Question

A bullet of mass 0.005kg moving with a speed of 200m/s enters a heavy wooden block and is stopped after a distance of 50 cm. What is the average force extended by the block on the bullet ? Also find impulse ?

Answer 1

Work done by force applied by block on the bullet =K.E of bullet

$F_{avg}\times$displacement=$\frac{1}{2}m{v}^2$

$F_{avg}\times 50\times {10}^{-2}=\frac{1}{2}0.005{\left(200\right)}^{2}F=0.005\times 40000=200N$

Deacceleration of bullet is =$\frac{-F}{m}=\frac{-200}{0.005}=40\times {10}^{3}m/se{c}^2$

Time taken to stop the bullet is $t$.

$v-u=at$

$0-200=-40\times {10}^{3}t$

$t=5\times {10}^{-3}$s

Impulse= F.t =$200\times 5\times {10}^{-3}=1$Ns

Also Impulse=change in Momentum

$=mv-0=0.005\times 200=1$Ns

We can clearly see that both are equal.