Question

A bullet of mass $0.01kg$ is fired horizontally into a $4kg$ wooden block at rest on a horizontal surface. The coefficient of kinetic friction between the block and surface is $0.25$. The bullet remains embedded in the block and combination moves $20m$ before coming to rest. Find the speed of the bullet when it strike the block.

Answer 1

$a=\mu g=0.25\times 10=2.5m/s^2$

$v^2=u^2+2as$

$0=u^2-2\times (-2.5)\times 20$

$u^2=100\therefore u=10m/s$

Now momentum balance for collision

$(0.01\times u^{\prime })+0=4.01\times 4$

$\therefore u^{\prime }=\frac{4.01\times 10}{0.01}$

$=4010m/s$