Question

A bullet of mass 0.01 kg is fired horizontally into a 4kg wooden block at rest on horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.25. The bullet gets embedded in the block and the combination moves 20 m before coming to rest. With what speed did the bullet strike the block?

Answer 1

Given,

Mass of bullet is $m=0.01kg$

Mass of the block is $M=4kg$

$Coefficient=0.25,distance=20m$

So, let the speed of the block just after the bullet embedded in it be $V$ and $v$ be the speed of bullet before striking the block,

By applying conservation of momentum,

$mv=(m+M)V$

$V=\frac{mv}{M+m}$

So, By the work-energy theorem,

$\frac{1}{2}(m+M)V^2=Force\times displacement=\mu Nd$

$\Rightarrow \frac{1}{2}(m+M)(\frac{mv}{M+m}{)}^2=\mu (m+M)gd$

$\frac{mv}{M+m}=2\sqrt{\mu gd}$

$V=\frac{2\sqrt{\mu gd}(m+M)}{m}$

$\frac{\sqrt{0.25\times 10\times 20(0.01+4)}}{0.01}$

$=567.09m/s$

Thus, the required velocity is $=567.09m/s$