JEE Physics Centre of Mass MCQs Set 03

Practice MCQs for JEE Physics Centre of Mass

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Multiple Choice Questions

Question 1: A man is sitting in a moving train. Then:
(a) his kinetic energy may be zero (as measured in the train frame)
(b) his momentum must be zero
(c) his kinetic energy must be zero
(d) his kinetic energy is never zero
Show Answer & Explanation

Answer: (a) his kinetic energy may be zero (as measured in the train frame)

Explanation:
1. In the train frame his velocity is zero, so kinetic energy is zero
2. In the ground frame it is non-zero

Teacher's Note:
1. Kinetic energy depends on the frame of reference.

Question 2: A 50 g ball dropped from 10 m rebounds after losing 75 percent of its kinetic energy. If the contact lasts 0.01 s, the impulse of the impact force (taking g = 10 m/s\(^2\), neglecting weight during the impact) is approximately:
(a) 1.3 N s
(b) 1.06 N s
(c) 13 N s
(d) 0.5 N s
Show Answer & Explanation

Answer: (b) 1.06 N s

Explanation:
1. \( v_1=\sqrt{2gh}=10\sqrt2 \) m/s
2. Rebound KE is one quarter, so \( v_2=\frac{v_1}2=5\sqrt2 \) m/s
3. \( J=m(v_1+v_2)=0.05\times15\sqrt2\approx1.06 \) N s

Teacher's Note:
1. Reversing direction adds the two speeds.

Question 3: The area under the F-t curve of a collision is A. If the target of mass M was initially at rest, its speed just after the collision is:
(a) \( \frac MA \)
(b) \( AM \)
(c) \( \sqrt{\frac{2A}{M}} \)
(d) \( \frac AM \)
Show Answer & Explanation

Answer: (d) \( \frac AM \)

Explanation:
1. Area \( =J=\Delta p=Mv-0 \)
2. \( v=\frac AM \)

Teacher's Note:
1. Impulse equals the area under the force-time graph.

Question 4: A force on a particle rises uniformly from 0 to 2 N between t = 0 and t = 1 s and then falls uniformly back to zero at t = 2 s. The total impulse over 2 s is:
(a) 0
(b) 1 N s
(c) 3 N s
(d) 2 N s
Show Answer & Explanation

Answer: (d) 2 N s

Explanation:
1. The graph is a triangle of base 2 s and height 2 N
2. Area \( =\frac12(2)(2)=2 \) N s

Teacher's Note:
1. Impulse is the area under the force-time graph.

Question 5: A ball of mass 0.15 kg moving at 20 m/s is hit back along the same line by a bat at 20 m/s. If contact lasts 0.01 s, the average force on the ball is:
(a) 600 N
(b) 300 N
(c) 60 N
(d) 1200 N
Show Answer & Explanation

Answer: (a) 600 N

Explanation:
1. \( \Delta p=0.15(20)-0.15(-20)=6 \) kg m/s
2. \( F=\frac{6}{0.01}=600 \) N

Teacher's Note:
1. Reversal of velocity doubles the momentum change.

Question 6: A ball of mass m falling at speed u hits the ground with coefficient of restitution e. The impulse exerted by the ground on the ball is:
(a) \( mu(1-e) \)
(b) \( mue \)
(c) \( mu(1+e) \)
(d) \( 2mu \)
Show Answer & Explanation

Answer: (c) \( mu(1+e) \)

Explanation:
1. Rebound speed \( =eu \)
2. \( J=mv-(-mu)=m(eu+u)=mu(1+e) \)

Teacher's Note:
1. Take up as positive and subtract the initial momentum.

Question 7: A super ball bounces elastically between two rigid walls a distance d apart at speed v0 (gravity neglected). The average force exerted by the ball on one wall is:
(a) \( \frac{mv_0^2}{2d} \)
(b) \( \frac{2mv_0^2}{d} \)
(c) \( \frac{mv_0^2}{d} \)
(d) \( \frac{4mv_0^2}{d} \)
Show Answer & Explanation

Answer: (c) \( \frac{mv_0^2}{d} \)

Explanation:
1. Time between successive hits on the same wall: \( \frac{2d}{v_0} \)
2. Impulse per hit \( =2mv_0 \)
3. \( F=\frac{2mv_0}{2d/v_0}=\frac{mv_0^2}d \)

Teacher's Note:
1. Average force is impulse per hit divided by the time between hits.

Question 8: A 50 g bullet strikes a 5 kg sand bag hanging from a fixed point with a horizontal velocity and sticks to it. The ratio of the final to the initial kinetic energy of the bullet is approximately:
(a) \( 10^{-2} \)
(b) \( 10^{-3} \)
(c) \( 10^{-6} \)
(d) \( 10^{-4} \)
Show Answer & Explanation

Answer: (d) \( 10^{-4} \)

Explanation:
1. \( 0.05v=5.05V\Rightarrow\frac Vv\approx10^{-2} \)
2. Kinetic energy ratio \( =\left(\frac Vv\right)^2\approx10^{-4} \)

Teacher's Note:
1. The bullet's final speed is about one hundredth of the initial, so its KE ratio is the square.

