Question

A block of mass 1kg is suspended freely by a thread. A bullet of mass 3kg is fired horizontally at the block and it becomes embedded in it. The block swings to one side using a vertical distance of 50cm. With what speed did the bullet hit the block?

Answer 1

Given:

$m=1\:\rm kg$

$M=3\:\rm kg$

$h=0.5\:\rm m$

Let the initial speed of a bullet is denoted as $u$. The law of conservation of momentum gives

$$Mu=(m+M)v$$

The law of conservation of energy gives

$$\frac{(M+m)v^2}{2}=(M+m)gh$$

Hence, we get

$$u=\frac{M+m}{M}v=\frac{M+m}{M}\sqrt{2gh}$$

$$u=\frac{3+1}{3}\sqrt{2*9.8*0.5}=4.17\:\rm m/s$$