question_id stringlengths 8 35 | subject stringclasses 3
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values | topic stringclasses 459
values | question stringlengths 17 24.5k | options stringlengths 2 4.26k | correct_option stringclasses 6
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1l6f5hcxe | physics | center-of-mass | center-of-mass-of-discrete-particles | <p>Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $$\sqrt x$$ m. The value of... | [] | null | 2 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6ummxzq/f571ac43-99e4-45d3-856a-a0bab6271d11/f49d5360-1c86-11ed-b633-f353ad3cb8e4/file-1l6ummxzr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6ummxzq/f571ac43-99e4-45d3-856a-a0bab6271d11/f49d5360-1c86-11ed-b633-f353ad3cb8e4... | integer | jee-main-2022-online-25th-july-evening-shift | 9,415 |
1l6p4a8py | physics | center-of-mass | center-of-mass-of-discrete-particles | <p>Two bodies of mass $$1 \mathrm{~kg}$$ and $$3 \mathrm{~kg}$$ have position vectors $$\hat{i}+2 \hat{j}+\hat{k}$$ and $$-3 \hat{i}-2 \hat{j}+\hat{k}$$ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :</p> | [{"identifier": "A", "content": "$$\\hat{i}+2 \\hat{j}+\\hat{k}$$"}, {"identifier": "B", "content": "$$-3 \\hat{i}-2 \\hat{j}+\\hat{k}$$"}, {"identifier": "C", "content": "$$-2 \\hat{i}+2 \\hat{k}$$"}, {"identifier": "D", "content": "$$2 \\hat{i}-\\hat{j}+2 \\hat{k}$$"}] | ["A"] | null | <p>$${\overline r _{com}} = {{{m_1}{{\overline r }_1} + {m_2}{{\overline r }_2}} \over {{m_1} + {m_2}}}$$</p>
<p>$$ = {{(1 - 9)\widehat i + (2 - 6)\widehat j + (1 + 3)\widehat k} \over 4}$$</p>
<p>$$ = {{ - 8\widehat i - 4\widehat j + 4\widehat k} \over 4}$$</p>
<p>$${\overline r _{com}} = - 2\widehat i - \widehat j +... | mcq | jee-main-2022-online-29th-july-morning-shift | 9,416 |
lv2eryu7 | physics | center-of-mass | center-of-mass-of-discrete-particles | <p>In a system two particles of masses $$m_1=3 \mathrm{~kg}$$ and $$m_2=2 \mathrm{~kg}$$ are placed at certain distance from each other. The particle of mass $$m_1$$ is moved towards the center of mass of the system through a distance $$2 \mathrm{~cm}$$. In order to keep the center of mass of the system at the original... | [] | null | 3 | <p>To solve this problem, we can make use of the concept of center of mass. The center of mass (CM) of a system remains unchanged if the internal forces act within the system without any external force. When one mass moves toward the CM, to keep the CM at the same position, the other mass must move in a way that the pr... | integer | jee-main-2024-online-4th-april-evening-shift | 9,418 |
KftxWbaB1sYJEq96 | physics | center-of-mass | collision | The block of mass $$M$$ moving on the frictionless horizontal surface collides with the spring of spring constant $$k$$ and compresses it by length $$L.$$ The maximum momentum of the block after collision is
<img src="data:image/png;base64,UklGRowjAABXRUJQVlA4IIAjAACwvwCdASpcAoUBP4HA12O2Mawmo1r54sAwCWlu9II/26weQvAGM7/... | [{"identifier": "A", "content": "$${{k{L^2}} \\over {2M}}$$ "}, {"identifier": "B", "content": "$$\\sqrt {Mk} \\,\\,L$$ "}, {"identifier": "C", "content": "$${{M{L^2}} \\over k}$$ "}, {"identifier": "D", "content": "Zero"}] | ["B"] | null | Elastic energy stored in the spring = $${1 \over 2}k{L^2}$$
<br><br>And kinetic energy of the block = $${1 \over 2}M{v^2}$$
<br><br>$$\therefore$$ $${1 \over 2}M{v^2} = {1 \over 2}k{L^2}$$
<br><br>$$ \Rightarrow v = \sqrt {{k \over M}} .L$$
<br><br>$$\therefore$$ Momentum $$ = M \times v = M \times \sqrt {{k \over M}} ... | mcq | aieee-2005 | 9,419 |
ZWPIk1c72I8EcZvd | physics | center-of-mass | collision | A mass $$'m'$$ moves with a velocity $$'v'$$ and collides inelastically with another identical mass. After collision the $${1^{st}}$$ mass moves with velocity $${v \over {\sqrt 3 }}$$ in a direction perpendicular to the initial direction of motion. Find the speed of the $${2^{nd}}$$ mass after collision.
<img src="data... | [{"identifier": "A", "content": "$${\\sqrt 3 v}$$ "}, {"identifier": "B", "content": "$$v$$ "}, {"identifier": "C", "content": "$${v \\over {\\sqrt 3 }}$$ "}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}v$$ "}] | ["D"] | null | Assume speed of second mass = $${v_1}$$
<br><br>As momentum is conserved,
<br><br>In $$x$$-direction, $$mv = m{v_1}\cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(1)$$
<p>In $$y$$-direction, $${{mv} \over {\sqrt 3 }} = m{v_1}\,\sin \theta \,\,\,\,\,\,\,\,\,\,...(2)$$
<br><img class="question-image" src="https://image... | mcq | aieee-2005 | 9,420 |
pnDHZ821HAXsBq7V | physics | center-of-mass | collision | A block of mass $$0.50$$ $$kg$$ is moving with a speed of $$2.00$$ $$m{s^{ - 1}}$$ on a smooth surface. It strike another mass of $$1.0$$ $$kg$$ and then they move together as a single body. The energy loss during the collision is : | [{"identifier": "A", "content": "$$0.16J$$ "}, {"identifier": "B", "content": "$$1.00J$$ "}, {"identifier": "C", "content": "$$0.67J$$ "}, {"identifier": "D", "content": "$$0.34$$ $$J$$ "}] | ["C"] | null | Let $$m$$ = 0.50 kg and $$M$$ = 1.0 kg
<br><br>Initial kinetic energy of the system when 1 kg mass is at rest,
<br><br>$$K.{E_i} = {1 \over 2}m{u^2} + {1 \over 2}M{\left( 0 \right)^2}$$
<br><br>$$ = {1 \over 2} \times 0.5 \times 2 \times 2 + 0 = 1J$$
<br><br>For collision, applying conservation of linear momentum
<br><... | mcq | aieee-2008 | 9,421 |
xnNKC9egwPygXHiy | physics | center-of-mass | collision | <b>Statement - 1 : </b> Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
<br/><br/><b>Statement - 2 :</b> Principle of conservation of momentum holds true for all kinds of collisions. | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is the correct explanation of Statement - 1 "}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is <b>not</b> the correct explanation of Statement - 1 "}, {"identifier": "C", "con... | ["A"] | null | In completely inelastic collision,
<br><br>$${m_1}{v_1} + {m_2}{v_2} = {m_1}v + {m_2}v$$
<br><br>after collision both particle have common velocity $$v$$, so all energy is not lost
<br><br>$$\therefore$$ Statement - $$1$$ is true
<br><br>The principle of conservation of momentum applicable for all kinds of collision... | mcq | aieee-2010 | 9,422 |
ItHLBj1wfhKlo13J | physics | center-of-mass | collision | This question has statement $${\rm I}$$ and statement $${\rm I}$$$${\rm I}$$. Of the four choices given after the statements, choose the one that best describes the two statements.
<p><b>Statement - $${\rm I}$$:</b> A point particle of mass $$m$$ moving with speed $$\upsilon $$ collides with stationary point particle o... | [{"identifier": "A", "content": "Statement - $${\\rm I}$$ is true, Statement - $${\\rm II}$$ is true; Statement - $${\\rm II}$$ is the correct explanation of Statement - $${\\rm I}$$."}, {"identifier": "B", "content": "Statement - $${\\rm I}$$ is true, Statement - $${\\rm II}$$ is true; Statement - $${\\rm II}$$ is not... | ["D"] | null | Initial energy = $${{{P^2}} \over {2m}}$$, where $$P$$ is the momentum and m is the mass of the moving particle.
<br><br>Loss of energy is maximum when collision is inelastic means when the particles get stuck together as a result of the collision.
<br><br>So after collision energy = $${{{P^2}} \over {2\left( {m + M} ... | mcq | jee-main-2013-offline | 9,423 |
zAeid4s9T1Uzjbdo | physics | center-of-mass | collision | A particle of mass $$m$$ moving in the $$x$$ direction with speed $$2v$$ is hit by another particle of mass $$2m$$ moving in the $$y$$ direction with speed $$v.$$ If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to: | [{"identifier": "A", "content": "$$56\\% $$ "}, {"identifier": "B", "content": "$$62\\% $$"}, {"identifier": "C", "content": "$$44\\% $$"}, {"identifier": "D", "content": "$$50\\% $$"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91oguxy/85d124a8-cb21-4c9b-ad8d-99746c3176de/08f77b70-4800-11ed-8757-0f869593f41f/file-1l91oguxz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l91oguxy/85d124a8-cb21-4c9b-ad8d-99746c3176de/08f77b70-4800-11ed-8757-0f869593f41f/fi... | mcq | jee-main-2015-offline | 9,424 |
oEOoq5RyTV9C2IA9TiqfN | physics | center-of-mass | collision | Two particles A and B of equal mass M are moving with the same speed $$\upsilon $$ as shown in the figure. They collide completely inelastically and move as a single particle C. The angle $$\theta $$ that the path of C makes with the X-axis is given by :
<br/><br/><img src="data:image/png;base64,UklGRvQLAABXRUJQVlA4IOg... | [{"identifier": "A", "content": "tan $$\\theta $$ = $${{\\sqrt 3 + \\sqrt 2 } \\over {1 - \\sqrt 2 }}$$ "}, {"identifier": "B", "content": "tan $$\\theta $$ = $${{\\sqrt 3 - \\sqrt 2 } \\over {1 - \\sqrt 2 }}$$"}, {"identifier": "C", "content": "tan $$\\theta $$ = $${{1 - \\sqrt 2 } \\over {\\sqrt 2 \\left( {1 + \\sq... | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263585/exam_images/xytseqdiorrfffockqn1.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Online) 9th April Morning Slot Physics - Center of Mass and Collision Question 98 English Explanation">
... | mcq | jee-main-2017-online-9th-april-morning-slot | 9,426 |
MX7f8MTbmBWWmGKZdy3BO | physics | center-of-mass | collision | A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90<sup>o</sup> with respect to each other. The mass of unknown particle is : | [{"identifier": "A", "content": "$${m \\over 2}$$"}, {"identifier": "B", "content": "m"}, {"identifier": "C", "content": "$${m \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "2 m"}] | ["B"] | null | <p>In figure (i) before collision, m' is mass of unknown particle; m is mass of proton; v<sub>1</sub> is initial velocity.</p>
<p>(i) Before collision :</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l32mth8k/035418a3-4c67-48c4-8a2a-6921067bdf76/12040440-d1bd-11ec-9218-efb2cf12c71b/file-1l32mt... | mcq | jee-main-2018-online-15th-april-evening-slot | 9,428 |
VQNAfJ0JwW46dhyKi6RYv | physics | center-of-mass | collision | If 10<sup>22</sup> gas molecules each of mass 10<sup>–26</sup> kg
collide with a surface (perpendicular to it)
elastically per second over an area 1 m<sup>2</sup> with
a speed 10<sup>4</sup> m/s, the pressure exerted by the gas
molecules will be of the order of : | [{"identifier": "A", "content": "10<sup>8</sup> N/m<sup>2</sup>"}, {"identifier": "B", "content": "10<sup>16</sup> N/m<sup>2</sup>"}, {"identifier": "C", "content": "10<sup>4</sup> N/m<sup>2</sup>"}, {"identifier": "D", "content": "2 N/m<sup>2</sup>"}] | ["D"] | null | <p>Momentum imparted to the surface in one collision,</p>
<p>$$\Delta p = ({p_i} - {p_f}) = mv - ( - mv) = 2mv$$ ..... (i)</p>
<p>Force on the surface due to n collision per second, $$F = {n \over t}(\Delta p) = n\Delta p$$ ($$\because$$ $$t = 1s$$)</p>
<p>= 2 mnv [from Eq. (i)]</p>
<p>So, pressure on the surface,</p>
... | mcq | jee-main-2019-online-8th-april-morning-slot | 9,430 |
g0cXRENY5yb9VpJCw118hoxe66ijvztw64l | physics | center-of-mass | collision | Two particles, of masses M and 2M, moving,
as shown, with speeds of 10 m/s and 5 m/s,
collide elastically at the origin. After the
collision, they move along the indicated
directions with speeds u<sub>1</sub> and u<sub>2</sub>, respectively.
