Question

# A lead bullet of mass $2kg$, traveling with a velocity of $20m·{s}^{-1}$, comes to rest after penetrating $20m$ in a still target. Find the resistive force applied by the target and acceleration of the bullet.

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Solution

## Step 1: Given dataThe initial velocity of the bullet, $u=20m·{s}^{-1}$.As the bullet comes to rest thus the final velocity of the bullet, $v=0m·{s}^{-1}$.Displacement before the bullet stops, $s=20m$.Step 2: Determine the acceleration of the bulletLet the acceleration of the bullet be $a$.It is known that, ${v}^{2}={u}^{2}+2as$.$⇒a=\frac{{v}^{2}-{u}^{2}}{2s}\phantom{\rule{0ex}{0ex}}⇒a=\frac{{\left(0\right)}^{2}-{\left(20\right)}^{2}}{2·20}\phantom{\rule{0ex}{0ex}}⇒a=-\frac{400}{40}\phantom{\rule{0ex}{0ex}}⇒a=-10$Thus, the acceleration is $-10m·{s}^{-2}$, negative sign implies that the velocity is decreasing.Step 3: Determine the resistive force applied by the targetThe acceleration of the bullet is $a=-10m·{s}^{-2}$.The mass of the bullet is $m=2kg$.Thus, the force applied by the target can be given by, $F=ma$ $⇒F=2×\left(-10\right)N\phantom{\rule{0ex}{0ex}}⇒F=-20N$A negative sign indicates that the force is resistive.Therefore, the resistive force is $20N$.Hence, the resistive force applied by the target is $20N$ and the acceleration of the bullet is $-10m·{s}^{-2}$.

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