JEE Physics System Of Particles and Rotational Motion MCQs Set 56

Answer: (b) linear momentum of the system does not change in time

Explanation:
1. Zero net external force means the total linear momentum is constant (\( d\vec P/dt=\vec F_{ext}=0 \)).
2. The external torques can still be non-zero and internal forces can change the kinetic energy, so the other quantities need not be constant.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (c) \( 90^\circ \)

Explanation:
1. \( \vec F=\dfrac{d\vec p}{dt}=-Ak\left[\hat i\sin(kt)+\hat j\cos(kt)\right] \).
2. \( \vec F\cdot\vec p=-A^2k[\sin kt\cos kt-\cos kt\sin kt]=0 \), so the angle is \( 90^\circ \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) (b) and (d) only

Explanation:
1. Equal masses in an elastic collision exchange velocities: C stops and A moves with v.
2. At maximum compression A and B move together at \( v/2 \): \( KE=\tfrac12(2m)\left(\dfrac v2\right)^2=\dfrac{mv^2}4 \), so (b) is true and (a) false.
3. Spring energy \( \tfrac12kx^2=\tfrac12mv^2-\dfrac{mv^2}4=\dfrac{mv^2}{4} \), so \( x=v\sqrt{\dfrac m{2k}} \): (d) true.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) \( \dfrac{v_0}3 \)

Explanation:
1. C stops and A moves with \( v_0 \) (equal masses, elastic).
2. Then \( mv_0=(m+2m)V\Rightarrow V=\dfrac{v_0}3 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) \( \dfrac{2mv_0^2}{3x_0^2} \)

Explanation:
1. Energy: \( \tfrac12mv_0^2=\tfrac12(3m)\dfrac{v_0^2}{9}+\tfrac12kx_0^2 \).
2. \( \tfrac12kx_0^2=\tfrac12mv_0^2\cdot\tfrac23=\dfrac{mv_0^2}3\Rightarrow k=\dfrac{2mv_0^2}{3x_0^2} \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (c) \( \dfrac{m(R-r)}{M+m} \)

Explanation:
1. The cylinder's centre moves horizontally by \( R-r \) relative to the block.
2. No external horizontal force: \( m(R-r-d)=Md \), so \( d=\dfrac{m(R-r)}{M+m} \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) \( m\sqrt{\dfrac{2g(R-r)}{M(M+m)}} \)

Explanation:
1. Horizontal momentum: \( mv=MV \). Energy: \( mg(R-r)=\tfrac12mv^2+\tfrac12MV^2 \).
2. \( v^2=\dfrac{2gM(R-r)}{M+m} \), so \( V=\dfrac mM v=m\sqrt{\dfrac{2g(R-r)}{M(M+m)}} \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) 0.20 m s⁻¹

Explanation:
1. Momentum conservation: \( 50v_m=20v_s \) and \( v_s+v_m=0.7 \).
2. \( v_m=\dfrac{20}{70}(0.7)=0.2 \) m/s.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) \( 2\sqrt2\,v \)

Explanation:
1. Initial momentum \( =m(2v)+2m(v)=4mv \).
2. After: \( 2\,m\,u\cos45^\circ=4mv\Rightarrow u=2\sqrt2\,v \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (c) 1.2 kg

Explanation:
1. \( v_1'=\dfrac{m_1-m_2}{m_1+m_2}u=\dfrac u4 \Rightarrow4(2-m_2)=2+m_2 \).
2. \( 8-4m_2=2+m_2\Rightarrow m_2=1.2 \) kg.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) \( v_4-v_2 \)

Explanation:
1. Momentum: \( m_1v_1+0.5m_1v_2=m_1(0.5v_1)+0.5m_1v_4 \).
2. \( v_1+0.5v_2=0.5v_1+0.5v_4\Rightarrow0.5v_1=0.5(v_4-v_2)\Rightarrow v_1=v_4-v_2 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) 4m

Explanation:
1. After the collision \( KE=0.36 \) of initial, so \( |v'|=0.6u \) and it is reversed: \( \dfrac{m-M}{m+M}=-0.6 \).
2. \( m-M=-0.6m-0.6M\Rightarrow0.4M=1.6m\Rightarrow M=4m \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) in an elliptical orbit

Explanation:
1. The satellite moves tangentially with the orbital speed \( v \); the meteorite moves radially with speed \( v \). Momentum conservation gives a combined velocity with tangential component \( v/2 \) and radial component \( v/2 \).
2. The speed \( v/\sqrt2 \) is less than the escape speed (\( \sqrt2 v \)), and the tangential part is below the circular value, so the body moves in an elliptical orbit.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (c) \( m\left(\dfrac{\theta_0+\theta_1}{\theta_0-\theta_1}\right) \)

Explanation:
1. For small angles, speed at the bottom \( \propto\theta \). The bob rebounds: \( \dfrac{v'}{v}=\dfrac{M-m}{M+m}=\dfrac{\theta_1}{\theta_0} \).
2. \( M(\theta_0-\theta_1)=m(\theta_0+\theta_1)\Rightarrow M=m\dfrac{\theta_0+\theta_1}{\theta_0-\theta_1} \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (d) 40 m

Explanation:
1. They meet at \( t=1 \) s at height \( 100-5=95 \) m. Wood speed 10 m/s down, bullet \( 100-10=90 \) m/s up.
2. Combined: \( V=\dfrac{0.02(90)-0.03(10)}{0.05}=30 \) m/s up; extra height \( =\dfrac{30^2}{20}=45 \) m.
3. Above the top: \( 95+45-100=40 \) m.

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (a) 4

Explanation:
1. After A and B stick: speed \( v/2 \). After sticking to C: \( V=\dfrac{2m(v/2)}{2m+M}=\dfrac{mv}{2m+M} \).
2. Final KE \( =\dfrac{m^2v^2}{2(2m+M)} \) must be \( \dfrac16\cdot\tfrac12mv^2 \), so \( 2m+M=6m \).
3. \( M/m=4 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.

Answer: (b) \( \sqrt2\,v_0 \)

Explanation:
1. \( v_1+v_2=v_0 \) and \( v_1^2+v_2^2=1.5v_0^2 \), so \( 2v_1v_2=-0.5v_0^2 \).
2. \( (v_1-v_2)^2=1.5v_0^2+0.5v_0^2=2v_0^2 \), so \( |v_1-v_2|=\sqrt2\,v_0 \).

Teacher's Note:
1. Apply conservation of linear momentum along each axis separately when no external force acts in that direction.
2. For collisions use momentum conservation first, then the energy condition (elastic or inelastic) to fix the remaining unknown.