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A particle of mass m, kinetic energy K and momentum P collides head on elastically with another particle of mass 2m at rest. After collision
List-I List-II
I) Momentum of first particle P) 4P/3
II) Momentum of second particle Q) –P/3
III) K.E of first particle R) 8K/9
IV) K.E of second particle S) K/9
T) K/3

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$(I) \to (S); (II) \to (Q); (III) \to (P); (IV) \to (R)$

$(I) \to (Q); (II) \to (P); (III) \to (S); (IV) \to (R)$

$(I) \to (Q); (II) \to (P); (III) \to (S); (IV) \to (T)$

$(I) \to (S); (II) \to (Q); (III) \to (R); (IV) \to (P)$

answer is B.

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Detailed Solution

$\begin{aligned} v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) v_{1} + (\frac{2 m_{1}}{m_{1} + m_{2}}) v_{2} - (1) \\ v_{2} = (\frac{m_{1} - m_{1}}{m_{1} + m_{2}}) v_{2} + (\frac{2 m_{1}}{m_{1} + m_{2}}) v_{1} - (2) \end{aligned}$

$v_{1} = - \frac{v_{1}}{3} a n d v_{2} = \frac{2}{3} v_{1}$

Momentum of first particle is –P/3.

Momentum of second particle is 4P/3

K.E of first particle is K/9

K.E of second particle is 8K/9

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