The values of u<sub>1</sub> and u<sub>2</sub> are nearly :
<img src="data:imag... | [{"identifier": "A", "content": "6.5 m/s and 3.2 m/s"}, {"identifier": "B", "content": "6.5 m/s and 6.3 m/s"}, {"identifier": "C", "content": "3.2 m/s and 6.3 m/s"}, {"identifier": "D", "content": "3.2 m/s and 12.6 m/s"}] | ["B"] | null | 2MV, cos 30° + Mv<sub>2</sub> cos 45° = 10 M cos 30° + 10 cos 45°<br><br>
$$ \Rightarrow {v_1}\sqrt 3 + {{{v_2}} \over {\sqrt 2 }} = 5\sqrt 3 + 5\sqrt 2 $$ ...(i)<br><br>
2MV, sin 30° – MV<sub>2</sub> sin 45° = –10 M sin 30° + 10 M sin 45°<br><br>
$${V_1} - {{{V_2}} \over {\sqrt 2 }} = - 5 + 5\sqrt 2 $$&... | mcq | jee-main-2019-online-10th-april-morning-slot | 9,431 |
RdT8iUaE0q90yW6aBNDmT | physics | center-of-mass | collision | A particle of mass 'm' is moving with speed '2v'
and collides with a mass '2m' moving with
speed 'v' in the same direction. After collision,
the first mass is stopped completely while the
second one splits into two particles each of
mass 'm', which move at angle 45° with respect
to the origianl direction.
The speed of ... | [{"identifier": "A", "content": "2 $$\\sqrt2$$v"}, {"identifier": "B", "content": "v / (2 $$\\sqrt2$$ )"}, {"identifier": "C", "content": "v / $$\\sqrt2$$"}, {"identifier": "D", "content": "$$\\sqrt2$$v"}] | ["A"] | null | Initial momentum · P<sub>i</sub>
= 2mv + 2mv = 4 mv<br>
Let v<sup>'</sup> be the speed of $$l$$ particle
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265484/exam_images/vr7aeezmxuya5yrhjcrk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (... | mcq | jee-main-2019-online-9th-april-evening-slot | 9,432 |
BQVxm3CHxJjC3vPE0HQaI | physics | center-of-mass | collision | A body of mass 2 kg makes an eleastic collision
with a second body at rest and continues to move
in the original direction but with one fourth of its
original speed. What is the mass of the second
body ? | [{"identifier": "A", "content": "1.2 kg"}, {"identifier": "B", "content": "1.0 kg"}, {"identifier": "C", "content": "1.8 kg"}, {"identifier": "D", "content": "1.5 kg"}] | ["A"] | null | By conservation of linear momentum:<br><br>
$$2{v_0} = 2\left( {{{{v_0}} \over 4}} \right) + mv \Rightarrow 2{v_0} = {{{v_0}} \over 2} + mv$$<br><br>
$$ \Rightarrow {{3{v_0}} \over 2} = mv\,\,...(1)$$<br><br>
Since collision is elastic<br><br>
$${V_{separation}} = {V_{approch}}$$<br><br>
$$ \Rightarrow v - {{{v_0}} \ov... | mcq | jee-main-2019-online-9th-april-morning-slot | 9,433 |
ZIcXfjv1T8bgwjEoT07tL | physics | center-of-mass | collision | A body of mass m<sub>1</sub> moving with an unknown
velocity of $${v_1}\mathop i\limits^ \wedge $$, undergoes a collinear collision
with a body of mass m<sub>2</sub> moving with a velocity
$${v_2}\mathop i\limits^ \wedge $$ . After collision, m<sub>1</sub> and m<sub>2</sub> move with
velocities of $${v_3}\mathop i\li... | [{"identifier": "A", "content": "$${v_4} - {{{v_2}} \\over 2}$$"}, {"identifier": "B", "content": "$${v_4} - {{{v_2}} \\over 4}$$"}, {"identifier": "C", "content": "$${v_4} - {v_2}$$"}, {"identifier": "D", "content": "$${v_4} + {v_2}$$"}] | ["C"] | null | Applying linear momentum conservation<br><br>
$${m_1}{v_1}\widehat i + {m_2}{v_2}\widehat i = {m_1}{v_3}\widehat i + {m_2}{v_4}\widehat i$$<br><br>
m<sub>1</sub>v<sub>1</sub> + 0.5 m<sub>1</sub>v<sub>2</sub> = m<sub>1</sub>(0.5 v<sub>1</sub>) + 0.5 m<sub>1</sub>v<sub>4</sub><br><br>
0.5 m<sub>1</sub>v<sub>1</sub> = 0.... | mcq | jee-main-2019-online-8th-april-evening-slot | 9,434 |
xOFjq4w5TNhYpRHQdk7JK | physics | center-of-mass | collision | A simple pendulum, made of a string of length $$\ell $$ and a bob of mass m, is released from a small angle $${{\theta _0}}$$. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It
bounces back and goes up to an angle $${{\theta _1}}$$. Then M is given by : | [{"identifier": "A", "content": "$${m \\over 2}\\left( {{{{\\theta _0} + {\\theta _1}} \\over {{\\theta _0} - {\\theta _1}}}} \\right)$$"}, {"identifier": "B", "content": "$${m \\over 2}\\left( {{{{\\theta _0} - {\\theta _1}} \\over {{\\theta _0} + {\\theta _1}}}} \\right)$$"}, {"identifier": "C", "content": "$$m\\left... | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264972/exam_images/pnpxc6iwgd5vdrx0lrey.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Physics - Center of Mass and Collision Question 89 English Explanation... | mcq | jee-main-2019-online-12th-january-morning-slot | 9,436 |
RyQ4T0YCkL3iEzw2JKL8p | physics | center-of-mass | collision | A body of mass 1 kg falls freely from a height of 100 m, on a platform mass 3 kg which is mounted on a spring having spring constant k = 1.25 $$ \times $$ 10<sup>6</sup> N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms<sup>–2</sup>
, the value of x will b... | [{"identifier": "A", "content": "8 cm"}, {"identifier": "B", "content": "4 cm"}, {"identifier": "C", "content": "40 cm"}, {"identifier": "D", "content": "80 cm"}] | ["B"] | null | velocity of 1 kg block just before it collides with 3kg block
<br><br>= $$\sqrt {2gh} = \sqrt {2000} $$ m/s
<br><br>Applying momentum conversation just before and just after collision.
<br><br>1 $$ \times $$ $$\sqrt {2000} $$ = 4v $$ \Rightarrow $$ v = $${{\sqrt {2000} } \over 4}$$ m/s
<br><br><img src="https:/... | mcq | jee-main-2019-online-11th-january-morning-slot | 9,437 |
yOqMeIm73bAqkicZ6LtId | physics | center-of-mass | collision | A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms<sup>–1</sup>, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the to... | [{"identifier": "A", "content": "30 m "}, {"identifier": "B", "content": "40 m"}, {"identifier": "C", "content": "20 m"}, {"identifier": "D", "content": "10 m"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264437/exam_images/luzo4llr3tdlts6frj7c.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Physics - Center of Mass and Collision Question 93 English Explanation... | mcq | jee-main-2019-online-10th-january-morning-slot | 9,438 |
A1zJwv7pYArMz6KnNqNNF | physics | center-of-mass | collision | Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M, Block A is given an initial speed $$\upsilon $$ towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically $$... | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["C"] | null | As in elastic or in elastic clollision, momentum is conserved.
<br><br>$$ \therefore $$ P<sub>i</sub> = P<sub>f</sub>
<br><br>P<sub>i</sub> = Initial momentum
<br><br>P<sub>f</sub> = Final Momentum
<br><br> mv = (2m + m) V<sub>f</sub>
<br><br>$$ \Rightarrow $$ ... | mcq | jee-main-2019-online-9th-january-morning-slot | 9,439 |
ZS4gWWYemjATkd9Cr9jgy2xukfrtrg65 | physics | center-of-mass | collision | Two bodies of the same mass are moving with the same speed, but in different directions in a
plane. They have a completely inelastic collision and move together thereafter with a final speed
which is half of their initial speed. The angle between the initial velocities of the two bodies (in
degree) is ________. | [] | null | 120 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267164/exam_images/d9xgpotrwc6q37lnbm13.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Physics - Center of Mass and Collision Question 62 English Explanation"... | integer | jee-main-2020-online-6th-september-morning-slot | 9,440 |
VKhgwb5YmY7b9fMoRCjgy2xukf3ufgfj | physics | center-of-mass | collision | A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg
collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal
direction just before the collision then the kinetic energy just before the combined system strikes
the floor, is [Take g =... | [{"identifier": "A", "content": "23 J"}, {"identifier": "B", "content": "21 J"}, {"identifier": "C", "content": "20 J"}, {"identifier": "D", "content": "19 J"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267066/exam_images/oubyomaxlt5cjmbgbozw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Evening Slot Physics - Center of Mass and Collision Question 64 English Explanation"... | mcq | jee-main-2020-online-3rd-september-evening-slot | 9,442 |
9nLXMWfGmnJhXCL0uDjgy2xukf23ezdn | physics | center-of-mass | collision | A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and
collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings
as a result of the collision making angle $$\theta $$ before momentarily coming to rest. If the rod has mass
M = 2 kg, an... | [{"identifier": "A", "content": "63<sup>o</sup>"}, {"identifier": "B", "content": "69<sup>o</sup>"}, {"identifier": "C", "content": "55<sup>o</sup>"}, {"identifier": "D", "content": "49"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264594/exam_images/xmx13c3lol11dcx0m9ew.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266554/exam_images/pnao7grq3pktokedegyq.webp"><source media="(max-wid... | mcq | jee-main-2020-online-3rd-september-morning-slot | 9,443 |
KYVJx4P027p8LPWO8yjgy2xukexyzmwo | physics | center-of-mass | collision | A particle of mass m is moving along the x-axis
