# Bullets of mass 40 g each, are fired from a machine gun at a rate of 5 per second towards a firmly fixed hard surface of area 10 cm^2.

27 views
in Physics
closed

Bullets of mass 40 g each, are fired from a machine gun at a rate of 5 per second towards a firmly fixed hard surface of area 10 cm2. Each bullet hits normal to the surface at 400 m/s and rebounds in such a way that the coefficient of restitution for the collision between bullet and the surface is 0.75. Calculate average force and average pressure experienced by the surface due to this firing.

+1 vote
by (38.6k points)
selected

For the collision,

u1 = 400 m s-1, e = 0.75

For the firmly fixed hard surface, u2 = v2 = 0

e = 0.75 = $\frac{v_1-v_2}{u_2-u_1}=\frac{v_1-0}{0-400}$

∴ v1 = -300 m/s.

Negative sign indicates that the bullet rebounds in exactly opposite direction. Change in momentum of each bullet = m(v1 – u1)

The same momentum is transferred to the surface per collision in opposite direction.

∴ Momentum transferred to the surface, per collision,

p = m (u1 – v1) = 0.04(400 – [-300]) = 28 Ns

The rate of collision is same as rate of firing.

∴ Momentum received by the surface per second, $\frac{dp}{dt}$ = average force experienced by the surface = 28 × 5 = 140 N

This is the average force experienced by the surface of area A = 10 cm2 = 10-3 m2

∴ Average pressure experienced,

P = $\frac{F}{A}=\frac{140}{10^{-3}}$ = 1.4 × 105 N m-2

∴ P ≈ 1.4 times the atmospheric pressure.

The average force and average pressure experienced by the surface due to the firing is 140 N and 1.4 × 105 N m-2 respectively.