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Q.

For the system shown in the figure, a small block of mass m and smooth irregular shaped block of mass M, both free to move are placed on a smooth horizontal plane. The minimum velocity v imparted to block so that it will overcome the highest point of M is

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a

$\sqrt{\frac{2 m}{M} gh}$

b

$\sqrt{2 (1 + \frac{M}{m}) gh}$

c

$\sqrt{2 gh}$

d

$\sqrt{2 (1 + \frac{m}{M}) gh}$

answer is B.

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Detailed Solution

For the given situation, block should be able to reach just at highest point of M, and their relative velocity is to become zero. Let their common velocity is $v_{C}$ .

$\begin{array}{l} COLM \to (M + m) v_{C} = mv \Rightarrow v_{C} = \frac{mv}{M + m} \\ COME : \frac{1}{2} (M + m) v_{C}^{2} + mgh = \frac{1}{2} {mv}^{2} \end{array}$

Solve to get $v = \sqrt{2 (1 + \frac{m}{M}) gh}$

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