with initial velocity $$u\widehat i$$. It collides elastically
with a particle of mass 10 m at rest and then
moves with half its initial kinetic energy (see
figure). If $$\sin {\theta _1} = \sqrt n \sin {\theta _2}$$ then value of n is
________.
<img src="data:image/png;b... | [] | null | 10 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265061/exam_images/jqfkxnr60syvzkgyq5ke.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Evening Slot Physics - Center of Mass and Collision Question 66 English Explanation"... | integer | jee-main-2020-online-2nd-september-evening-slot | 9,444 |
kMziWJoYrtTekb2JBj7k9k2k5hhc2hz | physics | center-of-mass | collision | A particle of mass m is dropped from a height
h above the ground. At the same time another
particle of the same mass is thrown vertically
upwards from the ground with a speed of $$\sqrt {2gh} $$. If they collide head-on completely
inelastically, the time taken for the combined
mass to reach the ground, in units of $$\s... | [{"identifier": "A", "content": "$$\\sqrt {{1 \\over 2}} $$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$$\\sqrt {{3 \\over 2}} $$"}, {"identifier": "D", "content": "$$\\sqrt {{3 \\over 4}} $$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266619/exam_images/sypjio5qe7rurgxwnyps.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 8th January Evening Slot Physics - Center of Mass and Collision Question 72 English Explanation">
... | mcq | jee-main-2020-online-8th-january-evening-slot | 9,447 |
63epoaj6Pbm53RjYHOjgy2xukev2td6f | physics | center-of-mass | collision | A particle of mass m with an initial velocity $$u\widehat i$$
collides perfectly elastically with a mass 3 m at
rest. It moves with a velocity $$v\widehat j$$ after collision,
then, v is given by : | [{"identifier": "A", "content": "$$v = \\sqrt {{2 \\over 3}} u$$"}, {"identifier": "B", "content": "$$v = {u \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$$v = {u \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$$v = {1 \\over {\\sqrt 6 }}u$$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263788/exam_images/poqdjc1xnmmdiphatyez.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263934/exam_images/utvlhagwigwe4k8ngxgy.webp"><img src="https://res.c... | mcq | jee-main-2020-online-2nd-september-morning-slot | 9,449 |
LJLtggkvzeGoqA59qc1kltj841t | physics | center-of-mass | collision | Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.<br/><br/>Assertion A : Body 'P' having mass M moving with speed 'u' has head-on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, body 'Q' will have a maximum speed equal ... | [{"identifier": "A", "content": "A is correct but R is not correct."}, {"identifier": "B", "content": "A is not correct but R is correct."}, {"identifier": "C", "content": "Both A and R are correct and R is the correct explanation of A."}, {"identifier": "D", "content": "Both A and R are correct but R is NOT the correc... | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267752/exam_images/hynytorlgqmtqikmstpj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Morning Shift Physics - Center of Mass and Collision Question 55 English Explanation... | mcq | jee-main-2021-online-26th-february-morning-slot | 9,451 |
naiuXl1mNoIsJ5RfbI1kmhpn0q5 | physics | center-of-mass | collision | A ball of mass 10 kg moving with a velocity $$10\sqrt 3 $$ m s<sup>$$-$$1</sup> along X-axis, hits another ball of mass 20 kg which is at rest. After collision, the first ball comes to rest and the second one disintegrates into two equal pieces. One of the pieces starts moving along Y-axis at a speed of 10 m/s. The sec... | [] | null | 30 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267398/exam_images/vr5sah2jfxu5n90qkht1.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Physics - Center of Mass and Collision Question 54 English Explanation">
... | integer | jee-main-2021-online-16th-march-morning-shift | 9,452 |
lyk0mQOQje3YjZEInA1kmiou8ta | physics | center-of-mass | collision | A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance h = 9.8 cm before the (block + bullet) pendulum comes momentarily to rest... | [{"identifier": "A", "content": "831.4 m/s"}, {"identifier": "B", "content": "811.4 m/s"}, {"identifier": "C", "content": "841.4 m/s"}, {"identifier": "D", "content": "821.4 m/s"}] | ["A"] | null | $${P_i} = 0.01 \times u + 0 = {P_f} = 6 \times v$$<br><br>$$v = {{0.01u} \over 6}$$<br><br>using energy conservation<br><br>$${1 \over 2} \times 6 \times {\left( {{u \over {600}}} \right)^2} = 6 \times 9.8 \times 9.8 \times {10^{ - 2}}$$<br><br>$$ \Rightarrow $$ $$u = 6 \times 98 \times \sqrt 2 = 588\sqrt 2 $$ m/s | mcq | jee-main-2021-online-16th-march-evening-shift | 9,453 |
tCalfbeSCZn9oM6hxq1kmkbql7g | physics | center-of-mass | collision | Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant K. A third block C of mass m moving with a speed v along the line joining A and B collides with A. The maximum compression in the spring is <br/><br/><img src="data:... | [{"identifier": "A", "content": "$$\\sqrt {{{mv} \\over K}} $$"}, {"identifier": "B", "content": "$$\\sqrt {{m \\over {2K}}} $$"}, {"identifier": "C", "content": "$$\\sqrt {{{mv} \\over {2K}}} $$"}, {"identifier": "D", "content": "$$v\\sqrt {{m \\over {2K}}} $$"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267032/exam_images/xwtwr7zepeuurxwx4yqw.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264876/exam_images/xkwedjow334ui7oao1k3.webp"><img src="https://res.c... | mcq | jee-main-2021-online-17th-march-evening-shift | 9,454 |
zrYmXIT3MzxdqqLXSJ1kmkryme8 | physics | center-of-mass | collision | A ball of mass 10 kg moving with a velocity 10$$\sqrt 3 $$ m/s along the x-axis, hits another ball of mass 20 kg which is at rest. After the collision, first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along y-axis with a speed of 10 m/s. The second piece starts... | [] | null | 20 | Velocity of 10 kg ball = $${v_{10}} = 10\sqrt {3\widehat i} $$<br><br>initial total momentum of system = $$10 \times 10\sqrt {3\widehat i} $$<br><br>Final total momentum of system<br><br>$$ = 10 \times 10\widehat j + 10 \times x(\cos 30^\circ \widehat i - \sin 30^\circ \widehat j)$$<br><br>Now by conservation of moment... | integer | jee-main-2021-online-18th-march-morning-shift | 9,455 |
fvjQ8RvFfiVc4SVLEM1kmlvgugm | physics | center-of-mass | collision | An object of mass m<sub>1</sub> collides with another object of mass m<sub>2</sub>, which is at rest. After the collision the objects move with equal speeds in opposite direction. The ratio of the masses m<sub>2</sub> : m<sub>1</sub> is : | [{"identifier": "A", "content": "1 : 1"}, {"identifier": "B", "content": "3 : 1"}, {"identifier": "C", "content": "2 : 1"}, {"identifier": "D", "content": "1 : 2"}] | ["B"] | null | Before collision<br><br> <picture><source media="(max-width: 2321px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263574/exam_images/qndypg8zbkjbtu5wxz22.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267376/exam_images/ou7eyc6dkyg7qew4uu9j.web... | mcq | jee-main-2021-online-18th-march-evening-shift | 9,456 |
1krpqef5y | physics | center-of-mass | collision | A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $$\lef... | [] | null | 4 | The given situation can be shown as<br><br> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kypoz6o1/630aca9f-74a0-4ed4-a616-de2652c7aedb/75019620-7b6e-11ec-92f3-61d366101705/file-1kypoz6o2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kypoz6o1/630aca9f-74a0-4ed4-a616-de2652c7... | integer | jee-main-2021-online-20th-july-morning-shift | 9,457 |
1krumn9t7 | physics | center-of-mass | collision | A body of mass 2 kg moving with a speed of 4 m/s. makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial peed. The speed of the two body centre of mass is $${x \over {10}}$$ m/s. Then the value of x is ___________. | [] | null | 25 | p<sub>i</sub> = p<sub>f</sub><br><br>2 $$\times$$ 4 = 2 $$\times$$ 1 + m<sub>2</sub> $$\times$$ v<sub>2</sub><br><br>m<sub>2</sub>v<sub>2</sub> = 6 ..... (i)<br><br>by coefficient of restitution<br><br>$$1 = {{{v_2} - 1} \over 4} \Rightarrow {v_2} = 5$$ m/s<br><br>by (i)<br><br>m<sub>2</sub> $$\times$$ 5 = 6<br><br>m<s... | integer | jee-main-2021-online-25th-july-morning-shift | 9,458 |
1ktfo9htl | physics | center-of-mass | collision | A bullet of 10 g, moving with velocity v, collides head-on with the stationary bob of a pendulum and recoils with velocity 100 m/s. The length of the pendulum is 0.5 m and mass of the bob is 1 kg. The minimum value of v = ____________ m/s so that the pendulum describes a circle. (Assume the string to be inextensible an... | [] | null | 400 | $$V' = \sqrt {5gR} = \sqrt {5 \times 10 \times 0.5} $$<br><br>V' = 5 m/s<br><br>m<sub>1</sub> V = m<sub>2</sub> $$\times$$ 5 $$-$$ m<sub>1</sub> $$\times$$ 100<br><br>$${{10} \over {1000}} \times V = 5 - {{10} \over {1000}} \times 100$$<br><br>V = 400 m/s | integer | jee-main-2021-online-27th-august-evening-shift | 9,460 |
1kth2w92x | physics | center-of-mass | collision | A body of mass M moving at speed V<sub>0</sub> collides elastically with a mass 'm' at rest. After the collision, the two masses move at angles $$\theta$$<sub>1</sub> and $$\theta$$<sub>2</sub> with respect to the initial direction of motion of the body of mass M. The largest possible value of the ratio M/m, for which ... | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264319/exam_images/mcdijut1xrwp3d57papi.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Physics - Center of Mass and Collision Question 38 English Explanation">... | mcq | jee-main-2021-online-31st-august-morning-shift | 9,461 |
1l57pusel | physics | center-of-mass | collision | <p>What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass?</p>
<p>(Assume the collision to be head-on elastic collision)</p> | [{"identifier": "A", "content": "50.0%"}, {"identifier": "B", "content": "66.6%"}, {"identifier": "C", "content": "55.6%"}, {"identifier": "D", "content": "33.3%"}] | ["C"] | null | <p>For a head on elastic collision</p>
<p>$${v_2} = {{m{u_1}} \over {m + 5m}} + {{m{u_1}} \over {m + 5m}}$$</p>
<p>$$ = {{2{u_1}} \over 6}$$ or $${{{u_1}} \over 3}$$</p>
<p>Initial kinetic energy of first mass $$ = {1 \over 2}mu_1^2$$</p>
<p>Final kinetic energy of second mass</p>
<p>$$ = {1 \over 2} \times 5m{\left( {... | mcq | jee-main-2022-online-27th-june-morning-shift | 9,463 |
1l6f4foje | physics | center-of-mass | collision | <p>Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 ms<sup>$$-$$1</sup> collide and rebound with the same speed. If the time duration of contact is t = 0.005 s, then what is the force exerted on the ball due to each other?</p> | [{"identifier": "A", "content": "100 N"}, {"identifier": "B", "content": "200 N"}, {"identifier": "C", "content": "300 N"}, {"identifier": "D", "content": "400 N"}] | ["B"] | null | <p>Change in momentum of one ball</p>
<p>= 2 $$\times$$ (0.05)(10) kg m/s</p>
<p>= 1 kg m/s</p>
<p>$$ \Rightarrow {F_{avg}} = {1 \over {\Delta t}} = {1 \over {0.005}}$$ N</p>
<p>= 200 N</p> | mcq | jee-main-2022-online-25th-july-evening-shift | 9,464 |
1ldr156ec | physics | center-of-mass | collision | <p>As per the given figure, a small ball P slides down the quadrant of a circle and hits the other ball Q of equal mass which is initially at rest. Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball Q after collision will be :</p>
<p>(g = 10 m/s<sup>2</sup>)</p>
<p><img src="... | [{"identifier": "A", "content": "0.25 m/s"}, {"identifier": "B", "content": "4 m/s"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2 m/s"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lepw79yz/2ad5304d-47e6-4f48-b88c-84d28d7a6a00/2bff60b0-b84e-11ed-a412-5b325e9b0d2e/file-1lepw79z0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lepw79yz/2ad5304d-47e6-4f48-b88c-84d28d7a6a00/2bff60b0-b84e-11ed-a412-5b325e9b0d2e... | mcq | jee-main-2023-online-30th-january-morning-shift | 9,466 |
1ldu00t6f | physics | center-of-mass | collision | <p>A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2 m/s. The initial speed of the smaller body before collision is ___________ ms$$^{-1}$$.</p> | [] | null | 4 | <p><b>Before collision</b></p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1les28d5t/a0b69fe4-7a4d-42f1-b0f1-c67f1d5a4141/52583110-b97f-11ed-981f-89b0ac23a016/file-1les28d5u.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1les28d5t/a0b69fe4-7a4d-42f1-b0f1-c67f1d5a4141/52583... | integer | jee-main-2023-online-25th-january-evening-shift | 9,467 |
jaoe38c1lse6t7e4 | physics | center-of-mass | collision | <p>A body starts falling freely from height $$H$$ hits an inclined plane in its path at height $$h$$. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of $$\frac{H}{h}$$ for which the body will take the maximum time to reach the ground is __________.<... | [] | null | 2 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lslupoi3/cc2aecc8-c56e-451b-bdf9-359e235929a2/3e3c51b0-cb40-11ee-ad47-a16d1086e690/file-6y3zli1lslupoi4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lslupoi3/cc2aecc8-c56e-451b-bdf9-359e235929a2/3e3c51b0-cb40-11ee... | integer | jee-main-2024-online-31st-january-morning-shift | 9,469 |
lv5gsdz6 | physics | center-of-mass | collision | <p>A stationary particle breaks into two parts of masses $$m_A$$ and $$m_B$$ which move with velocities $$v_A$$ and $$v_B$$ respectively. The ratio of their kinetic energies $$\left(K_B: K_A\right)$$ is :</p> | [{"identifier": "A", "content": "$$v_B: v_A$$\n"}, {"identifier": "B", "content": "$$1: 1$$\n"}, {"identifier": "C", "content": "$$m_B v_B: m_A v_A$$\n"}, {"identifier": "D", "content": "$$m_B: m_A$$"}] | ["A"] | null | <p>A stationary particle breaks into two parts with masses $$m_A$$ and $$m_B$$, which then move with velocities $$v_A$$ and $$v_B$$, respectively. We need to determine the ratio of their kinetic energies $$K_A$$ and $$K_B$$.</p>
<p>Since the initial momentum of the particle is zero, the momentum of the two parts must ... | mcq | jee-main-2024-online-8th-april-morning-shift | 9,470 |
lNQBy7v8fsuRC7Kx | physics | center-of-mass | impulse-momentum-theorem | Consider the following two statements :
<br/><br>$$A.$$ Linear momentum of a system of particles is zero
<br/><br>$$B.$$ Kinetic energy of a system of particles is zero.
