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A sphere of mass $2 k g$ moving with a velocity of $4 m s^{- 1}$ collides head–on with another sphere of mass $2 k g$ moving in opposite direction with a velocity of $2 m s^{- 1} .$ After collision if they stick together and move with a velocity of v then the total initial momentum and velocity v are

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$2 k g m s^{- 1}, \frac{2}{7} m / s$

$1 k g m s^{- 1}, \frac{1}{7} m / s$

$3 k g m s^{- 1}, \frac{3}{7} m / s$

$4 k g m s^{- 1}, 1 m / s$

answer is D.

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Detailed Solution

$P = (m_{1} u_{1} - m_{2} u_{2}) = 2 \times 4 - 2 \times 2 = 4 k g m s^{- 1}$

$v = \frac{m_{1} u_{1} - m_{2} u_{2}}{m_{1} + m_{2}} = \frac{2 \times 4 - 2 \times 2}{2 + 2} = \frac{4}{4} = 1 m s^{- 1}$

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