Question

A bullet of mass 10 g travelling at a speed of 300 m/s penetrates deeply into a fixed target and comes to rest after 0.1 s. Calculate the magnitude of the average force exerted on the bullet.

• A. 3 N
• B. 30 N
• C. 300 N
• D. 30,000 N

Answer 1

The correct option is B 30 N

Here, Initial velocity of the bullet, u = 300 m/s
Final velocity, v = 0 m/s Time taken to stop, t = 0.1 s

From the first equation of motion,
v = u + at
a = $\frac{v-u}{t}$
⇒ a = $\frac{0-300}{0.1}$
⇒ a = $-3000m/s^2$
From Newton’s second law, F = ma
Here, m = 10 g = 0.01 kg
⇒ F = 0.01 ✕ (-3000) = -30 N
Here, the negative sign implies that the force on the bullet is acting in the opposite direction of the initial velocity.
The magnitude of force is 30 N.