Question

A bullet of mass 20 g moving horizontally at a speed of 300 m/s is fired into a wooden block of mass 500 g suspended by a long string.The bullet crosses the block and emerges on the other side.If the centre of mass of the block rises through a height of 20.0 cm,find the speed of the bullet as it emerges from the block.

Answer 1

Mass of bullet,m=20 gm=0.02 kg

Mass of wooden block,M=500 gm=0.5 kg

Velocity of the bullet with which it strikes u=300 m/sec

Let the bullet emerges out with velocity V and the velocity of block $v^1$

As per law of conservation of momentum,mu=$M{v}^1+M_{1}v....\left(i\right)$

Again applying work-energy principle for the block after the collision,

$0-\left(\frac{1}{2}\right)\left(M\right)\times \left(v^1\right)=-M.g.h$

(where h=0.2 m),

$\Rightarrow (v^1{)}^2=2gh$

$v^1=\sqrt{2gh}$

$=\sqrt{20\times 2.0}=2m/sec$

Substituting the value of $v^1$ in equation (i),we get

$0.02\times 300=0.5\times 2+0.2\times v$

$\Rightarrow v=\frac{6-1}{0.02}$

$=\frac{5}{0.02}=250m/sec$