While firing a bullet of mass 8 g, the recoil velocity of the gun of mass 5 kg becomes 64 cm`cdot s^(-1)`. The bullet penetrates 50 cm through a target and then stops. Express the average resistance on the bullet in newton.

Explanation

Initial momenta of the gun and the bullet were zero, as both were at rest. Let the masses of the gun and the bullet be M and m respectively and the respective velocities after firing be V and v.
From the law of conservation of linear momentum,
0 = MV + mv
`therefore " " v = - (M)/(m)V = (5)/(8 xx 10^(-3))xx 0.64 = - 400 m cdot s^(-1)`
The bullet comes to rest after penetrating a distance of 50 cm or 0.5 m. If the retardation is a due to the resistance of the material of the target,
0 = `(400)^(2) - 2a cdot 0.5 " " or, a = 16 xx 10^(4) " "m cdot s^(-2)`
Hence, average resistance,
F = ma = 0.008 `xx 16 xx 10^(4) = 1280`N.