Question

A bullet of mass 0.0 12 kg and horizontal speed 70 m s1 ? strikes a block of wood of 0.4 kg and instantly comes to rest with respect to the block. T he block is mass suspended from the ceiling by means of thin wires. Calculate the height to which the block rises.

Answer 1

Let

$m=0.012\:\rm kg$

$M=0.4\:\rm kg$

$v=70\:\rm m/s$

The law of conservation of momentum gives

$$mv=(m+M)u$$

So, the velocity of the block of wood when it strikes by a bullet

$$u=\frac{mv}{m+M}=\frac{0.012\times 70}{0.012+0.4}=2.04\:\rm m/s$$

Since

$$\Delta E_{kinetic}=\Delta E_{potential}$$

we get

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

So

$$h=\frac{u^2}{2g}=\frac{2.04^2}{2\times9.8}=0.212\:\rm m$$