Question

A bullet of mass 50 g moving with a speed of 100 m s^–1 enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).

Answer 1

Given:

Mass of the bullet (m) = 50 g

Convert g into kg

1000 g = 1 kg

So, 50 g = `50/1000 = 0.05` kg

Initial velocity (u) = 100 ms⁻¹

Final velocity (v) = 0

Distance of penetration (s) = 50 cm

Convert cm into m

100 cm = 1 m

So, 50 cm = `50/100 = 0.5` m

According to the third equation of motion,

2as = v² − u²

or

a = `(v^2 - u^2)/(2s)`

Substituting, we get,

`a = ((0)^2 - (100)^2)/(2 xx 0.5) = (-10000)/1 = -10000` ms⁻²

Now, since force = mass × acceleration

When we substitute, we obtain

F = 0.05 × (−10000) = −500 N

The stopping force acts in the opposite direction of the bullet’s speed, as shown by the negative sign.

As a result, 500 N of stopping force is applied to the bullet.