<br/><br>then </br></br></br> | [{"identifier": "A", "content": "$$A$$ does not imply $$B$$ and $$B$$ does not imply $$A$$"}, {"identifier": "B", "content": "$$A$$ implies $$B$$ but $$B$$ does not imply $$A$$ "}, {"identifier": "C", "content": "$$A$$ does not imply $$B$$ but $$B$$ implies $$A$$ "}, {"identifier": "D", "content": "$$A$$ implies $$B$$ ... | ["A"] | null | <p>The correct answer is <strong>Option A : $$A$$ does not imply $$B$$ and $$B$$ does not imply $$A$$.</strong></p>
<p>Here's why :</p>
<p>Statement $$A:$$ The linear momentum of a system of particles being zero does not mean that the kinetic energy is also zero. For example, consider two equal mass particles movin... | mcq | aieee-2003 | 9,471 |
amEaHEve22kHcp18 | physics | center-of-mass | impulse-momentum-theorem | A player caught a cricket ball of mass $$150$$ $$g$$ moving at a rate of $$20$$ $$m/s.$$ If the catching process is completed in $$0.1s,$$ the force of the blow exerted by the ball on the hand of the player is equal to | [{"identifier": "A", "content": "$$150$$ $$N$$ "}, {"identifier": "B", "content": "$$3$$ $$N$$ "}, {"identifier": "C", "content": "$$30$$ $$N$$ "}, {"identifier": "D", "content": "$$300$$ $$N$$ "}] | ["C"] | null | We know, Force$$ \times $$ time = Impulse = Change in momentum
<br><br>$$\therefore$$ $$F \times t = m\left( {v - u} \right)$$
<br><br>$$ \Rightarrow $$ $$F = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N$$ | mcq | aieee-2006 | 9,473 |
0vCAeFI4lhK6kSEF | physics | center-of-mass | impulse-momentum-theorem | A bomb of mass $$16kg$$ at rest explodes into two pieces of masses $$4$$ $$kg$$ and $$12$$ $$kg.$$ The velocity of the $$12$$ $$kg$$ mass is $$4\,\,m{s^{ - 1}}.$$ The kinetic energy of the other mass is | [{"identifier": "A", "content": "$$144$$ $$J$$ "}, {"identifier": "B", "content": "$$288$$ $$J$$ "}, {"identifier": "C", "content": "$$192$$ $$J$$ "}, {"identifier": "D", "content": "$$96$$ $$J$$ "}] | ["B"] | null | Here linear momentum is conserved as no external force is acting on the bomb.
<br><br>Let the velocity and mass of $$4$$ $$kg$$ piece be $${v_1}$$ and $${m_1}$$ and that of $$12$$ $$kg$$ piece be $${v_2}$$ and $${m_2}$$.
<br><img class="question-image" src="https://imagex.cdn.examgoal.net/3T7YjPhEZDsqaIC1J/y2vNdG9oZcJC... | mcq | aieee-2006 | 9,474 |
YYvRuUKiW44jidck | physics | center-of-mass | impulse-momentum-theorem | The figure shows the position$$-$$time $$(x-t)$$ graph of one-dimensional motion of body of mass $$0.4$$ $$kg.$$ The magnitude of each impulse is
<img src="data:image/png;base64,"/> | [{"identifier": "A", "content": "$$0.4$$ $$Ns$$ "}, {"identifier": "B", "content": "$$0.8$$ $$Ns$$ "}, {"identifier": "C", "content": "$$1.6$$ $$Ns$$ "}, {"identifier": "D", "content": "$$0.2$$ $$Ns$$ "}] | ["B"] | null | Motion is uniform as graph is straight line. In x -t graph, velocity is $${x \over t}$$ and the slope of the graph tells the direction of the velocity. Here because of impulse the direction of velocity or slope of the graph changes.
<br><br>During each collision
<br><br>Initial velocity $${v_1} = {2 \over 2} = 1\,m{s^{... | mcq | aieee-2010 | 9,475 |
nyIfvm2HihAnNQ2gQlWVL | physics | center-of-mass | impulse-momentum-theorem | A ball is thrown vertically up (taken as +z-axis)
from the ground. The correct momentum-height
(p-h) diagram is : | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264949/exam_images/yl0xsa3ras6dkfeprn2j.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 9th April Morning Slot Physics - Center of Mass and Coll... | ["D"] | null | Momentum p = mv …(1)<br><br>
and for motion under gravity $$h = {{{u^2} - {v^2}} \over {2g}}$$ ...(2)<br><br>
$$h = {{{u^2} - {p^2}/m} \over {2g}}$$ | mcq | jee-main-2019-online-9th-april-morning-slot | 9,477 |
Xzv5EQELrubBDoHCOLTW1 | physics | center-of-mass | impulse-momentum-theorem | A wedge of mass M = 4m lies on a frictionless
plane. A particle of mass m approaches the
wedge with speed v. There is no friction
between the particle and the plane or between
the particle and the wedge. The maximum
height climbed by the particle on the wedge is
given by :- | [{"identifier": "A", "content": "$${{{v^2}} \\over {g}}$$"}, {"identifier": "B", "content": "$${{2{v^2}} \\over {7g}}$$"}, {"identifier": "C", "content": "$${{{v^2}} \\over {2g}}$$"}, {"identifier": "D", "content": "$${{2{v^2}} \\over {5g}}$$"}] | ["D"] | null | Initial condition can be shown in the figure
below
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lk25nvxa/233cb955-9742-46f3-9b57-ec5058f9932c/a8c077e0-2209-11ee-92b9-c1cea67c1a68/file-6y3zli1lk25nvxb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lk25nvxa... | mcq | jee-main-2019-online-9th-april-evening-slot | 9,478 |
kIjnqKIosJlYQlKYU63rsa0w2w9jx3pnzgb | physics | center-of-mass | impulse-momentum-theorem | A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The
man pushes his son so that he starts moving at a speed of 0.70 ms<sup>–1</sup> with respect to the man. The speed of the
man with respect to the surface is : | [{"identifier": "A", "content": "0.28 ms<sup>\u20131</sup>"}, {"identifier": "B", "content": "0.47 ms<sup>\u20131</sup>"}, {"identifier": "C", "content": "0.20 ms<sup>\u20131</sup>"}, {"identifier": "D", "content": "0.14 ms<sup>\u20131</sup>"}] | ["C"] | null | 50 V<sub>1</sub> = 20 V<sub>2</sub><br><br>
V<sub>1</sub> + V<sub>2</sub> = 0.70<br><br>
V<sub>1</sub> = 0.20 | mcq | jee-main-2019-online-12th-april-morning-slot | 9,479 |
l13Df72BzP6e5PtSKUjgy2xukfyy8d0f | physics | center-of-mass | impulse-momentum-theorem | Particle A of mass m<sub>1</sub>
moving with velocity $$\left( {\sqrt3\widehat i + \widehat j} \right)m{s^{ - 1}}$$ collides with another particle B of mass m<sub>2</sub>
which is at rest initially. Let $$\overrightarrow {{V_1}} $$
and $$\overrightarrow {{V_2}} $$
be the velocities of particles A and B after collisi... | [{"identifier": "A", "content": "105<sup>o</sup>"}, {"identifier": "B", "content": "15<sup>o</sup>"}, {"identifier": "C", "content": "-45<sup>o</sup>"}, {"identifier": "D", "content": "60<sup>o</sup>"}] | ["A"] | null | Given m<sub>1</sub>
= 2m<sub>2</sub>
<br>So let, m<sub>2</sub> = m and m<sub>1</sub> = 2m
<br><br>From momentum conservation
<br><br>$${\overrightarrow p _i}$$ = $${\overrightarrow p _f}$$
<br><br>$$ \Rightarrow $$ (2m)$$\left( {\sqrt 3 \widehat i + \widehat j} \right)$$ + 0 = 2m$$\left( {\widehat i + \sqrt 3 \widehat... | mcq | jee-main-2020-online-6th-september-evening-slot | 9,480 |
eQjN0E1T8XZN1OX0PR1klrouc8s | physics | center-of-mass | impulse-momentum-theorem | Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.)<sub>A</sub> : (K.E.)<sub>B</sub> will be $${{A \over 1}}$$, so the value of A will be ________. | [] | null | 2 | Given that, $${{{M_1}} \over {{M_2}}} = {1 \over 2}$$<br><br>Also, p<sub>1</sub> = p<sub>2</sub> = p<br><br>$$ \Rightarrow $$ M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> = p<br><br>Also, we know that<br><br>$$K = {{{p^2}} \over {2M}} \Rightarrow {K_1} = {{{p^2}} \over {2{M_1}}}$$ & $${K_2} = {{{p^2}} \o... | integer | jee-main-2021-online-24th-february-evening-slot | 9,481 |
T09mnlLnVs7w7irjhv1klt3s3yq | physics | center-of-mass | impulse-momentum-theorem | Two particles having masses 4 g and 16 g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is n : 2. The value of n will be ___________. | [] | null | 1 | $$\because$$ relation b/w kinetic energy & momentum is<br><br>$$P = \sqrt {2mKE} $$ ($$\because$$ KE = same)<br><br>$$ \Rightarrow $$ $${{{p_1}} \over {{p_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}} $$<br><br>$$ \Rightarrow $$ $${n \over 2} = \sqrt {{4 \over {16}}} $$<br><br>$$ \Rightarrow $$ $$n = 1$$ | integer | jee-main-2021-online-25th-february-evening-slot | 9,482 |
og5cY0t2ZZVHEqHpO21kmlwxba7 | physics | center-of-mass | impulse-momentum-theorem | The projectile motion of a particle of mass 5 g is shown in the figure.<br/><br/><img src="data:image/png;base64,UklGRnoFAABXRUJQVlA4IG4FAADQIgCdASoYAWAAPm0ylkikIqIhIrOaWIANiWlu3WBpKV+jf8t7WP8j0PPsc/v/Afyby3/wHgD7hvjbzUf4nwwf3fuM6n+L79B/0n8y/bPzAdQLt50M/67wvI6P9x/VfUr/zP5L/aP3D9qfy5/x/6D/h/kM/ln9F/2n8r/rH6+fNj65v2O9mD9... | [] | null | 5 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267549/exam_images/kziwiwh5tsvlltamk0vj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Evening Shift Physics - Center of Mass and Collision Question 47 English Explanation"><... | integer | jee-main-2021-online-18th-march-evening-shift | 9,483 |
1krsuvhom | physics | center-of-mass | impulse-momentum-theorem | A bullet of '4 g' mass is fired from a gun of mass 4 kg. If the bullet moves with the muzzle speed of 50 ms<sup>$$-$$1</sup>, the impulse imparted to the gun and velocity of recoil of gun are : | [{"identifier": "A", "content": "0.2 kg ms<sup>$$-$$1</sup>, 0.1 ms<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "0.4 kg ms<sup>$$-$$1</sup>, 0.05 ms<sup>$$-$$1</sup>"}, {"identifier": "C", "content": "0.2 kg ms<sup>$$-$$1</sup>, 0.05 ms<sup>$$-$$1</sup>"}, {"identifier": "D", "content": "0.4 kg ms<sup>$$-$$1</su... | ["C"] | null | m<sub>Bullet</sub> = 4g, M<sub>Gun</sub> = 4 kg<br><br>v<sub>Bullet</sub> $$ \simeq $$ 50 m/s<br><br>Now, P<sub>B</sub> = P<sub>g</sub><br><br>P<sub>g</sub> = m $$\times$$ v<sub>Bullet</sub><br><br>= $${4 \over {1000}}$$ $$\times$$ 50<br><br>= 0.2 kg m/s<br><br>So impulse = 0.2 kg m/s<br><br>$${v_G} = {{0.2} \over {{M_... | mcq | jee-main-2021-online-22th-july-evening-shift | 9,485 |
