Question

A bullet weighing 4.5*10^-3 kg is fixed horizontally into a 1.8 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet comes to rest in the block which moves 1.8m. Find the speed of the bullet.

Answer 1

Given:

$m=4.5*10^{-3}\:\rm kg$

$M=1.8\:\rm kg$

$\mu=0.20$

$d=1.8\:\rm m$

Let the initial speed of the bullet we denoted as $u$.Then initial speed of the block is given by the law of conservation of momentum

$$v=u\frac{m}{M+m}\approx u\frac{m}{M}$$

The energy-work theorem gives

$$\frac{Mv^2}{2}=F_fd=\mu Mgd$$

So,

$$v=\sqrt{2\mu g d}=u\frac{m}{M}$$

$$u=\sqrt{2\mu g d}\frac{M}{m}$$

$$u=\sqrt{2*0.20*9.8*1.8}*\frac{1.8}{4.5*10^{-3}}\\=1063\:\rm m/s$$