JEE Physics System Of Particles and Rotational Motion MCQs Set 13

Answer: (c) \(\frac{m}{M+m}\,v\)

Explanation:
1. Horizontal momentum is conserved during the short impact
2. \(mv=(M+m)V\)
3. \(V=\frac{m}{M+m}v\)

Teacher's Note:
1. In a perfectly inelastic collision momentum is conserved but kinetic energy is not. The string is vertical so it exerts no horizontal impulse.

Answer: (b) \(2\hat k\) m/s

Explanation:
1. \(\vec v_{cm}=\frac{m_1\vec v_1+m_2\vec v_2}{m_1+m_2}\)
2. \(10(2,-7,3)+2(-10,35,-3)=(0,0,24)\)
3. Dividing by 12 gives \(2\hat k\) m/s

Teacher's Note:
1. Add the momentum vectors component by component and divide by the total mass.

Answer: (d) \(20\sqrt2\) m/s

Explanation:
1. The fragments have masses 3 kg, 1 kg and 1 kg
2. The two 1 kg fragments have a combined momentum of magnitude \(\sqrt{60^2+60^2}=60\sqrt2\) kg m/s
3. The initial momentum is zero, so the 3 kg fragment has the same magnitude: \(v=\frac{60\sqrt2}{3}=20\sqrt2\) m/s

Teacher's Note:
1. In an explosion from rest the total momentum remains zero. The lightest pair's resultant is balanced by the heavy fragment.

Answer: (a) 30 degrees

Explanation:
1. The surface is smooth, so the component parallel to it is unchanged: \(v\cos\phi=u\cos45^\circ\)
2. The normal component is reduced by the factor \(e\): \(v\sin\phi=e\,u\sin45^\circ\)
3. \(\tan\phi=e=\frac1{\sqrt3}\), so \(\phi=30^\circ\)

Teacher's Note:
1. For an oblique collision with a smooth surface, the tangential component is unchanged and only the normal component is multiplied by \(e\).

Answer: (b) along the original parabolic path

Explanation:
1. The forces of the explosion are internal, so they cannot change the motion of the centre of mass
2. Only gravity acts externally, as it did on the shell
3. The centre of mass therefore keeps following the original parabola

Teacher's Note:
1. Internal forces never change the motion of the centre of mass. This applies to explosions, collisions and breaking of bodies.

Answer: (a) zero

Explanation:
1. There is no external horizontal force on the system (still water, no friction)
2. The centre of mass of the system, initially at rest, therefore stays where it is
3. The boat shifts, but the centre of mass of the whole system does not move: shift = 0

Teacher's Note:
1. The boat and the men move relative to each other, but the system's centre of mass stays fixed when there is no external horizontal force.

Answer: (b) \(\frac23\) kg and \(\frac68v\)

Explanation:
1. For an elastic collision with a body at rest: \(v_1'=\frac{m_1-m_2}{m_1+m_2}v=\frac v2\)
2. \(\frac{2-m_2}{2+m_2}=\frac12\) gives \(m_2=\frac23\) kg
3. \(v_{cm}=\frac{m_1v}{m_1+m_2}=\frac{2v}{8/3}=\frac34v=\frac68v\)

Teacher's Note:
1. The speed of the centre of mass is unaffected by the collision and can be found from the initial momentum.

Answer: (b) 5.2 m/s

Explanation:
1. Equal masses and equal angles on both sides mean equal final speeds \(v'\)
2. Momentum along the initial direction: \(m\times9=2mv'\cos30^\circ\)
3. \(v'=\frac{9}{2\times0.866}=5.2\) m/s

Teacher's Note:
1. Use momentum conservation along the original line of motion. The components perpendicular to it cancel by symmetry.

Answer: (c) \(-2\hat i-\hat j+\hat k\)

Explanation:
1. \(\vec r_{cm}=\frac{1(\hat i+2\hat j+\hat k)+3(-3\hat i-2\hat j+\hat k)}{4}\)
2. \(=\frac{-8\hat i-4\hat j+4\hat k}{4}=-2\hat i-\hat j+\hat k\)

Teacher's Note:
1. The centre of mass lies closer to the heavier body. Weigh each position vector by its mass.

Answer: (c) 9 m/s

Explanation:
1. Take the ball's direction as positive: ball \(+3\) m/s, wall \(-3\) m/s (approaching)
2. In an elastic collision the relative speed of separation equals the relative speed of approach, which is 6 m/s
3. The wall is very heavy and keeps its velocity, so the ball's velocity is \(-3-6=-9\) m/s, a speed of 9 m/s

Teacher's Note:
1. A ball bouncing off a moving wall gains twice the wall's speed. Always work with relative velocities in elastic collisions.

Answer: (c) 20 m/s

Explanation:
1. Kinetic energy just before impact: \(\tfrac12mv_0^2+mgh\)
2. After losing half: \(\tfrac12\left(\tfrac12mv_0^2+mgh\right)\), and this must lift the ball back to height \(h\), so it equals \(mgh\)
3. \(\tfrac12mv_0^2+mgh=2mgh\), so \(v_0^2=2gh=400\) and \(v_0=20\) m/s

Teacher's Note:
1. Track energy at the ground: before impact, after impact and at the top of the rebound.