Question

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.

Answer 1

Step 1: Given that:

Mass of bullet($m_{bullet}$) = $10g$ = $\frac{10}{1000}kg=0.01kg$

Velocity of the bullet($u_{bullet}$) = $400m{s}^{-1}$

Mass of freely suspended wooden block($M$) = $900g$ = $\frac{900}{1000}kg=0.9kg$

Initial velocity of the wooden block($u_{block}$)= 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:

$m_{bullet}{u}_{bullet}+M{u}_{block}=(m_{bullet}+M)v_{block}$

$\Rightarrow 0.01kg\times 400m{s}^{-1}+0.9kg\times 0=(0.01kg+0.9kg)v_{block}$

$\Rightarrow 0.91v_{block}=4$

$\Rightarrow v_{block}=\frac{4}{0.91}$

$\Rightarrow v_{block}=4.40m{s}^{-1}$

Thus,

The velocity of the wooden block is $4.40m{s}^{-1}$ .