For the collision,

u_{1} = 400 m s^{-1}, e = 0.75

For the firmly fixed hard surface, u_{2} = v_{2} = 0

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

∴ v_{1} = -300 m/s.

Negative sign indicates that the bullet rebounds in exactly opposite direction. Change in momentum of each bullet = m(v_{1} – u_{1})

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

∴ Momentum transferred to the surface, per collision,

p = m (u_{1} – v_{1}) = 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 cm^{2} = 10^{-3 }m^{2}

∴ Average pressure experienced,

P = \(\frac{F}{A}=\frac{140}{10^{-3}}\) = 1.4 × 10^{5} 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 × 10^{5} N m^{-2} respectively.