Question 9: One hundred identical sliders sit at equal spacing on a frictionless track. Slider 1 is pushed with velocity v toward slider 2, and every collision is perfectly inelastic. The final velocity of the stuck-together set of 100 sliders is:
(a) \( \frac v{99} \)
(b) \( \frac v{100} \)
(c) zero
(d) \( v \)
Show Answer & Explanation

Answer: (b) \( \frac v{100} \)

Explanation:
1. \( mv=100\,mV \)
2. \( V=\frac v{100} \)

Teacher's Note:
1. Momentum is conserved throughout; only the total mass matters.

Question 10: A very massive ball moving with speed v collides head-on elastically with a tiny ball at rest. Immediately after the impact the tiny ball moves with speed approximately:
(a) v
(b) v/2
(c) very large
(d) 2v
Show Answer & Explanation

Answer: (d) 2v

Explanation:
1. The heavy ball's velocity is unchanged
2. \( \frac{v_2-v}{0-v}=-1\Rightarrow v_2=2v \)

Teacher's Note:
1. Elastic: relative speed of separation equals relative speed of approach.

Question 11: During the head-on collision of masses 1 kg and 2 kg the maximum deformation energy is \( \frac{100}3 \) J. If they move in opposite directions before the collision, their velocity of approach is:
(a) 5 m/s
(b) 20 m/s
(c) \( 10\sqrt2 \) m/s
(d) 10 m/s
Show Answer & Explanation

Answer: (d) 10 m/s

Explanation:
1. \( U=\frac12\mu v_{rel}^2 \), \( \mu=\frac{1\times2}{3}=\frac23 \) kg
2. \( \frac{100}3=\frac12\cdot\frac23v_{rel}^2 \)
3. \( v_{rel}^2=100 \), so \( v_{rel}=10 \) m/s

Teacher's Note:
1. Maximum deformation energy equals the kinetic energy lost relative to the centre of mass.

Question 12: Block A of mass m moves with speed v on a frictionless track and collides elastically head-on with block B of mass m/2 moving toward it at 2v. The final speed of A is:
(a) \( v \)
(b) \( \frac{5v}3 \)
(c) \( \frac{2v}3 \)
(d) \( 3v \)
Show Answer & Explanation

Answer: (a) \( v \)

Explanation:
1. Take A's direction as positive: \( mv-\frac m2(2v)=mv_1+\frac m2v_2\Rightarrow 2v_1+v_2=0 \)
2. Elastic: \( v_2-v_1=3v \)
3. Solving: \( v_1=-v \), so A returns at speed v

Teacher's Note:
1. Use momentum and the elastic condition together.

Question 13: In a collision between two solid spheres (no external force), the velocity of separation along the line of impact:
(a) cannot be greater than the velocity of approach
(b) cannot be less than the velocity of approach
(c) cannot be equal to the velocity of approach
(d) can be anything
Show Answer & Explanation

Answer: (a) cannot be greater than the velocity of approach

Explanation:
1. \( e=\frac{v_{sep}}{v_{app}} \) with \( 0\le e\le1 \)
2. So \( v_{sep}\le v_{app} \)

Teacher's Note:
1. e = 1 is elastic; e < 1 means some energy is lost.

Question 14: A particle of mass m moving at 20 m/s hits a wall moving toward it at 5 m/s and collides elastically. The speed of the particle after the collision is:
(a) 20 m/s
(b) 30 m/s
(c) 25 m/s
(d) 22 m/s
Show Answer & Explanation

Answer: (b) 30 m/s

Explanation:
1. Wall velocity is unchanged
2. \( 20+5=V-5 \)
3. \( V=30 \) m/s

Teacher's Note:
1. In the wall's frame the ball approaches at 25 m/s and leaves at 25 m/s.

Question 15: A ball moves at 2 m/s toward a wall that moves toward the ball at 1 m/s. After an elastic collision the ball's speed is:
(a) 3 m/s
(b) 4 m/s
(c) 2 m/s
(d) 1 m/s
Show Answer & Explanation

Answer: (b) 4 m/s

Explanation:
1. Relative approach \( =2+1=3 \) m/s
2. After the collision separation is also 3 m/s
3. The wall is at 1 m/s (toward the ball), so ball speed \( =3+1=4 \) m/s

Teacher's Note:
1. Work in the frame of the heavy wall.

