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.