A bullet of mass 10 g travelling 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.

Solution

Given, mass of the bullet (m) = 10g (or 0.01 kg)

Initial velocity of the bullet (u) = 150 m/s

Terminal velocity of the bullet (v) = 0 m/s

Time period (t) = 0.03 s

To find the distance of penetration, the acceleration of the bullet must be calculated

Let the distance of penetration be s

As per the first law of motion

v = u + at

0 = 150 + a (0.03)

a = -5000 ms^-2

v² = u² + 2as

0 = 150² + 2 x (-5000)s

s = 2.25 m

As per the second law of motion, F = ma

F = 0.01kg × (-5000 ms^-2)

F = -50 N