A long plank of mass M lies on a horizontal surface , on the surface of which there is a ball of mass m . At the initial moment of time, the plank is given initial velocity $v_{o}$ as shown, and the ball hits the plank with some velocity downwards. Make estimate of the distance travelled by the plank until it stops. Consider that the ball manages to hit the surface of the plank many times without jumping off it. Assume the impact is absolutely elastic, the coefficient of friction between the plank and the horizontal surface is equal to $μ$ and there is no friction between the ball and plank.

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$\frac{v_{0}^{2}}{2 μ g} \frac{M}{m + M}$

$\frac{v_{0}^{2}}{2 μ g} \frac{m}{m + M}$

$\frac{v_{0}^{2}}{μ g} \frac{M}{m + M}$

$\frac{v_{0}^{2}}{μ g} \frac{m}{m + M}$

answer is B.

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Detailed Solution

Let v be the velocity of projection of ball and t is the time taken by plank to stop is t
The frictional impulse given by ball to plank in time t is $μ 2 m v n = μ 2 m v \frac{t}{2 v / g} = μ m g t$
When n is the number of collisions made by ball before plank stops
Hence it can be approximated that the effective acceleration of the plank is $a = - \frac{μ M g + μ m g}{M}$
and the path travelled by the block to the stop will be $s = \frac{v_{0}^{2}}{2 a} = \frac{v_{0}^{2}}{2 μ g} \frac{M}{m + M}$