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`. 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.

Explanation

For the collision,
`u_1=400 m s^-1,e=0.75,v_1,= v1 ?`
For the firmly fixed hard surface, `u_2= v_2=0`
`e=0.75=(v_1-v_2)/(u_2-u_1)=(v_1-0)/(0-400) :. v_1=-300 m//s.`
-ve sign indicates that the bullet rebounds in
exactly opposite direction.
Change in momentum of each bullet `= m (v_1-u_1)`
Equal and opposite will be the momentum transferred to the surface, per collision. ∴ Momentum transferred to the surface,
per collision
`p=m(u_1-v_1)=0.04 (400-[-300])=28 N s`
The rate of collision is same as rate of firing.
∴ Momentum received by the surface per second, `"dp"/"dt" xx 28 xx 5 =140 N`
This must be the average force experienced by the surface of area `A= 10 cm^2=10^-3 m^2`
∴ Average pressure experienced,
`p=F/A=140/10^-3=1.4 xx 10^5 N m^-2`
`~~1.4` times the atmospheric pressure.