JEE Physics Collisions MCQs Set 02

Answer: (c) 4 m

Explanation:
1. Energy retained is 40%, and energy is proportional to height: h = 0.4 × 10 = 4 m.

Teacher's Note:
1. Height is proportional to energy.
2. Retained = 100% − 60%.

Answer: (c) 3 m/s

Explanation:
1. Equal masses with equal and opposite speeds: the centre of mass is at rest, so after impact the speeds are equal.
2. Separation speed = e × approach speed = 0.5 × 12 = 6 m/s, so each ball moves at 3 m/s.

Teacher's Note:
1. Use the relative-speed definition of e.
2. Total momentum is zero.

Answer: (a) 6.6 cm/s towards the west

Explanation:
1. Take east positive: \( 10(10)+20(-15) = 30v \Rightarrow v = -\frac{200}{30} = -6.67 \) cm/s.
2. Negative sign means towards the west.

Teacher's Note:
1. Opposite directions have opposite signs.
2. Momentum conservation.

Answer: (d) 6 m/s and 0 m/s

Explanation:
1. For a head-on elastic collision of equal masses the velocities are exchanged.
2. So the second ball moves at 6 m/s and the first stops.

Teacher's Note:
1. Equal masses exchange velocities.
2. Momentum and energy are both conserved.

Answer: (b) \( 3\sqrt3 \) m/s

Explanation:
1. Momentum along the initial direction: \( m(9) = 2mv\cos30^\circ \).
2. \( v = \frac{9}{2\cos30^\circ} = \frac{9}{\sqrt3} = 3\sqrt3 \) m/s.

Teacher's Note:
1. Perpendicular components cancel by symmetry.
2. Both balls have equal speeds.

Answer: (b) 1 m/s

Explanation:
1. Momenta: 30 kg m/s and 40 kg m/s at right angles, resultant \( \sqrt{30^2+40^2} = 50 \) kg m/s.
2. \( v = \frac{50}{50} = 1 \) m/s.

Teacher's Note:
1. Add momenta as vectors.
2. Total mass is 50 kg.

Answer: (c) \( e^n = \sqrt{\frac{h_n}{h_0}} \)

Explanation:
1. Each collision multiplies the speed by e and the height by \( e^2 \). So \( h_n = e^{2n}h_0 \).

Teacher's Note:
1. Height ∝ speed².
2. Repeated collisions multiply.

Answer: (a) \( \frac{1-e}{1+e} \)

Explanation:
1. \( v_1+v_2 = u \) and \( v_2-v_1 = eu \), so \( v_1 = \frac{(1-e)u}2 \), \( v_2 = \frac{(1+e)u}2 \).
2. Ratio \( \frac{v_1}{v_2} = \frac{1-e}{1+e} \).

Teacher's Note:
1. Use momentum and the restitution definition.
2. Equal masses.

Answer: (d) \( u\sqrt{\sin^2\theta+e^2\cos^2\theta} \)

Explanation:
1. The horizontal component \( u\sin\theta \) is unchanged (no friction); the vertical component becomes \( eu\cos\theta \).
2. \( v = \sqrt{u^2\sin^2\theta+e^2u^2\cos^2\theta} \).

Teacher's Note:
1. Only the normal component is reversed and reduced.
2. Angle is with the vertical.

Answer: (b) \( \frac u2 \)

Explanation:
1. At contact the line joining the centres makes angle φ with the velocity where \( \sin\varphi = \frac{\sqrt3/2\,R}{R} \), so \( \varphi = 60^\circ \).
2. For equal masses in an elastic collision the sphere takes the velocity component along the line of centres: \( v = u\cos60^\circ = \frac u2 \).

Teacher's Note:
1. Impulse acts along the line of centres.
2. Equal elastic masses exchange normal components.