Question

A person fires a 30 g bullet towards the target in a shooting range with a velocity of 30 m/s. The bullet embeds into the target up to a depth of 10 cm before coming to a stop. What will be the magnitude of the force on the bullet?

• A. 165 N
• B. 200 N
• C. 175 N
• D. 135 N

Answer 1

The correct option is D 135 N

Mass, $m=30g=0.03kg$
Initial velocity, $u=30m/s$
Displacement, $s=10cm=0.1m$
The bullet stops in target, hence final velocity is zero.
$v=0m/s$

By applying the 3rd equation of motion, we have:
$v^2=u^2+2as$
$0^2={30}^2+2\times a\times 0.1$
$a=-900/(2\times 0.1)=-4500m/s^2$
Negative acceleration shows that acceleration is acting in the direction opposite to that of the initial velocity.

Now using Newton’s second law,
$F=ma$
$F=0.03kg\times -4500m/s^2$
$F=-135N$
Negative value shows that the force is acting in direction opposite to the initial velocity.