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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 wave...
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 a...
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...
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...
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 ree...
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...
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 ...
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 ...
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 ray...
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 ...
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 th...
.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 ree...
(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...
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-u...
/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 dia...
��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...
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...
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). Som...
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 ...
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 veloci...
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...
��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 p...
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 displace...
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 general...
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 ...
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 i...
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 constructi...
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 ske...
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 afte...
: 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. lef...
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 trigonome...
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 (...
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 a...
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 ...
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...
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...
�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...
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 stationa...
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 i...
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 th...
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 th...
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 ...
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??. Ac...
: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 i...
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 accelera...
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 rock...
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 (Coulom...
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...
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 ...
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 gr...
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 ca...
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...
��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 ...
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 t...
=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 ob...
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 velocit...
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 e...
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: ...
, 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...
/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 2...
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: ...
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...
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 conve...
: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 i...
. 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 Questio...
�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 ...
�!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 col...
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 i...
¡!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...
. 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 £ 1...
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 ...
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 giv...
� 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 (forwa...
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...
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 d...
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 w...
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 t...
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 ...
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 an...
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 source...
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...
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 th...
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 Autho...
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 ...
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....
]. 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...
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 conser...
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 + ...
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 p1...
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 ast...
!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...
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, the...
�!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 ...
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...
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 = 3...
!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 u...
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) S...
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 =...
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 ...
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...

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