JEE Physics System Of Particles and Rotational Motion MCQs Set 06

Answer: (c) \( \dfrac{PQ}{P+R} \)

Explanation:
1. Perfectly inelastic: \( PQ = (P + R)V \Rightarrow V = \dfrac{PQ}{P+R} \).

Teacher's Note:
1. The bullet embeds in the body, so the two share a common velocity.

Answer: (d) \( \dfrac{1-e}{1+e} \)

Explanation:
1. Equal masses: \( v_1 = \dfrac{(1-e)u}{2} \) and \( v_2 = \dfrac{(1+e)u}{2} \).
2. \( \dfrac{v_1}{v_2} = \dfrac{1-e}{1+e} \).

Teacher's Note:
1. For e = 1 the first sphere stops and for e = 0 both move together. Check both limits.

Answer: (c) 10 and 16

Explanation:
1. Equal masses in a head-on elastic collision exchange their velocities.
2. So P ends with 10 m/s and Q with 16 m/s.

Teacher's Note:
1. The relative speed of approach is 6 m/s and equals the relative speed of separation. The exchange rule satisfies both conservation laws.

Answer: (c) 0.6

Explanation:
1. \( e = \sqrt{h_2/h_1} = \sqrt{9/25} = 0.6 \).

Teacher's Note:
1. The speed before and after the collision are \( \sqrt{2gh_1} \) and \( \sqrt{2gh_2} \). Their ratio is e.

Answer: (d) 95%

Explanation:
1. Final speed \( V = \dfrac{0.05\times 10}{1} = 0.5 \) m/s.
2. \( K_i = \tfrac{1}{2}(0.05)(100) = 2.5 \) J; \( K_f = \tfrac{1}{2}(1)(0.25) = 0.125 \) J.
3. Loss \( = \dfrac{2.375}{2.5} = 95\% \).

Teacher's Note:
1. In a perfectly inelastic collision with a stationary target the fractional loss is M/(m + M).

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

Explanation:
1. \( mv = (M + m)V \Rightarrow V = \dfrac{m}{M+m}v \).

Teacher's Note:
1. Momentum is conserved, but kinetic energy is not, so do not use the square-root options.

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

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

Teacher's Note:
1. Height scales as the square of the speed, so each bounce multiplies the height by e².

Answer: (d) Zero

Explanation:
1. Total momentum \( = mV - mV = 0 \), so the combined mass has zero velocity.

Teacher's Note:
1. Equal and opposite momenta cancel.

Answer: (c) mgL

Explanation:
1. The bob's speed at the lowest point is \( \sqrt{2gL} \), with kinetic energy mgL.
2. Equal masses exchange velocities in an elastic collision, so the block gets all of it: mgL.

Teacher's Note:
1. The bob stops and the block moves off with the bob's velocity.

Answer: (d) \( \tfrac{1}{4}m(u_1 - u_2)^2 \)

Explanation:
1. Common velocity \( V = \dfrac{u_1 + u_2}{2} \).
2. Loss \( = \tfrac{1}{2}m(u_1^2 + u_2^2) - \tfrac{1}{2}(2m)V^2 = \tfrac{1}{4}m(u_1 - u_2)^2 \).

Teacher's Note:
1. The loss depends only on the relative velocity. For equal masses it is m(Δu)²/4.

Answer: (a) 5.2 m/s

Explanation:
1. By symmetry the two balls have equal speeds v. Momentum along the original direction: \( 9m = 2mv\cos30^\circ \).
2. \( v = \dfrac{9}{\sqrt3} \approx 5.2 \) m/s.

Teacher's Note:
1. The perpendicular components cancel by symmetry, so only the component along the initial direction is needed.

Answer: (d) 8

Explanation:
1. A body that comes to rest after an elastic head-on collision must have the same mass as the target.
2. \( \rho_1\tfrac{4}{3}\pi R^3 = \rho_2\tfrac{4}{3}\pi\left(\tfrac{R}{2}\right)^3 \Rightarrow \dfrac{\rho_2}{\rho_1} = 8 \).

Teacher's Note:
1. The velocity exchange only happens for equal masses. Then compare volumes.

Answer: (b) 10 kg

Explanation:
1. \( 5u = (5 + m)\dfrac{u}{3} \Rightarrow 15 = 5 + m \).
2. \( m = 10 \) kg.

