Question

A bullet of mass $0.04kg$ moving with a speed of $90m/s$ enters a heavy wooden block and is stopped after a distance of $60cm$ . What is the average force exerted by the block on the bullet?

Answer 1

Given,

$m=0.04kg$

$u=90m/s$

$v=0m/s$

$S=60cm$

From 3rd equation of motion,

$2aS=v^2-u^2$

$a=-\frac{u^2}{2S}$

$a=-\frac{90\times 90}{2\times 60\times {10}^{-2}}=-6750m/s^2$

The resistive force, $f=ma$

$f=-0.04\times 67.5$

$f=-270N$