Question

A lead bullet of mass 20g, travelling with a velocity of 350 ms^-1, comes to rest after penetrating 40 cm in a still target. Find:

1) the resistive force offered by the target and 

2) the retardation caused by it

Answer 1

Mass of bullet = 20g = 0.020 kg

Intial velocity, u = 350 m/s

Final velocity, v = 0 m/s

Distance, s = 40 cm = 0.4 m

From equation `v^2 - u^2 = 2as`

`0^2 - (350)^2 = 2a xx 0.4`

Acceleration `a = - (350 xx 350)/(2 xx 0.4) = -1.53 xx 10^5 ms^(-2)`

1) Resistive force, F = mass x acceleration

`F = 0.020 xx 153 xx 10^5`

= 3062.5 N

2) Retardation = -acceleration = `1.53 xx 10^5 ms^(-2)`