Question

A lead bullet of mass $2kg$, travelling with a velocity of $20m{s}^{-1}$, comes to rest after penetrating $20m$ in a still target. Find the average acceleration $\left(a\right)$ of the bullet and resistive force $\left(F\right)$ applied by the target.

• A. $a=20m{s}^{-2}$​​
  $F=40N$
• B. $a=40m{s}^{-2}$
  $F=80N$
• C. $a=-10m{s}^{-2}$
  $F=20N$
• D. $a=0m{s}^{-2}$
  $F=0N$

Answer 1

The correct option is C $a=-10m{s}^{-2}$

$F=20N$
Given:
Initial velocity, $u=20m{s}^{-1}$
Final velocity, $v=0$
Distance travelled, $s=20m$
Mass, $m=2kg$

From third equation of motion, we have
$v^2=u^2+2as$​
$0^2={20}^2+(2\times a\times 20)$
$\Rightarrow a=-10m{s}^{-2}$

Using Newton's second law of motion,
$F=ma=2\times (-10)$
$F=-20N$
So, resistive force is $20N$.