Question 16: A sphere of mass m moving at constant velocity hits another stationary sphere of the same mass with coefficient of restitution e. The ratio of the speed of the first sphere to that of the second after the collision is:
(a) \( \frac{1+e}{1-e} \)
(b) \( \frac{1-e}{1+e} \)
(c) \( \frac{e+1}{e-1} \)
(d) \( \frac{e-1}{e+1} \)
Show Answer & Explanation

Answer: (b) \( \frac{1-e}{1+e} \)

Explanation:
1. \( mu=mv_1+mv_2\Rightarrow u=v_1+v_2 \)
2. \( v_2-v_1=eu \)
3. \( v_1=\frac{(1-e)u}2,\ v_2=\frac{(1+e)u}2 \)
4. \( \frac{v_1}{v_2}=\frac{1-e}{1+e} \)

Teacher's Note:
1. Add and subtract the two equations.

Question 17: A 2 kg block moving at 6 m/s collides with a 4 kg block at rest and they stick together. The kinetic energy lost is:
(a) 24 J
(b) 12 J
(c) 36 J
(d) 6 J
Show Answer & Explanation

Answer: (a) 24 J

Explanation:
1. \( V=\frac{2(6)}{6}=2 \) m/s
2. \( K_i=36 \) J, \( K_f=\frac12(6)(4)=12 \) J
3. Loss = 24 J

Teacher's Note:
1. Compute kinetic energy before and after.

Question 18: A ball of 2 kg moving at 4 m/s collides head-on with a 4 kg ball moving at 2 m/s in the opposite direction. After the collision the 2 kg ball returns at 2 m/s. The velocity of the 4 kg ball and the coefficient of restitution are:
(a) 1 m/s in the opposite direction, and 0.5
(b) 2 m/s in the original direction, and 1
(c) 1 m/s in the original direction of the 2 kg ball, and 0.5
(d) 0, and 0.25
Show Answer & Explanation

Answer: (c) 1 m/s in the original direction of the 2 kg ball, and 0.5

Explanation:
1. \( 2(4)-4(2)=2(-2)+4v_2\Rightarrow v_2=1 \) m/s
2. \( e=\frac{1-(-2)}{4-(-2)}=\frac36=0.5 \)

Teacher's Note:
1. Take the 2 kg ball's initial direction as positive.

Question 19: For the collision in which a 2 kg ball at 4 m/s meets a 4 kg ball at 2 m/s (opposite), with the 2 kg ball returning at 2 m/s, the maximum potential energy of deformation is:
(a) 12 J
(b) 48 J
(c) 24 J
(d) 36 J
Show Answer & Explanation

Answer: (c) 24 J

Explanation:
1. At maximum deformation both move with the same velocity, 0 m/s
2. \( U=\frac12(2)(16)+\frac12(4)(4)-0=16+8=24 \) J

Teacher's Note:
1. At maximum deformation the common velocity follows from momentum conservation.

Question 20: A 2 kg block moving at 10 m/s meets a very heavy object moving at 2 m/s in the opposite direction. After an elastic collision, the velocity of the 2 kg block is:
(a) 12 m/s in the reverse direction
(b) 10 m/s in the reverse direction
(c) 6 m/s in the reverse direction
(d) 14 m/s in the reverse direction
Show Answer & Explanation

Answer: (d) 14 m/s in the reverse direction

Explanation:
1. The heavy object keeps 2 m/s
2. In its frame the block approaches at 12 m/s and leaves at 12 m/s
3. So the block leaves at 12 + 2 = 14 m/s relative to the ground

Teacher's Note:
1. Heavier object velocity is essentially unchanged.

Question 21: Three identical balls A, B and C lie in a line on a smooth surface. A moves with speed u and strikes B elastically, and B then strikes C elastically. The final velocities are:
(a) A at rest, B and C moving at u/2
(b) A and B at rest, C moving at u
(c) A, B and C each at u/3
(d) A moving back, B and C moving forward
Show Answer & Explanation

Answer: (b) A and B at rest, C moving at u

Explanation:
1. Equal masses exchange velocities in each elastic head-on collision
2. A stops and B moves at u
3. Then B stops and C moves at u

Teacher's Note:
1. Velocity is passed along the line like Newton's cradle.

Question 22: A ball of mass m collides elastically with a stationary ball of mass 2m. The fraction of the incident ball's kinetic energy transferred to the 2m ball is:
(a) \( \frac19 \)
(b) \( \frac12 \)
(c) \( \frac49 \)
(d) \( \frac89 \)
Show Answer & Explanation

Answer: (d) \( \frac89 \)

Explanation:
1. \( v_2=\frac{2m}{3m}u=\frac{2u}3 \)
2. Energy gained \( =\frac12(2m)\frac{4u^2}9=\frac{4mu^2}9 \)
3. Fraction \( =\frac{4/9}{1/2}=\frac89 \)

Teacher's Note:
1. Use the elastic collision formulas for a target at rest.

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Practice MCQs for JEE Physics Centre of Mass

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