an
explosion, total kinetic energy is not conserved. There is a deﬂnite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So: ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter ¡¡¡¡!pBef ore = ¡!p1 + ¡!p2 But the object was initially at rest so: 0 = ¡!p1 + ¡!p2 0 = m1¡!v1 + m2¡!v2 (A) Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) Now we know that m1= 2.3 kg but we do not know what the mass of m2 is. However, we do know that: mT ot = m1 + m2 m2 = mT ot ¡ = 4 kg ¡ = 1:7 kg m1 2:3 kg Step 4 : (NOTE TO SELF: step is deprecated, use westep instead.) Now we can substitute all the values we know into equation (A) and solve for ¡!v2. Let’s choose the positive x-direction to have a postive sign and the negative x-direction to have a negative sign: 0 = m1¡!v1 + m2¡!v2 0 = (2:3)( 17) + 1:7¡!v2 (A) ¡ 39:1 + 1:7¡!v2 0 = 1:7¡!v2 = 39:1 ¡ ¡!v2 = 23 m:s¡1 † ¡!v2 = 23 m:s¡1 in the positive x -direction Step 5 : (NOTE TO SELF: step is deprecated, use westep instead.) Now we need to calculate the energy that the second piece carries: Ek2 = = 2 m2¡!v2 (1:7)(23)2 1 2 1 2 = 449:65 J 174 The kinetic energy of the second piece is Ek2 = 449:65 J † Step 6 : (NOTE TO SELF: step is deprecated, use westep instead.) Now the amount of energy used in a non-conservative way in the explosion, is the diﬁerence between the amount
of energy released in the explosion and the total kinetic energy of the exploded pieces: We know that: E ¡ EkT ot = 800 EkT ot ¡ EkT ot = Ek1 + Ek2 = = 1 2 1 2 m1¡!v1 2 + 449:65 (2:3)(17)2 + 449:65 = 332:35 + 449:65 = 782 J Step 7 : (NOTE TO SELF: step is deprecated, use westep instead.) So going back to: E ¡ EkT ot = 800 = 800 EkT ot 782 ¡ ¡ = 18 J 18 J of energy was used in a non-conservative way in the explosion † 9.4 Important Equations and Quantities Quantity velocity momentum energy Symbol Unit \| \| J ¡!v ¡!p E Units m s kg:m s kg:m s2 2 S.I. Units or m:s¡1 or kg:m:s¡1 or kg:m2s¡2 Direction X X \| Table 9.1: Units commonly used in Collisions and Explosions Momentum: Kinetic energy: ¡!p = m¡!v Ek = 1 2 m¡!v 2 175 (9.13) (9.14) Chapter 10 Newtonian Gravitation 10.1 Properties Gravity is a force and therefore must be described by a vector - so remember magnitude and direction. Gravity is a force that acts between any two objects with mass. To determine the magnitude of the force we use the following equation: F = Gm1m2 r2 This equation describes the force between two bodies, one of mass m1, the other of mass m2 (both have units of Kilogrammes, or Kg for short). The G is Newton’s ‘Gravitational Constant’ 10¡11 [Nm2kg¡2]) and r is the straight line distance between the two bodies in meters. (6:673 This means the bigger the masses, the greater the force between them. Simply put, big things matter big with gravity. The 1=r2 factor (or you may prefer to say r¡2) tells us that the distance between the two bodies plays a role as well. The closer two bodies are, the stronger the gravitational force between them is. We feel the gravitational attraction of the Earth most at the surface since that is the closest
we can get to it, but if we were in outer-space, we would barely even know the Earth’s gravity existed! (10.1) £ Remember that F = ma (10.2) which means that every object on the earth feels the same gravitational acceleration! That means whether you drop a pen or a book (from the same height), they will both take the same length of time to hit the ground... in fact they will be head to head for the entire fall if you drop them at the same time. We can show this easily by using the two equations above (10.1 and 10.2). The force between the Earth (which has the mass me) and an object of mass mo is F = Gmome r2 and the acceleration of an object of mass mo (in terms of the force acting on it) is So we substitute equation (10.3) into equation (10.4), and we ﬂnd that ao = F mo ao = Gme r2 176 (10.3) (10.4) (10.5) Since it doesn’t matter what mo is, this tells us that the acceleration on a body (due to the Earth’s gravity) does not depend on the mass of the body. Thus all objects feel the same gravitational acceleration. The force on diﬁerent bodies will be diﬁerent but the acceleration will be the same. Due to the fact that this acceleration caused by gravity is the same on all objects we label it diﬁerently, instead of using a we use g which we call the gravitational acceleration. 10.2 Mass and Weight Weight is a force which is measured in Newtons, it is the force of gravity on an object. People are always asking other people \What is your weight?" when in fact they should be asking \What is your mass?". Mass is measured in Kilograms (Kg) and is the amount of matter in an object, it doesn’t change unless you add or remove matter from the object (if you continue to study physics through to university level, you will ﬂnd that Einstein’s theory of relativity means that mass can change when you travel as fast as light does, but you don’t need to worry about that right now). There are 1000g in 1Kg and 1000Kg in a Tonne. To change mass into weight we use Newton’s 2nd Law
which is F = Ma. The weight is the force and gravity the acceleration, it can be rewritten as: W is the Weight, measured in Newtons. M is the Mass, measured in Kg and g is the acceleration due to gravity, measured in m=s2 it is equal to 10 on the Earth. W = mg (10.6) 10.2.1 Examples 1. A bag of sugar has a mass of 1Kg, what is it’s weight? (Acceleration due to gravity = 10m=s2) - Step 1: Always write out the equation, it helps you to understand the question, and you will get marks as well. W = M g (10.7) - Step 2: Fill in all the values you know. (remember to make sure the mass is in Kg and NOT in grams or Tonnes!) W = 1 10 £ W = 10 (10.8) - Step 3: Write out the answer remembering to include the units! You will lose marks if you don’t W = 10Newtons (10.9) 2. A space-man has a mass of 90Kg, what is his weight (a) on the earth? (b) on the moon? (c) in outer space? (The acceleration due to gravity on the earth is 10m=s2, on the moon gravity is 1/6 of the gravity on earth). (a) W = M g 10 = 900 W = 90 £ W = 900Newtons 177 (10.10) (b) W = 90 W = M g 1=6 = 150 10 £ £ W = 150Newtons (10.11) (c) Weightless in outer space because g = 0. So now when somebody asks you your weight, you know to reply \Anything!! But my mass is a diﬁerent matter!" 10.3 Normal Forces If you put a book on a table it does not accelerate it just lies on the table. We know that gravity is acting on it with a force F = G mEmbook r2 (10.12) but if there is a net force there MUST be an acceleration and there isn’t. This means that the gravitational force is being balanced by another force1. This force we call the normal force. It is the reaction force between the book and the table. It is equal to the force of gravity on the book. This is also the force we measure

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