Teacher's Note:
1. Moving together means a perfectly inelastic collision, so a single momentum equation is enough.

Answer: (c) 0.5 m/s

Explanation:
1. \( V = \dfrac{0.01\times 500}{10.01} \approx 0.5 \) m/s.

Teacher's Note:
1. The bullet's mass is negligible compared with the ice, so the denominator is almost 10 kg.

Answer: (c) 40 m/s

Explanation:
1. For \( m_1 \gg m_2 \) and \( u_2 = 0 \), \( v_2 = 2u_1 = 40 \) m/s.

Teacher's Note:
1. A light body hit by a very heavy one at rest moves off with twice the heavy body's speed. This is the 'tennis ball against a truck' situation.

Answer: (c) 2.86 cm/s towards east

Explanation:
1. Taking east as positive: \( 5(10) + 2(-15) = 7V \Rightarrow V = \dfrac{20}{7} \approx 2.86 \) cm/s towards the east.

Teacher's Note:
1. Choose a positive direction and attach the sign to each velocity before applying conservation of momentum.

Answer: (b) 800 J

Explanation:
1. \( V = \dfrac{20\times 10}{25} = 8 \) m/s.
2. \( K = \tfrac{1}{2}(25)(8)^2 = 800 \) J.

Teacher's Note:
1. The initial kinetic energy is 1000 J, so 200 J is lost to deformation and heat.

Answer: (d) m = M

Explanation:
1. \( v_1 = \dfrac{m - M}{m + M}u = 0 \Rightarrow m = M \).

Teacher's Note:
1. Complete transfer of velocity needs equal masses, as in billiards.

Answer: (c) \( \dfrac{1}{3} \)

Explanation:
1. \( v_A = \dfrac{m_A - m_B}{m_A + m_B}u \) and \( v_B = \dfrac{2m_A}{m_A + m_B}u \). Equal speeds in opposite directions: \( v_A = -v_B \).
2. \( m_A - m_B = -2m_A \Rightarrow m_B = 3m_A \), so \( m_A/m_B = \tfrac{1}{3} \).

Teacher's Note:
1. Use the standard elastic-collision formulae and set the velocities equal in magnitude, opposite in sign.

Answer: (b) \( h\left(\dfrac{1+e^2}{1-e^2}\right) \)

Explanation:
1. \( s = h + 2e^2h + 2e^4h + \cdots = h + 2e^2h\cdot\dfrac{1}{1-e^2} \).
2. \( = h\,\dfrac{1 - e^2 + 2e^2}{1 - e^2} = h\,\dfrac{1+e^2}{1-e^2} \).

Teacher's Note:
1. Each rebound adds a 'up and down' trip of 2e²h, 2e⁴h, ... which form a geometric series.

Answer: (b) 12 m/s, 0 m/s

Explanation:
1. Identical balls in a head-on elastic collision exchange velocities.
2. The second ball moves at 12 m/s and the first comes to rest.

Teacher's Note:
1. Read the order in the question: second ball first, then first ball.

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

Explanation:
1. Momentum: \( \vec v_i = \vec v_1 + \vec v_2 \). Energy: \( v_i^2 = v_1^2 + v_2^2 \).
2. Squaring the first and comparing with the second gives \( \vec v_1\cdot\vec v_2 = 0 \), so the angle is 90°.

Teacher's Note:
1. Identical bodies, one at rest, elastic collision, not head-on: the paths are always at right angles.

Answer: (b) 6 m

Explanation:
1. Energy retained \( = 60\% \), and height is proportional to energy.
2. \( h' = 0.6\times 10 = 6 \) m.

Teacher's Note:
1. Potential energy at the top equals the kinetic energy just before the bounce. The loss is a fraction of this.

Answer: (d) \( \left(\dfrac{m}{m+M}\right)v \)

Explanation:
1. \( mv = (m + M)V \Rightarrow V = \dfrac{m}{m+M}v \).

Teacher's Note:
1. Remember the denominator is the total mass after sticking.

Answer: (c) 0

Explanation:
1. Momenta: \( 2\times 3 = 6 \) and \( 3\times 2 = 6 \) in opposite directions, so the total is zero.
2. Therefore the combination is at rest.

Teacher's Note:
1. Compare the magnitudes of the momenta first. If they are equal and opposite the result is zero.