A particle of mass 1kg with kinetic energy of 8 Joule makes an elastic head on collision with a stationary particle which has twice its mass. Then the fraction of energy transferred to the second body is

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$\frac{3}{4}$

$\frac{17}{36}$

$\frac{25}{72}$

$\frac{9}{25}$

answer is C.

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Detailed Solution

$\frac{1}{2} {mu}^{2} = 8 \Rightarrow u = \sqrt{\frac{16}{1}} m / s = 4 m / s$

$2 \times v_{2} + 1 \times v_{1} = 1 \times 4 \begin{array}{l} v_{1} = (4 - 2 v_{2})________(1) \\ Also, v_{2} - v_{1} = 1 - 0 \Rightarrow v_{2} - 4 + 2 v_{2} = 1 \Rightarrow v_{2} = \frac{5}{3} m / s \end{array} Fraction of energy transferred = \frac{\frac{1}{2} \times 2 \times {(\frac{5}{3})}^{2}}{\frac{1}{2} \times 1 \times {(4)}^{2}} = \frac{25}{72}$

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