JEE Physics Centre of Mass MCQs Set 04

Answer: (c) 0.5

Explanation:
1. \( e^2=\frac{h'}h=\frac14 \)
2. \( e=\frac12 \)

Teacher's Note:
1. h' = e^2 h.

Answer: (a) \( \tan\beta=\frac{\tan\alpha}{e} \)

Explanation:
1. Tangential component \( v_0\sin\alpha \) is unchanged
2. Normal component becomes \( ev_0\cos\alpha \)
3. \( \tan\beta=\frac{v_0\sin\alpha}{ev_0\cos\alpha}=\frac{\tan\alpha}e \)

Teacher's Note:
1. Impulse acts only along the normal.

Answer: (a) 2

Explanation:
1. \( \tan\beta=\frac{\tan45^\circ}{1/2}=2 \)

Teacher's Note:
1. Use tan(beta) = tan(alpha)/e.

Answer: (a) 90 degrees

Explanation:
1. Along the common normal the velocities are exchanged
2. The tangential components remain unchanged
3. So the final velocities are perpendicular

Teacher's Note:
1. Valid when one equal-mass body is initially at rest.

Answer: (d) \( \frac V{\sqrt3} \)

Explanation:
1. Each velocity makes 30 degrees with the original direction
2. \( mV=2mv'\cos30^\circ \)
3. \( v'=\frac{V}{2\cos30^\circ}=\frac V{\sqrt3} \)

Teacher's Note:
1. By symmetry each ball moves at half of 60 degrees to the original line.

Answer: (b) \( \frac13 \)

Explanation:
1. Line of impact is along the final velocity of the initially stationary ball
2. Approach speed along it \( =V\cos30^\circ=\frac{\sqrt3V}2 \)
3. Separation speed \( =v'-v'\cos60^\circ=\frac{V}{2\sqrt3} \)
4. \( e=\frac{V/(2\sqrt3)}{\sqrt3V/2}=\frac13 \)

Teacher's Note:
1. The target ball moves along the line of impact.

Answer: (a) -2 m/s and 0.75 m/s

Explanation:
1. \( 12-12=3V_1+8V_2\Rightarrow3V_1+8V_2=0 \)
2. \( V_2-V_1=\frac12(5.5)=2.75 \)
3. \( 3V_1+8(2.75+V_1)=0\Rightarrow V_1=-2 \) m/s, \( V_2=0.75 \) m/s

Teacher's Note:
1. Total momentum is zero here.

Answer: (c) 12 N s and 6 N s

Explanation:
1. At maximum deformation the common velocity is 0
2. \( J_D=3(0-4) \), magnitude 12 N s
3. \( J_R=eJ_D=6 \) N s

Teacher's Note:
1. Common velocity from zero total momentum.

Answer: (c) 33 J

Explanation:
1. \( U=\frac12(3)(16)+\frac12(8)(2.25)-0=24+9=33 \) J

Teacher's Note:
1. Loss of kinetic energy up to the maximum-deformation instant.

Answer: (d) the momentum is mv

Explanation:
1. \( mv=(M+m)u\Rightarrow u=\frac{mv}{M+m} \)
2. Momentum just after \( =mv \)
3. \( K=\frac{m^2v^2}{2(M+m)} \)

Teacher's Note:
1. Momentum is conserved in the impact (thread tension is vertical).

Answer: (b) 60 m/s

Explanation:
1. \( V=\frac{0.05v}{0.5}=\frac v{10} \)
2. \( V^2=2gh=36\Rightarrow V=6 \) m/s
3. \( v=60 \) m/s

Teacher's Note:
1. The bob decelerates at g after the impact.

Answer: (b) 0.2 m

Explanation:
1. \( V=\frac{0.02(200)}{2}=2 \) m/s
2. \( h=\frac{V^2}{2g}=\frac4{20}=0.2 \) m

Teacher's Note:
1. Momentum for the impact, then energy for the swing.

Answer: (c) 24 J

Explanation:
1. \( \mu=\frac{1\times3}{4}=0.75 \) kg
2. Loss \( =\frac12\mu v_{rel}^2=\frac12(0.75)(64)=24 \) J

Teacher's Note:
1. Only the relative motion contributes to the loss.

Answer: (a) the final kinetic energy equals the initial kinetic energy

Explanation:
1. Momentum is always conserved with no external force
2. Elastic means total kinetic energy before and after is the same

Teacher's Note:
1. Elastic collisions conserve both momentum and kinetic energy.

Answer: (a) both balls move after the collision

Explanation:
1. Momentum must be conserved, so both cannot be at rest
2. The stationary ball must receive some momentum when momentum changes

Teacher's Note:
1. Each ball gets equal and opposite impulse.

Answer: (b) 0.7 kg/s

Explanation:
1. \( F=v_{rel}\left
2. \frac{dm}{dt}\right
3. \)
4. \( 210=300\times\frac{dm}{dt} \)
5. \( \frac{dm}{dt}=0.7 \) kg/s

Teacher's Note:
1. Thrust = relative speed times rate of mass ejection.

Answer: (d) 3000 N

Explanation:
1. \( F=v_{rel}\frac{dm}{dt}=1500\times2=3000 \) N

Teacher's Note:
1. Thrust depends on the exhaust speed relative to the rocket.

Answer: (b) 10 m/s\(^2\)

Explanation:
1. Thrust \( =2000\times10=20000 \) N
2. Weight \( =10000 \) N
3. \( a=\frac{20000-10000}{1000}=10 \) m/s\(^2\)

Teacher's Note:
1. Net force = thrust - weight.

Answer: (b) 2773 m/s

Explanation:
1. \( v=v_r\ln\frac{m_0}m=2000\ln4 \)
2. \( =2000\times1.386\approx2773 \) m/s

Teacher's Note:
1. With no gravity v = v_r ln(m0/m).

Answer: (c) \( m_0e^{-at/u} \)

Explanation:
1. \( ma=-u\frac{dm}{dt} \)
2. \( \int_{m_0}^m\frac{dm}m=-\frac aut\Rightarrow\ln\frac m{m_0}=-\frac{at}u \)
3. \( m=m_0e^{-at/u} \)

Teacher's Note:
1. Separate variables and integrate.

Answer: (c) \( \frac{Ft}{m_0+\mu t} \)

Explanation:
1. The sand has zero horizontal velocity, so \( F=\frac{d(mv)}{dt} \)
2. \( F\,dt=d[(m_0+\mu t)v] \)
3. \( v=\frac{Ft}{m_0+\mu t} \)

Teacher's Note:
1. Momentum rate equals the applied force because sand is added at rest.

Answer: (d) 0.5 m/s upward

Explanation:
1. Take up as positive: \( 2(-3)+4(4)=2(4)+4v \)
2. \( 10=8+4v\Rightarrow v=0.5 \) m/s

Teacher's Note:
1. Gravity is non-impulsive, so momentum is conserved.