1kruko215 | physics | center-of-mass | impulse-momentum-theorem | Two billiard balls of equal mass 30g strike a rigid wall with same speed of 108 kmph (as shown) but at different angles. If the balls get reflected with the same speed then the ratio of the magnitude of impulses imparted to ball 'a' and ball 'b' by the wall along 'X' direction is :<br/><br/><img src="data:image/png;bas... | [{"identifier": "A", "content": "1 : 1"}, {"identifier": "B", "content": "$$\\sqrt 2 $$ : 1"}, {"identifier": "C", "content": "2 : 1"}, {"identifier": "D", "content": "1 : $$\\sqrt 2 $$"}] | ["B"] | null | Impulse = change in momentum<br><br>Ball (a) $$\left| {\overrightarrow {\Delta p} } \right|$$ = 2mu = J<sub>1</sub><br><br>Ball (b) $$\left| {\overrightarrow {\Delta p} } \right|$$ = 2mu cos45$$^\circ$$ = J<sub>2</sub><br><br>$${{{J_1}} \over {{J_2}}} = {1 \over {\cos 45^\circ }} = \sqrt 2 $$ | mcq | jee-main-2021-online-25th-july-morning-shift | 9,486 |
1l56aadtz | physics | center-of-mass | impulse-momentum-theorem | <p>A man of 60 kg is running on the road and suddenly jumps into a stationary trolly car of mass 120 kg. Then, the trolly car starts moving with velocity 2 ms<sup>$$-$$1</sup>. The velocity of the running man was ___________ ms<sup>$$-$$1</sup>, when he jumps into the car.</p> | [] | null | 6 | Total momentum of (man + trolley) system is always conserved
<br/><br/>Initially man was moving with velocity v<sub>1</sub> and trolley was at rest, finally both were moving with velocity 2 ms<sup>$$-$$1</sup> after man jumps on the trolley.
<br/><br/>So,
<br/><br/>$
\Rightarrow \quad m_{1} v_{1}+0=\left(m_{1}+m_{2}\... | integer | jee-main-2022-online-28th-june-morning-shift | 9,487 |
1l58bcavr | physics | center-of-mass | impulse-momentum-theorem | <p>An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero?</p> | [{"identifier": "A", "content": "Momentum"}, {"identifier": "B", "content": "Potential Energy"}, {"identifier": "C", "content": "Acceleration"}, {"identifier": "D", "content": "Force"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5jhmqf4/db9636cb-1bea-4bce-8bbd-a227247c026f/8a98c400-029a-11ed-a9b8-43edceee002f/file-1l5jhmqf5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5jhmqf4/db9636cb-1bea-4bce-8bbd-a227247c026f/8a98c400-029a-11ed-a9b8-43edceee002f... | mcq | jee-main-2022-online-26th-june-morning-shift | 9,488 |
1l58idx58 | physics | center-of-mass | impulse-momentum-theorem | <p>A batsman hits back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15 ms<sup>$$-$$1</sup>. The impulse imparted to the ball is ___________ Ns.</p> | [] | null | 12 | <p>$$l = m\Delta v$$</p>
<p>$$ = 0.4 \times 2 \times 15 = 12$$ Ns</p> | integer | jee-main-2022-online-26th-june-evening-shift | 9,489 |
1l5w28aiq | physics | center-of-mass | impulse-momentum-theorem | <p>Two bodies A and B of masses 5 kg and 8 kg are moving such that the momentum of body B is twice that of the body A. The ratio of their kinetic energies will be :</p> | [{"identifier": "A", "content": "4 : 5"}, {"identifier": "B", "content": "2 : 5"}, {"identifier": "C", "content": "5 : 4"}, {"identifier": "D", "content": "5 : 2"}] | ["B"] | null | <p>Given,</p>
<p>Mass of body A = 5 kg</p>
<p>Mass of body B = 8 kg</p>
<p>Momentum of body B is twice that of body A,</p>
<p>$$\therefore$$ $${P_B} = 2{P_A}$$</p>
<p>We know,</p>
<p>Kinetic Energy $$(K) = {{{P^2}} \over {2m}}$$</p>
<p>$$\therefore$$ $${{{K_A}} \over {{K_B}}} = {\left( {{{{P_A}} \over {{P_B}}}} \right)... | mcq | jee-main-2022-online-30th-june-morning-shift | 9,490 |
1l6i0cv9h | physics | center-of-mass | impulse-momentum-theorem | <p>A ball of mass $$0.15 \mathrm{~kg}$$ hits the wall with its initial speed of $$12 \mathrm{~ms}^{-1}$$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is $$100 \mathrm{~N}$$, calculate the time duration of the contact of ball with the wall.</p> | [{"identifier": "A", "content": "0.018 s"}, {"identifier": "B", "content": "0.036 s"}, {"identifier": "C", "content": "0.009 s"}, {"identifier": "D", "content": "0.072 s"}] | ["B"] | null | <p>F = 100 N</p>
<p>$$\Delta$$P = 2 $$\times$$ 0.15 $$\times$$ 12</p>
<p>= 3.6</p>
<p>$$\Rightarrow$$ t = $${{3.6} \over {100}}$$ = 0.036 s</p> | mcq | jee-main-2022-online-26th-july-evening-shift | 9,491 |
1l6i0e14v | physics | center-of-mass | impulse-momentum-theorem | <p>A body of mass $$8 \mathrm{~kg}$$ and another of mass $$2 \mathrm{~kg}$$ are moving with equal kinetic energy. The ratio of their respective momentum will be :</p> | [{"identifier": "A", "content": "1 : 1"}, {"identifier": "B", "content": "2 : 1"}, {"identifier": "C", "content": "1 : 4"}, {"identifier": "D", "content": "4 : 1"}] | ["B"] | null | <p>$$P = \sqrt {2m\,KE} $$</p>
<p>$$ \Rightarrow {{{P_1}} \over {{P_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}} $$</p>
<p>$$ = \sqrt {{8 \over 2}} = {2 \over 1}$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift | 9,492 |
1l6kmefbq | physics | center-of-mass | impulse-momentum-theorem | <p>A body of mass $$10 \mathrm{~kg}$$ is projected at an angle of $$45^{\circ}$$ with the horizontal. The trajectory of the body is observed to pass through a point $$(20,10)$$. If $$\mathrm{T}$$ is the time of flight, then its momentum vector, at time $$\mathrm{t}=\frac{\mathrm{T}}{\sqrt{2}}$$, is _____________.</p>
<... | [{"identifier": "A", "content": "$$\n100 \\hat{i}+(100 \\sqrt{2}-200) \\hat{j}$$"}, {"identifier": "B", "content": "$$100 \\sqrt{2} \\hat{i}+(100-200 \\sqrt{2}) \\hat{j}$$"}, {"identifier": "C", "content": "$$100 \\hat{i}+(100-200 \\sqrt{2}) \\hat{j}$$"}, {"identifier": "D", "content": "$$100 \\sqrt{2} \\hat{i}+(100 \\... | ["D"] | null | <p>m = 10 kg</p>
<p>$$\theta$$ = 45$$^\circ$$</p>
<p>$$y = x\tan \theta \left( {1 - {x \over R}} \right)$$</p>
<p>$$ \Rightarrow 10 = 20\left( {1 - {{20} \over R}} \right)$$</p>
<p>$$ \Rightarrow R = 40$$</p>
<p>$$40 = {{{u^2}} \over {10}} \Rightarrow u = 20$$</p>
<p>$$ \Rightarrow T = {{20 \times 20 \times {1 \over {\... | mcq | jee-main-2022-online-27th-july-evening-shift | 9,493 |
1l6m9py5f | physics | center-of-mass | impulse-momentum-theorem | <p>In two different experiments, an object of mass $$5 \mathrm{~kg}$$ moving with a speed of $$25 \mathrm{~ms}^{-1}$$ hits two different walls and comes to rest within (i) 3 second, (ii) 5 seconds, respectively. Choose the correct option out of the following :</p> | [{"identifier": "A", "content": "Impulse and average force acting on the object will be same for both the cases."}, {"identifier": "B", "content": "Impulse will be same for both the cases but the average force will be different."}, {"identifier": "C", "content": "Average force will be same for both the cases but the im... | ["B"] | null | <p>$$\Delta$$P = impulse = same since acceleration is different force acting will be different.</p> | mcq | jee-main-2022-online-28th-july-morning-shift | 9,494 |
1l6rhg0hg | physics | center-of-mass | impulse-momentum-theorem | <p>If momentum of a body is increased by 20%, then its kinetic energy increases by</p> | [{"identifier": "A", "content": "36%"}, {"identifier": "B", "content": "40%"}, {"identifier": "C", "content": "44%"}, {"identifier": "D", "content": "48%"}] | ["C"] | null | <p>Let, initial momentum of body $$({p_i}) = p$$</p>
<p>$$\therefore$$ Final momentum $$({p_f}) = {p_i} + 20\% $$ of $${p_i}$$</p>
<p>$$ = p + 0.2~p$$</p>
<p>$$ = 1.2~p$$</p>
<p>We know,</p>
<p>Kinetic energy $$(E) = {{{p^2}} \over {2m}}$$</p>
<p>$$\therefore$$ $${E_i} = {{{p^2}} \over {2m}}$$</p>
<p>and $${E_f} = {{{{... | mcq | jee-main-2022-online-29th-july-evening-shift | 9,495 |
1ldpkmvgy | physics | center-of-mass | impulse-momentum-theorem | <p>100 balls each of mass $$\mathrm{m}$$ moving with speed $$v$$ simultaneously strike a wall normally and reflected back with same speed, in time $$\mathrm{t ~s}$$. The total force exerted by the balls on the wall is</p> | [{"identifier": "A", "content": "$$\\frac{200 m v}{t}$$"}, {"identifier": "B", "content": "$$\\frac{100 m v}{t}$$"}, {"identifier": "C", "content": "$$\\frac{m v}{100 t}$$"}, {"identifier": "D", "content": "$$200 m v t$$"}] | ["A"] | null | When the balls strike the wall, the change in momentum of each ball is given by:
<br/><br/>$$\Delta p = mv - (-mv) = 2mv$$
<br/><br/> Since there are 100 balls, the total change in momentum of all the balls is $$\Delta P = 2m(100v) = 200mv.$$ The time taken for all the balls to strike the wall is $\mathrm{t}$ seconds. ... | mcq | jee-main-2023-online-31st-january-morning-shift | 9,496 |
