JEE Physics Collisions MCQs Set 01

Answer: (d) \( \frac dm\sqrt{kM} \)

Explanation:
1. Momentum conservation: gun recoil speed \( V = \frac{mv}{M} \).
2. Kinetic energy of the gun goes into the spring: \( \frac12MV^2 = \frac12kd^2 \Rightarrow V = d\sqrt{\frac kM} \)
3. \( v = \frac MmV = \frac dm\sqrt{kM} \)

Teacher's Note:
1. Use momentum conservation for the firing, then energy conservation for the spring compression.
2. The spring only starts acting after the bullet has left.

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

Explanation:
1. Perfectly inelastic collision: \( mv = (M+m)V \).
2. \( V = \frac{m}{m+M}v \)

Teacher's Note:
1. Momentum is conserved but kinetic energy is not when the bullet embeds.
2. The fraction of kinetic energy lost is M/(M + m).

Answer: (d) 6.67 cm/s towards the west

Explanation:
1. Take east as positive: \( 10(10) + 20(-15) = 30\,V \)
2. \( V = \frac{100-300}{30} = -6.67 \) cm/s, i.e. towards the west.

Teacher's Note:
1. Choose a positive direction and keep signs with every velocity.
2. The heavier and faster lump decides the direction after sticking.

Answer: (b) 6 m/s, 0

Explanation:
1. Equal masses colliding elastically exchange velocities.
2. The second ball moves at 6 m/s and the first comes to rest.

Teacher's Note:
1. Exchange of velocities needs equal masses and e = 1.
2. Check: momentum 6m before and after; kinetic energy 18m before and after.

Answer: (c) 4 m

Explanation:
1. Energy retained \( = 40\% \). Since energy \( = mgh \), the rebound height is \( 0.4\times10 = 4 \) m.

Teacher's Note:
1. Potential energy at the top and kinetic energy just after the bounce are proportional to height.
2. The coefficient of restitution here is the square root of 0.4.

Answer: (c) 3 m/s

Explanation:
1. By symmetry, total momentum is zero, so the balls rebound with equal speeds v.
2. Speed of separation \( = e\times \) speed of approach: \( 2v = 0.5(12) = 6 \)
3. \( v = 3 \) m/s

Teacher's Note:
1. Equal masses with equal and opposite velocities simply reverse and lose speed by the factor e.
2. Each ball's speed is multiplied by e.

Answer: (c) \( he^{2n} \)

Explanation:
1. After each bounce the speed becomes e times and the height \( e^2 \) times.
2. After n bounces: \( h_n = e^{2n}h \)

Teacher's Note:
1. Height is proportional to speed squared, so each bounce multiplies the height by e squared.
2. The sequence of heights is a geometric progression with ratio e squared.

Answer: (a) \( \frac{m_2}{m_1} \)

Explanation:
1. Momentum conservation: \( m_1\Delta v_1 = -m_2\Delta v_2 \).
2. \( \left|\frac{\Delta v_1}{\Delta v_2}\right| = \frac{m_2}{m_1} \)

Teacher's Note:
1. Equal and opposite changes of momentum mean the lighter body changes velocity more.
2. This holds for any collision, elastic or not.

Answer: (a) 0.25 m

Explanation:
1. At maximum compression both masses have the same velocity: \( V = \frac{2(10)+5(3)}{7} = 5 \) m/s.
2. Energy: \( \frac12(2)(100) + \frac12(5)(9) = \frac12(7)(25) + \frac12kx^2 \)
3. \( 122.5 = 87.5 + 560x^2 \Rightarrow x^2 = \frac{35}{560} = \frac1{16} \Rightarrow x = 0.25 \) m

Teacher's Note:
1. Maximum compression occurs when the relative velocity is zero.
2. Momentum and total mechanical energy are both conserved because the spring is ideal.

Answer: (b) 75% and 2 : 1

Explanation:
1. Kinetic energy just before and after the impact equals \( mgh_1 \) and \( mgh_2 \).
2. Loss \( = \frac{10-2.5}{10}\times100 = 75\% \)
3. Velocity ratio \( \frac{v_1}{v_2} = \sqrt{\frac{h_1}{h_2}} = \sqrt4 = 2 \)

Teacher's Note:
1. Velocity goes as the square root of height; energy goes as height.
2. The coefficient of restitution is 1/2.

Answer: (d) 4 m/s and 4.67 m/s

Explanation:
1. Momentum: \( 2(5)+3(4) = 2v_1+3v_2 \Rightarrow 2v_1+3v_2=22 \)
2. Restitution: \( v_2 - v_1 = \frac23(5-4) = \frac23 \)
3. Solving: \( v_1 = 4 \) m/s, \( v_2 = \frac{14}{3} = 4.67 \) m/s

Teacher's Note:
1. Two equations: momentum and restitution.
2. The heavier front ball speeds up, while the rear ball slows down but still moves forward.