A bullet of mass 10 gm moving with a velocity of 400 m/sec gets embedded in a freely suspended wooden block of mass 900 gm. Calculate the velocity of wooden block acquired.

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Solution

Step 1: Given that:
Mass of bullet( $m_{b u l l e t}$ ) = $10 g$ = $\frac{10}{1000} k g = 0.01 k g$
Velocity of the bullet( $u_{b u l l e t}$ ) = $400 m s^{- 1}$
Mass of freely suspended wooden block( $M$ ) = $900 g$ = $\frac{900}{1000} k g = 0.9 k g$
Initial velocity of the wooden block( $u_{b l o c k}$ )= 0
Step 2: Formula used:
According to law of conservation of momentum,
The total momentum of the bodies before collision= Total momentum of the bodies after collision
Mathematically; $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
Momentum= mass $\times$ velocity of the body
Step 3: Calculation of velocity of the block after collision:
$m_{b u l l e t} u_{b u l l e t} + M u_{b l o c k} = (m_{b u l l e t} + M) v_{b l o c k}$
$\Rightarrow 0.01 k g \times 400 m s^{- 1} + 0.9 k g \times 0 = (0.01 k g + 0.9 k g) v_{b l o c k}$
$\Rightarrow 0.91 v_{b l o c k} = 4$
$\Rightarrow v_{b l o c k} = \frac{4}{0.91}$
$\Rightarrow v_{b l o c k} = 4.40 m s^{- 1}$
Thus,
The velocity of the wooden block is $4.40 m s^{- 1}$ .

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