Question

A bullet of mass $40g$ moving with a speed of $90m{s}^{-1}$ enters a heavy wooden block and is stopped after a distance of $60cm$. The average resistive force exerted by the block on the bullet is

• A. $180N$
• B. $220N$
• C. $270N$
• D. $320N$

Answer 1

Answer: (C)

Here, $u=90m{s}^{-1},$ $v=0$
$m=40g$ $=\frac{40}{1000}kg$ $=0.04kg,$ $s=60cm$ $=0.6m$
using $v^2$ $-u^2$ =$2as$
$\therefore$ ${\left(0\right)}^2$ $-{\left(90\right)}^2$ $=2a\times 0.6$
$a=-\frac{{\left(90\right)}^2}{2\times 0.6}$ $=-6750m{s}^{-2}$
$-ve$ sign shows the retardation.
$\therefore$ The average resistive force exerted by block on the bullet is
$F=m\times a$ $=\left(0.04kg\right)$ $\left(6750m{s}^{-2}\right)$ $=270N$