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A steel sphere A of mass m, moving with velocity 4 m/s, collides elastically and obliquely with another identical stationary steel sphere B. After the impact sphere B starts moving with velocity V making an angle of $60^{o}$ with the original direction of sphere A. Find V.

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$2 \sqrt{3} m / s$

$2 \sqrt{2} m / s$

$2 m / s$

$3 m / s$

answer is C.

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Detailed Solution

After the impact, velocity vectors of the spheres will be mutually perpendicular,

Conserving Momentum,

$\begin{array}{l} {mV}^{'} \sin 30^{0} = mVsin 60^{0} \Rightarrow V^{'} = \sqrt{3} V . ...... (1) \\ and mVcos 60^{0} + {mV}^{'} \cos 30^{0} = m \times 4 \Rightarrow V + \sqrt{3} V^{'} = 8 . ...... (2) \end{array}$

Solving, we get V = 2 m/s.

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