Question

A bullet of mass 10 g traveling horizontally with a velocity of 150 m s–1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer 1

Mass of bullet, m = 10g = $\frac{10}{1000}$kg = 0.001kg
Initial velocity, u = 150 ms-1
Final velocity, v = 0 (since the bullet finally comes to rest)
Time, t = 0.03s
From the equation of motion, v = u + at
0 = 150 + a × 0.03
a =$-\frac{150}{0.03}$ = - 5000 ms-2
The magnitude of the force applied by the bullet on the block, F = ma = 0.01 × - 5000 = 50N