JEE Physics Centre of Mass MCQs Set 06

Answer: (d) the kinetic energy only

Explanation:
1. Internal forces occur in equal and opposite pairs, so their vector sum is zero and the total momentum cannot change.
2. They can do net work, e.g. an explosion or a spring releasing energy, so the kinetic energy can change.

Teacher's Note:
1. Internal work can change KE, but never the momentum of the system.
2. This distinction is the basis of explosion and collision problems.

Answer: (b) \( \tfrac12\vec a \)

Explanation:
1. \( \vec a_{cm}=\frac{m\vec a+m\cdot0}{2m}=\frac{\vec a}{2} \).

Teacher's Note:
1. The centre of mass acceleration is the mass-weighted mean of the individual accelerations.
2. The centre of mass is always defined for any system of particles.

Answer: (c) zero

Explanation:
1. No external force acts and the initial momentum is zero, so the total momentum remains zero.
2. Hence the centre of mass stays at rest (speed zero) at all times.

Teacher's Note:
1. Internal forces cannot move the centre of mass of a system that started at rest.
2. The different speeds just mean A is heavier than B (mass ratio 2 : 1).

Answer: (d) 100 percent

Explanation:
1. \( p=\sqrt{2mK} \), so \( p\propto\sqrt K \).
2. With K increased fourfold, p becomes \( \sqrt4=2 \) times: an increase of 100 percent.

Teacher's Note:
1. Momentum scales as the square root of kinetic energy.
2. Do not confuse 'twice as large' with 'twice more'; the increase is 100 percent.

Answer: (a) \( \tfrac32mv^2 \)

Explanation:
1. Momentum of the third fragment (mass 2m) balances the other two: \( 2m\,u=\sqrt2\,mv\Rightarrow u=\frac{v}{\sqrt2} \).
2. Energy released \( =\tfrac12mv^2+\tfrac12mv^2+\tfrac12(2m)\frac{v^2}{2}=\tfrac32mv^2 \).

Teacher's Note:
1. The mass of the third fragment is 4m - 2m = 2m.
2. Total initial energy is zero, so all kinetic energy comes from the explosion.

Answer: (b) \( \frac{100}{11} \) m/s

Explanation:
1. The man has no horizontal momentum, so horizontal momentum is conserved: \( 500\times10=550\,V \).
2. \( V=\frac{100}{11} \) m/s.

Teacher's Note:
1. Only the horizontal component matters because the water exerts only a vertical (buoyant) force.
2. The man's vertical momentum is absorbed by the water.

Answer: (c) \( \frac{MV}{M-m} \)

Explanation:
1. Momentum conservation: \( MV=m(0)+(M-m)v \).
2. \( v=\frac{MV}{M-m} \).

Teacher's Note:
1. The other piece has mass M - m.
2. The result exceeds V, as the second piece must carry all the momentum.

Answer: (d) each greater than v/2

Explanation:
1. Momentum along x: \( mv=2\,m\,u\cos\theta\Rightarrow u=\frac{v}{2\cos\theta} \).
2. Since \( \theta>0 \), \( \cos\theta<1 \) and each piece moves faster than v/2.

Teacher's Note:
1. The transverse components cancel by symmetry, so only the x-component is needed.
2. Dividing the momentum between two oblique paths requires larger speeds than a straight path would.

Answer: (c) 1

Explanation:
1. Speed of \( m_1 \) before impact: \( v=\sqrt{2gh} \). Combined speed after impact \( V=\frac{m_1v}{m_1+m_2} \).
2. Rise to height h/4: \( V=\sqrt{2g\,h/4}=\frac v2 \).
3. \( \frac{m_1}{m_1+m_2}=\frac12\Rightarrow m_1=m_2 \), ratio 1.

Teacher's Note:
1. Height is proportional to the square of speed, so a quarter of the height means half the speed.
2. Equal masses give half the speed after a perfectly inelastic collision.

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

Explanation:
1. Momentum is conserved horizontally: \( mv=100\,m\,V \).
2. \( V=\frac{v}{100} \).

Teacher's Note:
1. Each sticking collision adds one more slider to the moving cluster, but the total momentum never changes.
2. Count all 100 sliders in the final mass.

Answer: (b) 30 m/s

Explanation:
1. In the frame of the wall the particle approaches at 25 m/s and leaves at 25 m/s.
2. Back in the ground frame it leaves with \( 25+5=30 \) m/s.

Teacher's Note:
1. For a massive wall use the wall's frame, where an elastic bounce simply reverses the relative velocity.
2. The wall's motion adds twice to the rebound relative to a fixed wall: 20 + 2(5) = 30.

Answer: (c) \( \tfrac12m(u_1^2+u_2^2) \)

Explanation:
1. An elastic collision conserves total kinetic energy, no matter how many collisions occur.
2. Hence the final KE equals the initial KE: \( \tfrac12m(u_1^2+u_2^2) \).