ldquy4x0 | physics | center-of-mass | impulse-momentum-theorem | A machine gun of mass $10 \mathrm{~kg}$ fires $20 \mathrm{~g}$ bullets at the rate of 180 bullets per minute with a speed of $100 \mathrm{~m} \mathrm{~s}^{-1}$ each. The recoil velocity of the gun is | [{"identifier": "A", "content": "$ 0.02 \\mathrm{~m} / \\mathrm{s}$"}, {"identifier": "B", "content": "$1.5 \\mathrm{~m} / \\mathrm{s}$"}, {"identifier": "C", "content": "$2.5 \\mathrm{~m} / \\mathrm{s}$"}, {"identifier": "D", "content": "$0.6 \\mathrm{~m} / \\mathrm{s}$"}] | ["D"] | null | <p>Momentum of bullets per unit time</p>
<p>$$ = {{180 \times {{20} \over {1000}} \times 100} \over {60}}$$ kg m/s$$^2$$</p>
<p>= 6 N</p>
<p>$$\Rightarrow$$ Force on gun = 6 N</p>
<p>We cannot calculate recoil velocity with the given data.</p>
<p>If we consider recoil velocity at $$t = 1$$ s, then</p>
<p>$${V_{\mathrm{... | mcq | jee-main-2023-online-30th-january-evening-shift | 9,497 |
1ldr289os | physics | center-of-mass | impulse-momentum-theorem | <p>The figure represents the momentum time ($$\mathrm{p}-\mathrm{t}$$) curve for a particle moving along an axis under the influence of the force. Identify the regions on the graph where the magnitude of the force is maximum and minimum respectively?</p>
<p>If $$\left(t_{3}-t_{2}\right) < t_{1}$$</p>
<p><img src="da... | [{"identifier": "A", "content": "b and c"}, {"identifier": "B", "content": "c and a"}, {"identifier": "C", "content": "a and b"}, {"identifier": "D", "content": "c and b"}] | ["D"] | null | <p>$$F = {{dp} \over {dt}}$$</p>
<p>$$ \Rightarrow |F| = \left| {{{dp} \over {dt}}} \right| = |\mathrm{slope\,of\,p - t\,curve}|$$</p>
<p>As we can see from graph,</p>
<p>$$|{F_c}|$$ is maximum and $$|{F_b}|$$ is minimum.</p> | mcq | jee-main-2023-online-30th-january-morning-shift | 9,498 |
1ldr2fn46 | physics | center-of-mass | impulse-momentum-theorem | <p>A ball of mass $$200 \mathrm{~g}$$ rests on a vertical post of height $$20 \mathrm{~m}$$. A bullet of mass $$10 \mathrm{~g}$$, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance $$30 \mathrm{~m}$$ and the bullet at a dis... | [{"identifier": "A", "content": "120 m/s"}, {"identifier": "B", "content": "360 m/s"}, {"identifier": "C", "content": "400 m/s"}, {"identifier": "D", "content": "60 m/s"}] | ["B"] | null | <p>$$\because$$ Time of flight of each ball and bullet</p>
<p>$$ = \sqrt {{{2H} \over g}} = \sqrt {{{2 \times 20} \over {10}}} = 2$$ s</p>
<p>$$\Rightarrow$$ By applying linear momentum conservation</p>
<p>$$100u + 200(0) = 200\left( {{{30} \over 2}} \right) + 10\left( {{{120} \over 2}} \right)$$</p>
<p>$$u = 360$$ m... | mcq | jee-main-2023-online-30th-january-morning-shift | 9,499 |
1lgq1a943 | physics | center-of-mass | impulse-momentum-theorem | <p>A bullet of $$10 \mathrm{~g}$$ leaves the barrel of gun with a velocity of $$600 \mathrm{~m} / \mathrm{s}$$. If the barrel of gun is $$50 \mathrm{~cm}$$ long and mass of gun is $$3 \mathrm{~kg}$$, then value of impulse supplied to the gun will be :</p> | [{"identifier": "A", "content": "12 Ns"}, {"identifier": "B", "content": "3 Ns"}, {"identifier": "C", "content": "6 Ns"}, {"identifier": "D", "content": "36 Ns"}] | ["C"] | null | First, we need to find the velocity of the gun after the bullet is fired. We can use conservation of momentum to do this. The total momentum of the system of the gun and bullet is conserved before and after the bullet is fired. Therefore, we can write
<br/><br/>
$$m_g u_g + m_b u_b = m_g v_g + m_b v_b$$
<br/><br/>
wher... | mcq | jee-main-2023-online-13th-april-morning-shift | 9,500 |
1lguxz2s9 | physics | center-of-mass | impulse-momentum-theorem | <p>An average force of $$125 \mathrm{~N}$$ is applied on a machine gun firing bullets each of mass $$10 \mathrm{~g}$$ at the speed of $$250 \mathrm{~m} / \mathrm{s}$$ to keep it in position. The number of bullets fired per second by the machine gun is :</p> | [{"identifier": "A", "content": "25"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "100"}] | ["B"] | null | <p>To find the number of bullets fired per second, we can use the concept of momentum. When the machine gun fires bullets, it experiences a backward force due to the conservation of momentum. The force applied on the machine gun is used to balance this backward force.</p>
<p>First, let's find the momentum of each b... | mcq | jee-main-2023-online-11th-april-morning-shift | 9,501 |
1lh02ipzk | physics | center-of-mass | impulse-momentum-theorem | <p>The momentum of a body is increased by $$50 \%$$. The percentage increase in the kinetic energy of the body is ___________ $$\%$$.</p> | [] | null | 125 | <p>The momentum (p) and kinetic energy (K) of a body are related by the equations:</p>
<p>$p = mv$,</p>
<p>$K = \frac{1}{2}mv^2$,</p>
<p>where m is the mass and v is the velocity of the body.</p>
<p>We can express v in terms of p and m:</p>
<p>$v = \frac{p}{m}$,</p>
<p>and substitute this into the equation for K to get... | integer | jee-main-2023-online-8th-april-morning-shift | 9,502 |
jaoe38c1lsc3p10j | physics | center-of-mass | impulse-momentum-theorem | <p>A body of mass $$1000 \mathrm{~kg}$$ is moving horizontally with a velocity $$6 \mathrm{~m} / \mathrm{s}$$. If $$200 \mathrm{~kg}$$ extra mass is added, the final velocity (in $$\mathrm{m} / \mathrm{s}$$) is:</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["D"] | null | <p>Momentum will remain conserve</p>
<p>$$\begin{aligned}
& 1000 \times 6=1200 \times v \\
& v=5 \mathrm{~m} / \mathrm{s}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift | 9,504 |
jaoe38c1lse684ye | physics | center-of-mass | impulse-momentum-theorem | <p>An artillery piece of mass $$M_1$$ fires a shell of mass $$M_2$$ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:</p> | [{"identifier": "A", "content": "$$M_1 /\\left(M_1+M_2\\right)$$\n"}, {"identifier": "B", "content": "$$\\frac{M_2}{M_1}$$\n"}, {"identifier": "C", "content": "$$\\frac{M_1}{M_2}$$\n"}, {"identifier": "D", "content": "$$M_2 /\\left(M_1+M_2\\right)$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \left|\overrightarrow{\mathrm{p}_1}\right|=\left|\overrightarrow{\mathrm{p}_2}\right| \\
& \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{M}} ; \mathrm{p} \text { same } \\
& \mathrm{KE} \propto \frac{1}{\mathrm{~m}} \\
& \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{\mathrm{p}^2 / 2 \mathrm{M}_1}{\mathrm... | mcq | jee-main-2024-online-31st-january-morning-shift | 9,505 |
pgWvSIEXyKTkCNra | physics | center-of-mass | motion-of-center-of-mass | Two identical particles move towards each other with velocity $$2v$$ and $$v$$ respectively. The velocity of center of mass is | [{"identifier": "A", "content": "$$v$$ "}, {"identifier": "B", "content": "$$v/3$$ "}, {"identifier": "C", "content": "$$v/2$$ "}, {"identifier": "D", "content": "zero"}] | ["C"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/c84SriojYvqSujHdA/kNl0cD8ffQJBhayktDgBQx195q9Rx/7JEIpgVEtSC3j4ZogLIPh7/image.svg" loading="lazy" alt="AIEEE 2002 Physics - Center of Mass and Collision Question 111 English Explanation">
The velocity of center of mass of two particle system is
<br><br>$$... | mcq | aieee-2002 | 9,507 |
uMg3UsbBPZzhXKpNJgssy | physics | center-of-mass | motion-of-center-of-mass | Four particles A, B, C and D with masses
m<sub>A</sub> = m, m<sub>B</sub> = 2m, m<sub>C</sub> = 3m and m<sub>D</sub> = 4m are
at the corners of a square. They have
accelerations of equal magnitude with
directions as shown. The acceleration of the
centre of mass of the particles is :
<img src="data:image/png;base64,UklG... | [{"identifier": "A", "content": "$${a \\over 5}\\left( {\\mathop i\\limits^ \\wedge + \\mathop j\\limits^ \\wedge } \\right)$$"}, {"identifier": "B", "content": "$${a \\over 5}\\left( {\\mathop i\\limits^ \\wedge - \\mathop j\\limits^ \\wedge } \\right)$$"}, {"identifier": "C", "content": "$${a }\\left( {\\mathop... | ["B"] | null | $${\overrightarrow a _A} = - a\widehat i$$; $${\overrightarrow a _B} = - a\widehat j$$<br><br>
$${\overrightarrow a _C} = - a\widehat i$$; $${\overrightarrow a _D} = - a\widehat j$$<br><br>
$${\overrightarrow a _{cm}} = {{{m_a}{{\overrightarrow a }_a} + {m_b}{{\overrightarrow a }... | mcq | jee-main-2019-online-8th-april-morning-slot | 9,508 |
jaoe38c1lse6pfvb | physics | center-of-mass | motion-of-center-of-mass | <p>A solid circular disc of mass $$50 \mathrm{~kg}$$ rolls along a horizontal floor so that its center of mass has a speed of $$0.4 \mathrm{~m} / \mathrm{s}$$. The absolute value of work done on the disc to stop it is ________ J.</p> | [] | null | 6 | <p>Using work energy theorem</p>
<p>$$\begin{aligned}
& \mathrm{W}=\Delta \mathrm{KE}=0-\left(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2\right) \\
& \mathrm{W}=0-\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) \\
& =-\frac{1}{2} \times 50 \times 0.4^2\left(1+\frac{1}{2}\right)=-... | integer | jee-main-2024-online-31st-january-morning-shift | 9,509 |
4CFNbJNFYJKoZnxk | physics | circular-motion | non-uniform-circular-motion | A point $$P$$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $$P$$ is such that it sweeps out a length $$s = {t^3} + 5,$$ where $$s$$ is in metres and $$t$$ is in seconds. The radius of the path is $$20$$ $$m.$$ The acceleration of $$'P'$$ when $$t=2$$ $$s$$ is nearly.
