A ball of mass $4 kg$ moving with velocity $4 m/s$ collides with another identical ball at rest. The kinetic energy of the balls after collision is $\frac{7}{8}$ times of the original. Find the coefficient of restituton $(e)$ .

1

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0.866

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0.732

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0.239

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Solution

The correct option is B $0.866$

Given:
$(K . E)_{f i n a l} = \frac{7}{8} (K . E)_{i n i t i a l}$
$(K . E)_{l o s s} = (K . E)_{i n i t i a l} - (K . E)_{f i n a l}$
$\Rightarrow (K . E)_{i n i t i a l} - \frac{7}{8} (K . E)_{i n i t i a l}$
$= \frac{1}{8} (K . E)_{i n i t i a l}$ ... $(i)$

We know that
$(K E)_{l o s s} = \frac{1}{2} \times \frac{m_{1} m_{2}}{(m_{1} + m_{2})} (u_{a p p r o a c h})^{2} (1 - e^{2})$ ... $(i i)$
Equating $(i)$ and $(i i)$
$\frac{1}{2} \times \frac{m_{1} m_{2}}{(m_{1} + m_{2})} \times (u_{a p p r o a c h})^{2} (1 - e^{2}) = \frac{1}{8} (K . E .)_{i n i t i a l}$
$\Rightarrow \frac{1}{2} \times (\frac{m \times m}{m + m}) \times u_{1}^{2} \times (1 - e^{2}) = \frac{1}{8} (\frac{1}{2} m u_{1}^{2})$
$\Rightarrow \frac{1}{4} \times (1 - e^{2}) = \frac{1}{8} \times \frac{1}{2}$
$\Rightarrow e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$
or $e = \frac{\sqrt{3}}{2} = 0.866$

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