JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Conservation of Momentum

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What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass? (Assume the collision to be head-on elastic collision)

(A) $50.0%$
(B) $66.6%$
(C) $55.5%$ ✓ Correct
(D) $33.3%$

Answer: (C)

Solution

For a head-on elastic collision, the velocity of the second particle (initially at rest) after collision is:

v_{2} = \frac{2 m _{1} u _{1}}{m _{1} + m _{2}}

Given

m_{2} = 5 m_{1}

and

u_{1} = u_{0}

v_{2} = \frac{2 m _{1} u _{0}}{m _{1} + 5 m _{1}} = \frac{2 u _{0}}{6} = \frac{u _{0}}{3}

The kinetic energy transferred to the second particle is:

K E_{2} = \frac{1}{2} m_{2} v_{2}^{2} = \frac{1}{2} (5 m_{1}) (\frac{u _{0}}{3})^{2} = \frac{5 m _{1} u _{0}^{2}}{18}

The initial kinetic energy of the first particle is:

K E_{1} = \frac{1}{2} m_{1} u_{0}^{2}

Percentage of energy transferred:

\frac{K E _{2}}{K E _{1}} \times 100 = \frac{\frac{5 m _{1} u _{0}^{2}}{18}}{\frac{1}{2} m _{1} u _{0}^{2}} \times 100 = \frac{5}{9} \times 100 \approx 55.6%

Therefore, the answer is (C).

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Conservation of Momentum

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