... | [{"identifier": "A", "content": "$$13m/{s_2}$$ "}, {"identifier": "B", "content": "$$12m/{s^2}$$ "}, {"identifier": "C", "content": "$$7.2m{s^2}$$ "}, {"identifier": "D", "content": "$$14m/{s^2}$$ "}] | ["D"] | null | Given $$s = {t^3} + 5 $$
<br><br>$$\Rightarrow$$ Speed,$$\,\,\,v = {{ds} \over {dt}} = 3{t^2}$$
<br><br>Tangential acceleration $${a_t} = {{dv} \over {dt}} = 6t$$
<br><br>Radial acceleration $${a_c} = {{{v^2}} \over R} = {{9{t^4}} \over R}$$
<br><br>At $$\,\,\,\,t = 2s,\,\,{a_t} = 6 \times 2 = 12\,\,m/{s^2}$$
<br><br>... | mcq | aieee-2010 | 9,510 |
dz4yX3E7pMLYrED2lf1klryryhl | physics | circular-motion | non-uniform-circular-motion | A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ________ m/s. (Take g = 10 m/s<sup>2</sup>) | [] | null | 5 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265359/exam_images/frkpgatsgjuw3xpreu8d.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264192/exam_images/vwufcetwiqm4umoktpsh.webp"><img src="https://res.c... | integer | jee-main-2021-online-25th-february-morning-slot | 9,511 |
1l568h81v | physics | circular-motion | non-uniform-circular-motion | <p>A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k<sup>2</sup>rt<sup>2</sup>, where k is a constant. The power delivered to the particle by the force acting on it is given as</p> | [{"identifier": "A", "content": "zero"}, {"identifier": "B", "content": "mk<sup>2</sup>r<sup>2</sup>t<sup>2</sup>"}, {"identifier": "C", "content": "mk<sup>2</sup>r<sup>2</sup>t"}, {"identifier": "D", "content": "mk<sup>2</sup>rt"}] | ["C"] | null | <p>$${a_r} = {k^2}r{t^2} = {{{v^2}} \over r}$$</p>
<p>$$ \Rightarrow {v^2} = {k^2}{r^2}{t^2}$$ or $$v = krt$$</p>
<p>and $${{d|v|} \over {dt}} = kr$$</p>
<p>$$ \Rightarrow {a_t} = kr$$</p>
<p>$$ \Rightarrow |\overline F \,.\,\overline v | = (mkr)(krt)$$</p>
<p>$$ = m{k^2}{r^2}t = $$ power delivered</p> | mcq | jee-main-2022-online-28th-june-morning-shift | 9,512 |
1l56ukhcs | physics | circular-motion | non-uniform-circular-motion | <p>A stone tide to a spring of length L is whirled in a vertical circle with the other end of the spring at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is $$\sqrt {x({u^... | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "5"}] | ["B"] | null | <p>$$\overrightarrow v = \sqrt {{u^2} - 2gL} \widehat j$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5ifjvd9/19bdec24-cf90-42bb-9f11-b9b2b39390ea/a01583d0-0205-11ed-aa88-69f11483e075/file-1l5ifjvda.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5ifjvd9/19bdec24-c... | mcq | jee-main-2022-online-27th-june-evening-shift | 9,513 |
1l57r0ce7 | physics | circular-motion | non-uniform-circular-motion | <p>A pendulum of length 2 m consists of a wooden bob of mass 50 g. A bullet of mass 75 g is fired towards the stationary bob with a speed v. The bullet emerges out of the bob with a speed $${v \over 3}$$ and the bob just completes the vertical circle. The value of v is ___________ ms<sup>$$-$$1</sup>. (if g = 10 m/s<su... | [] | null | 10 | Let minimum velocity at the lowest point for the bob to compleate the circular path = v'
<br><br>And we know, v' = $\sqrt{5 r g}$
<br><br>From the conservation of momentum, we have
<br><br>$$
75 \times 10^{-3} \times v=50 \times 10^{-3} \times v^{\prime}+75 \times 10^{-3} \times \frac{v}{3}
$$
<br><br>$75 \times 10^{-... | integer | jee-main-2022-online-27th-june-morning-shift | 9,514 |
1ldsbqmho | physics | circular-motion | non-uniform-circular-motion | <p>A car is moving on a circular path of radius 600 m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of 54 km/hr is $$t(1-e^{-\pi/2})s$$. The value of t is __________... | [] | null | 40 | <p>$${{dv} \over {dt}} = {{{v^2}} \over R} \Rightarrow {{{v^2}} \over R} = v{{dv} \over {ds}}$$</p>
<p>$$ \Rightarrow {{dv} \over v} = {{ds} \over R} \Rightarrow \left. {\ln v} \right|_{15}^v = {s \over R}$$</p>
<p>$$ \Rightarrow v = 15{e^{\Delta /R}} = {{ds} \over {dt}} \Rightarrow dt = {1 \over {15}}{e^{ - \Delta /R}... | integer | jee-main-2023-online-29th-january-evening-shift | 9,515 |
jaoe38c1lsfmc9fh | physics | circular-motion | non-uniform-circular-motion | <p>A particle is moving in a circle of radius $$50 \mathrm{~cm}$$ in such a way that at any instant the normal and tangential components of it's acceleration are equal. If its speed at $$\mathrm{t}=0$$ is $$4 \mathrm{~m} / \mathrm{s}$$, the time taken to complete the first revolution will be $$\frac{1}{\alpha}\left[1-e... | [] | null | 8 | <p>$$\begin{aligned}
& \left|\vec{a}_c\right|=\left|\vec{a}_t\right| \\
& \frac{v^2}{r}=\frac{d v}{d t} \\
& \Rightarrow \int_\limits4^v \frac{d v}{v^2}=\int_\limits0^t \frac{d t}{r} \\
& \Rightarrow\left[\frac{-1}{v}\right]_4^v=\frac{t}{r} \\
& \Rightarrow \frac{-1}{v}+\frac{1}{4}=2 t
\end{aligned}$$</p>
<p>$$\begin{a... | integer | jee-main-2024-online-29th-january-evening-shift | 9,516 |
9XwDckxIh9qNOdxb | physics | circular-motion | uniform-circular-motion | The minimum velocity (in $$m{s^{ - 1}}$$) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction $$0.6$$ to avoid skidding is | [{"identifier": "A", "content": "$$60$$ "}, {"identifier": "B", "content": "$$30$$ "}, {"identifier": "C", "content": "$$15$$ "}, {"identifier": "D", "content": "$$25$$ "}] | ["B"] | null | For no skidding along curved track,
<br><br>The maximum velocity possible $${v_{\max }} = \sqrt {\mu rg} $$
<br><br>Here $$\mu = 0.6,\,r = 150m,\,g = 9.8$$
<br><br>$$\therefore$$ $${v_{\max }} = \sqrt {0.6 \times 150 \times 9.8} \simeq 30m/s$$ | mcq | aieee-2002 | 9,517 |
t24lssV0vCNG1oNS | physics | circular-motion | uniform-circular-motion | Which of the following statements is <b>FALSE</b> for a particle moving in a circle with a constant
angular speed? | [{"identifier": "A", "content": "The velocity vector is tangent to the circle."}, {"identifier": "B", "content": "The acceleration vector is tangent to the circle."}, {"identifier": "C", "content": "The acceleration vector points to the centre of the circle."}, {"identifier": "D", "content": "The velocity and accelerat... | ["B"] | null | Only option $$(b)$$ is false since acceleration vector acts along the radius of the circle or towards center of the circle for uniform circular motion and velocity vector always acts along the tangent of the circle. | mcq | aieee-2004 | 9,518 |
CYX8eg1ZwaGiHPvR | physics | circular-motion | uniform-circular-motion | For a particle in uniform circular motion the acceleration $$\overrightarrow a $$ at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis) | [{"identifier": "A", "content": "$$ - {{{v^2}} \\over R}\\cos \\theta \\widehat i + {{{v^2}} \\over R}\\sin \\theta \\widehat j$$"}, {"identifier": "B", "content": "$$ - {{{v^2}} \\over R}\\sin \\theta \\widehat i + {{{v^2}} \\over R}\\cos \\theta \\widehat j$$"}, {"identifier": "C", "content": "$$ - {{{v^2}} \\over R}... | ["C"] | null | For a particle in uniform circular motion,
<br><br>$${a_c} = {{{v^2}} \over R}$$ towards the center of the circle
<br><br>From figure, $$\overrightarrow a = {a_c}\cos \theta \left( { - \widehat i} \right) + {a_c}\sin \theta \left( { - \widehat j} \right)$$
<br><br>$$ = {{ - {v^2}} \over R}\cos \theta \widehat i - {{{v... | mcq | aieee-2010 | 9,519 |
QUTOe3ixMeVyWi9A | physics | circular-motion | uniform-circular-motion | Two cars of masses m<sub>1</sub> and m<sub>2</sub> are moving in circles of radii r<sub>1</sub> and r<sub>2</sub>, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is | [{"identifier": "A", "content": "m<sub>1</sub>r<sub>1</sub> : m<sub>2</sub>r<sub>2</sub>"}, {"identifier": "B", "content": "m<sub>1</sub> : m<sub>2</sub>"}, {"identifier": "C", "content": "r<sub>1</sub> : r<sub>2</sub>"}, {"identifier": "D", "content": "1 : 1"}] | ["C"] | null | We know, $$a = r\,{w^2} = r \times {\left( {{{2\pi } \over T}} \right)^2}$$
<br><br>Given, $${T_1} = {T_2} = T$$
<br><br>$${a_1} = {r_1} \times {\left( {{{2\pi } \over T}} \right)^2}$$
<br><br>$${a_2} = {r_2} \times {\left( {{{2\pi } \over T}} \right)^2}$$
<br><br>$$\therefore$$ $${{{a_1}} \over {{a_2}}} = {{{r_1}} \ov... | mcq | aieee-2012 | 9,520 |
2JlyHsCElIpON8bV24GxQ | physics | circular-motion | uniform-circular-motion | A conical pendulum of length 1 m makes an angle $$\theta $$ = 45<sup>o</sup> w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be: (Take g = 10 ms<sup>−2</sup> )
<br/><br/><img src="data:imag... | [{"identifier": "A", "content": "0.4 m/s"}, {"identifier": "B", "content": "4 m/s "}, {"identifier": "C", "content": "0.2 m/s"}, {"identifier": "D", "content": "2 m/s"}] | ["D"] | null | FBD of pendulum is :
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264331/exam_images/ytqjxk4jlytxrztqbnrp.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Online) 9th April Morning Slot Physics - Circular Motion Question 54 Englis... | mcq | jee-main-2017-online-9th-april-morning-slot | 9,521 |
PJBY3ec0cow284JnDLWaD | physics | circular-motion | uniform-circular-motion | A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of $$3.5$$ revolutions per second. A coin placed at a distnce of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is : (g = 10 m/s<sup>2</sup>) | [{"identifier": "A", "content": "0.5"}, {"identifier": "B", "content": "0.3"}, {"identifier": "C", "content": "0.7"}, {"identifier": "D", "content": "0.6"}] | ["D"] | null | <p>We have</p>
<p>$$m{\omega ^2}r = \mu mg$$ ...... (1)</p>
<p>Given : rate of ro<sup></sup>tation = 3.5 rev/s</p>
<p>$$\Rightarrow$$ 1 revolution = 2$$\pi$$ rad</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l32myl9a/2586242a-e474-4384-ab16-c8fa737101cc/a02ab3e0-d1bd-11ec-9218-efb2cf12c71b/fil... | mcq | jee-main-2018-online-15th-april-evening-slot | 9,522 |