Teacher's Note:
1. The number of collisions is irrelevant when each one is perfectly elastic.
2. For equal masses the velocities are simply exchanged in each head-on collision.

Answer: (d) \( 2v \)

Explanation:
1. \( v_2=\frac{2m_1}{m_1+m_2}v\to2v \) when \( m_2\ll m_1 \).
2. The heavy ball is almost unaffected, so the light ball leaves at twice its speed (relative speed v before and after).

Teacher's Note:
1. Think of a table-tennis ball hitting a moving bus: it rebounds with about twice the bus's speed.
2. The relative speed of separation equals the relative speed of approach.

Answer: (b) m

Explanation:
1. Fraction of energy transferred in an elastic head-on collision: \( \frac{4m\,M}{(m+M)^2} \).
2. This is a maximum (equal to 1) for \( M=m \): the incident ball stops and gives up all its energy.

Teacher's Note:
1. Equal masses give a complete transfer of kinetic energy in a head-on elastic collision.
2. For very light or very heavy targets the transfer tends to zero.

Answer: (c) \( e=\tfrac13 \)

Explanation:
1. For equal masses: \( v_1=\frac{(1-e)u}{2},\ v_2=\frac{(1+e)u}{2} \).
2. \( v_2=2v_1\Rightarrow1+e=2(1-e)\Rightarrow e=\frac13 \).

Teacher's Note:
1. Write both final velocities in terms of the initial speed and e.
2. Only the ratio of the final velocities is needed.

Answer: (d) 0.7 kg/s

Explanation:
1. Thrust \( F=v_r\frac{dm}{dt} \).
2. \( \frac{dm}{dt}=\frac{210}{300}=0.7 \) kg/s.

Teacher's Note:
1. Thrust is the exhaust speed times the mass flow rate.
2. Check units: (m/s)(kg/s) = N.

Answer: (b) of the box plus ball remains constant

Explanation:
1. There is no external horizontal force on the box-ball system, so its centre-of-mass velocity stays constant.
2. The box alone and the ball alone each change their velocity at every collision.

Teacher's Note:
1. Impacts with the walls are internal forces for the combined system.
2. The coefficient of restitution affects the energy loss but not the centre-of-mass velocity.

Answer: (c) an external force \( 3m\omega^2R \) must be applied, and it may act anywhere on the system

Explanation:
1. Total mass \( =3m \).
2. Centripetal force for the centre of mass: \( F_{ext}=M a_{cm}=3m\omega^2R \).
3. Newton's law for the centre of mass does not depend on where the external force is applied, so it may act anywhere (it only affects the rotation about the centre of mass).

Teacher's Note:
1. Newton's second law for a system relates the external force to the acceleration of the centre of mass.
2. The point of application matters for torque, not for translation.

Answer: (a) the right end of the plank

Explanation:
1. Take B's ground position as the origin. A starts 60 cm to its left and walks 60 cm to reach B.
2. Plank (mass 40) with A (mass 40) and B (mass 60) at rest: since B does not move, \( 40(60)+40\,X=0\Rightarrow X=-60 \) cm, so the plank slides 60 cm to the left.
3. B has stayed fixed, so relative to the plank B is now 60 cm to the right of the plank's centre, i.e. at its right end; A meets B there.

Teacher's Note:
1. B does not move relative to the ground, so B exerts no net horizontal effect on the plank's motion beyond the common momentum balance.
2. Convert the ground displacement of the plank into a shift of the meeting point.

Answer: (c) y = -5 cm

Explanation:
1. The centre of mass continues along the x-axis (y = 0), since the original motion was along the x-axis and no external force acts in the plane.
2. \( \frac m4(15)+\frac{3m}{4}\,y=0\Rightarrow y=-5 \) cm.

Teacher's Note:
1. The centre of mass keeps its straight-line path after the explosion.
2. The lighter fragment is farther from the line.

Answer: (d) \( \frac{p^2(m_1+m_2)}{2m_1m_2} \)

Explanation:
1. The fragments carry equal and opposite momenta of magnitude p.
2. \( E=\frac{p^2}{2m_1}+\frac{p^2}{2m_2}=\frac{p^2(m_1+m_2)}{2m_1m_2} \).

Teacher's Note:
1. Kinetic energy is expressed as p squared over 2m for each fragment.
2. With zero initial energy, this sum is the energy released.

Answer: (a) \( \frac{3mu}{5} \)

Explanation:
1. Let the common speed just after the impact be V. For B plus bullet (mass 2m): \( mu-J=2mV \); for A (mass 3m): \( J=3mV \).
2. \( mu=5mV\Rightarrow V=\frac u5 \).
3. \( J=3mV=\frac{3mu}{5} \).

Teacher's Note:
1. Gravity is negligible during the very short impact; the string is taut so both blocks get the same speed immediately.
2. The impulse of tension acts upward on A and upward on B (opposing the bullet).