First slide

Collisions

Question

A bullet of mass m moving with velocity v strikes a suspended wooden block of mass M and gets embedded in it. If the block rises to a height h, the initial velocity of the block will be

Moderate

A

$\sqrt{2 gh}$
B

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

$\frac{m}{M + m} 2 gh$
D

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

Solution

Initial K.E. of block when bullet strikes to it
$= \frac{1}{2} (m + M) V^{2}$
Due to this K.E. block will rise to a height h.
Its potential energy = $(m + M) gh .$
By the law of conservation of energy
$\frac{1}{2} (m + M) V^{2} = (m + M) gh ∴ V = \sqrt{2 gh}$

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