VdYTnxX0JjVZXsi6weoRM | physics | circular-motion | uniform-circular-motion | A particle is moving along a circular path with a constant speed of 10 ms<sup>–1</sup>. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60<sup>o</sup> around the centre of the circle? | [{"identifier": "A", "content": "zero "}, {"identifier": "B", "content": "10 m/s"}, {"identifier": "C", "content": "$$10\\sqrt 2 m/s$$"}, {"identifier": "D", "content": "$$10\\sqrt 3 m/s$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266911/exam_images/vpuji6pk4hocdy19ts4u.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Morning Slot Physics - Circular Motion Question 53 English Explanation">
<br>$$\lef... | mcq | jee-main-2019-online-11th-january-morning-slot | 9,523 |
NqvTe2S7LfrbBiP8jttgK | physics | circular-motion | uniform-circular-motion | A body is projected at t = 0 with a velocity 10 ms<sup>–1</sup>
at an angle of 60<sup>o</sup> with the horizontal. The radius of curvature of its trajectory at t = 1s is R. neglecting air resistance and taking acceleration due to gravity g = 10 ms<sup>–2</sup>, the value of R is : | [{"identifier": "A", "content": "2.8 m"}, {"identifier": "B", "content": "5.1 m"}, {"identifier": "C", "content": "2.5 m"}, {"identifier": "D", "content": "10.3 m"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266050/exam_images/j66lwsw8rlnrpkauyaww.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Morning Slot Physics - Circular Motion Question 52 English Explanation">
<br>v<sub>... | mcq | jee-main-2019-online-11th-january-morning-slot | 9,524 |
THBiRPAzVvdBokVoIA3rsa0w2w9jx6nytqs | physics | circular-motion | uniform-circular-motion | A smooth wire of length 2$$\pi $$r is bent into a circle and kept in a vertical plane. A bead can slide smoothly on
the wire. When the circle is rotating with angular speed $$\omega $$ about the vertical diameter AB, as shown in figure,
the bead is at rest with respect to the circular ring at position P as shown. Then ... | [{"identifier": "A", "content": "$${{\\sqrt 3 g} \\over {2r}}$$"}, {"identifier": "B", "content": "$${{2g} \\over {\\left( {r\\sqrt 3 } \\right)}}$$"}, {"identifier": "C", "content": "$${{\\left( {g\\sqrt 3 } \\right)} \\over r}$$"}, {"identifier": "D", "content": "$${{2g} \\over r}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264395/exam_images/gedopsomtx4e6qsf31kz.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Evening Slot Physics - Circular Motion Question 51 English Explanation">
<br>In $$\De... | mcq | jee-main-2019-online-12th-april-evening-slot | 9,525 |
SB95hGshHFXpQ73VH97k9k2k5gxsc7e | physics | circular-motion | uniform-circular-motion | A particle of mass m is fixed to one end of a
light spring having force constant k and
unstretched length $$\ell $$. The other end is fixed. The
system is given an angular speed $$\omega $$ about the
fixed end of the spring such that it rotates in a
circle in gravity free space. Then the stretch in
the spring is : | [{"identifier": "A", "content": "$${{m\\ell {\\omega ^2}} \\over {k - m{\\omega ^2}}}$$"}, {"identifier": "B", "content": "$${{m\\ell {\\omega ^2}} \\over {k - m{\\omega}}}$$"}, {"identifier": "C", "content": "$${{m\\ell {\\omega ^2}} \\over {k + m{\\omega ^2}}}$$"}, {"identifier": "D", "content": "$${{m\\ell {\\omega ... | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264526/exam_images/sv3rgdbszslmmxryuipw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 8th January Morning Slot Physics - Circular Motion Question 50 English Explanation">
<br>At elonga... | mcq | jee-main-2020-online-8th-january-morning-slot | 9,526 |
7DBJ4qBfeKvStWIpeOjgy2xukev0pl32 | physics | circular-motion | uniform-circular-motion | A bead of mass m stays at point P(a, b) on a
wire bent in the shape of a parabola y = 4Cx<sup>2</sup>
and rotating with angular speed $$\omega $$ (see figure).
The value of $$\omega $$ is (neglect friction) :
<img src="data:image/png;base64,UklGRvoHAABXRUJQVlA4IO4HAAAQOQCdASodAd0APm02mEikIyKhJfJ5AIANiWlu/kAItdzWM7Ouv9K... | [{"identifier": "A", "content": "$$2\\sqrt {2gC} $$"}, {"identifier": "B", "content": "$$2\\sqrt {gC} $$"}, {"identifier": "C", "content": "$$\\sqrt {{{2gC} \\over {ab}}} $$"}, {"identifier": "D", "content": "$$\\sqrt {{{2g} \\over C}} $$"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263326/exam_images/cqvpw0hdo7oxswrmarug.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263851/exam_images/sumpnvwwetw26z0zzqzs.webp"><source media="(max-wid... | mcq | jee-main-2020-online-2nd-september-morning-slot | 9,527 |
0sTInJTxtvGaHbK3TUjgy2xukfro4ykf | physics | circular-motion | uniform-circular-motion | A clock has a continuously moving second's hand of 0.1 m length. The average acceleration of the
tip of the hand (in units of ms<sup>–2</sup>) is of the order of : | [{"identifier": "A", "content": "10<sup>-3</sup>"}, {"identifier": "B", "content": "10<sup>-1</sup>"}, {"identifier": "C", "content": "10<sup>-2</sup>"}, {"identifier": "D", "content": "10<sup>-4</sup>"}] | ["A"] | null | R = 0.1 m
<br><br>$$\omega $$ = $${{2\pi } \over T}$$ = $${{2\pi } \over {60}}$$ = 0.105 rad/sec
<br><br>a = $${\omega ^2}R$$
<br><br>= (0.105)<sup>2</sup>(0.1)
<br><br>= 0.0011
<br><br>= 1.1 $$ \times $$ 10<sup>-3</sup>
<br><br>Average acceleration is of the order of 10<sup>–3</sup>. | mcq | jee-main-2020-online-6th-september-morning-slot | 9,528 |
yeOAPlBsskllmodDsi1kltipn9x | physics | circular-motion | uniform-circular-motion | A particle is moving with uniform speed along the circumference of a circle of radius R under the action of a central fictitious force F which is inversely proportional to R<sup>3</sup>. Its time period of revolution will be given by : | [{"identifier": "A", "content": "$$T \\propto {R^{{4 \\over 3}}}$$"}, {"identifier": "B", "content": "$$T \\propto {R^{{5 \\over 2}}}$$"}, {"identifier": "C", "content": "$$T \\propto {R^{{3 \\over 2}}}$$"}, {"identifier": "D", "content": "$$T \\propto {R^2}$$"}] | ["D"] | null | $$F \propto {1 \over {{R^3}}}$$<br><br>$$F = {K \over {{R^3}}}$$<br><br>$${{m{v^2}} \over R} = {K \over {{R^3}}}$$<br><br>$$m{(\omega R)^2} = {K \over {{R^2}}}$$<br><br>$$m{\omega ^2}{R^2} = {K \over {{R^2}}}$$<br><br>$${\omega ^2} = {K \over m}\left( {{1 \over {{R^4}}}} \right)$$<br><br>$${\left( {{{2\pi } \over T}} \... | mcq | jee-main-2021-online-26th-february-morning-slot | 9,529 |
EHIIk9UkGtVApYd4mN1kmiopld3 | physics | circular-motion | uniform-circular-motion | Statement I : A cyclist is moving on an unbanked road with a speed of 7 kmh<sup>$$-$$1</sup> and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s<sup>2</sup>)<br/><br/>Statement II : If... | [{"identifier": "A", "content": "Statement I is incorrect and statement II is correct"}, {"identifier": "B", "content": "Both statement I and statement II are true"}, {"identifier": "C", "content": "Statement I is correct and statement II is incorrect"}, {"identifier": "D", "content": "Both statement I and statement II... | ["B"] | null | On a horizontal ground,<br><br>$${v_{\max }} = \sqrt {\mu Rg} = \sqrt {0.2 \times 2 \times 9.8} = 1.97$$ m/s<br><br>= $$1.97 \times {{18} \over 5}$$<br><br>= 7.12 km/hr = 7.2 km/hr<br><br>Statement - 2<br><br>$${v_{\max }} = \sqrt {gr\left( {{{\tan \theta + \mu } \over {1 - \mu \tan \theta }}} \right)} = \sqrt {2 \... | mcq | jee-main-2021-online-16th-march-evening-shift | 9,531 |
7E3GUEg1rObPsh8wYl1kmj2dcpx | physics | circular-motion | uniform-circular-motion | A modern grand - prix racing car of mass m is travelling on a flat track in a circular arc of radius R with a speed v. If the coefficient of static friction between the tyres and the track is $$\mu$$<sub>s</sub>, then the magnitude of negative lift F<sub>L</sub> acting downwards on the car is : (Assume forces on the fo... | [{"identifier": "A", "content": "$$m\\left( {g - {{{v^2}} \\over {{\\mu _s}R}}} \\right)$$"}, {"identifier": "B", "content": "$$ - m\\left( {g + {{{v^2}} \\over {{\\mu _s}R}}} \\right)$$"}, {"identifier": "C", "content": "$$m\\left( {{{{v^2}} \\over {{\\mu _s}R}} - g} \\right)$$"}, {"identifier": "D", "content": "$$m\\... | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264573/exam_images/pteymc8s53mqvcycinfv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Morning Shift Physics - Circular Motion Question 43 English Explanation">
<br>$$f = {{m... | mcq | jee-main-2021-online-17th-march-morning-shift | 9,532 |
5MMZ73pL7eDQbr29qn1kmlvnwrq | physics | circular-motion | uniform-circular-motion | A particle of mass m moves in a circular orbit under the central potential field, $$U(r) = - {C \over r}$$, where C is a positive constant. The correct radius $$-$$ velocity graph of the particle's motion is : | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265122/exam_images/exjfg8llmiqwvmkvf008.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2021 (Online) 18th March Evening Shift Physics - Circular Motion Questio... | ["B"] | null | Given potential field U(r) = $${{ - C} \over r}$$<br><br>$$ \because $$ $$F = {{dU} \over {dr}} = {C \over {{r^2}}}$$<br><br>& $${F_c} = {{m{v^2}} \over r}$$<br><br>$$ \therefore $$ $${{m{v^2}} \over r} = {C \over {{r^2}}}$$<br><br>$$ \Rightarrow r = {C \over {m{v^2}}}$$<br><br>$$ \Rightarrow r$$ $$ \propto $$ $$ {... | mcq | jee-main-2021-online-18th-march-evening-shift | 9,533 |
S5lZvUCZYGi9u6LbmG1krpljzgn | physics | circular-motion | uniform-circular-motion | The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30$$^\circ$$ banked road at maximum possible speed without skidding is ____________ $$\times$$ 10<sup>3</sup> kg m/s<sup>2</sup>. [Given cos30$$^\circ$$ = 0.87, $$\mu$$<sub>s</sub> = 0.2] | [{"identifier": "A", "content": "12.4"}, {"identifier": "B", "content": "7.2"}, {"identifier": "C", "content": "6.96"}, {"identifier": "D", "content": "10.2"}] | ["D"] | null | The given situation can be represented as<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxofybix/398719e0-82dc-4102-981e-6996580a7696/2ac253a0-66f2-11ec-9866-0df874a38238/file-1kxofybiy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxofybix/398719e0-82dc-4102-981e-699... | mcq | jee-main-2021-online-20th-july-morning-shift | 9,534 |
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