A body of mass 1 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to $\frac{1}{4}$ of its original velocity. The mass of the second body is

$\frac{2}{5} k g$

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$\frac{3}{5} k g$

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$\frac{3}{2} k g$

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$\frac{2}{3} k g$

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Solution

The correct option is B
$\frac{3}{5} k g$

$1 \times V + m_{2} \times 0 = 1 \times \frac{V}{4} + m_{2} V_{2} m_{2} v_{2} = \frac{3 V}{4}$ For elastic collision, $V_{2} - V_{1} = u_{1} - u_{2} V_{2} - \frac{V}{4} = V - 0 V_{2} = \frac{5 V}{4}$ Substituting in $m_{2} v_{2} = \frac{3 v}{4} m_{2} \times \frac{5 v}{4} = \frac{3 v}{4} \Rightarrow m_{2} = \frac{3}{5} k g$

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