Average force of resistance:
A gun weighing 500 kN is utilized to fire a shell weighing 10 kN in a horizontal direction with a velocity of 500 metres/sec. With what initial velocity the gun shall recoil? What shall be the average force of resistance to bring it to rest in a distance of 1.25 metres?
Solution
Considering gravitational acceleration = 10 m/sec^{2}.
Mass of shell = 10 × 10^{3} /10
= 1000 kg
Mass of gun = 500 × 10^{3} /10
= 50 × 10^{3} kg or 50 tonne.
By applying the principle of conservation of momentum,
1 × 500 + 50 × V = 0
∴ V = - 10 metres / sec.
Negative sign of the velocity of gun mentions that the direction of velocity of the gun is opposite to that of the shell. This is for this reason, that motion of gun is called as recoil.
Consider F be the average retarding force causing retardation (- a) of the gun. As final velocity of gun is zero, through equation
V ^{2} = u ^{2} + 2 a s
we have
∴ 0 = 10^{2} + 2 a × 1.25
∴ a = - 100 / (1.25 × 2)
= - 40 metres / sec^{2}
∴ Retarding Force, F = m × a
= 50 × 40 = 2000 kN.