Question

A bullet of mass 50 g is moving with a velocity of 500 m s⁻¹. It penetrated 10 cm into a still target and comes to rest. Calculate:

1. the kinetic energy possessed by the bullet, and
2. the average retarding force offered by the target.

Answer 1

Given: Mass of bullet (m) = 50 g = 0.05 kg

Initial velocity (u) = 500 m/s

Final velocity (v) = 0 (comes to rest)

Penetration depth (s) = 10 cm = 0.1 m

To find: K.E. = ?

F = ?

(a) Mass of bullet = 50g = `50/1000 = 1/20 kg`

Velocity = 500 m/s

∴ K.E. of bullet = `1/2 mv^2`

= `1/2 xx 0.05 xx 500 xx 500`

K.E. = 6250 J

(b) u = 500 m/s,  v = 0 as bullet comes to rest
distance covered = 10 cm = `10/100 = 1/10 m`

v² − u² = 2aS

0 − 500 × 500 = `2a × 1/10`

∴ a = −1250000 m/s²

∴ Retardation = 1250000 m/s²

But F = m × a

∴ Retarding force offered by target = `1/20 xx 1250000`

F = 62500 N