First slide

Mechanical energy conservation

Question

A bullet of mass ‘m’ is fired with a velocity ‘v’ into a block of wood of mass ‘M’ suspended by a thread. If the bullet gets embedded in the block it rises to a height of

Moderate

A

$\frac{1}{2 g} {(\frac{m v}{M + m})}^{2}$
B

$\frac{m v}{M g}$
C

$\frac{v^{2}}{2 g}$
D

$\frac{m}{M + m} \frac{v^{2}}{2 g}$

Solution

Conserving the linear momentum just before and after the strike
$m v = (m + M) V_{C}$
$\Rightarrow V_{C} = \frac{m v}{m + M}$
Now, again from conservation of energy $\frac{1}{2} (m + M) V_{C}^{2} = (m + M) g h$
$\Rightarrow {(\frac{m V}{M + m})}^{2} = 2 g h$
$\Rightarrow h = \frac{1}{2 g} {(\frac{m V}{m + M})}^{2}$

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