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A particle of mass m moving towards east with a speed $V$ collides with another particle of same mass and same speed $V$ moving towards the north. If the two pieces stick to each other, the new particle of mass $2 m$ will have a speed

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$\frac{V}{\sqrt{2}}$

$\sqrt{2} V$

$\frac{V}{2}$

$V$

answer is C.

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Detailed Solution

$\sqrt{{(m v)}^{2} + {(m v)}^{2}} = (2 m) V'$

$∴ \frac{m V \sqrt{2}}{2 m} = V' \Rightarrow \frac{V}{\sqrt{2}} m / \sec$

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