A projectile weighing 200 kg is fired from a gun with a velocity of 500 m/s. If the velocity of recoil of the gun is 12.5 m/s. Calculate the mass of the gun.

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Solution

Given data:
Mass of the projectile, $m_{p} = 200 k g$
Initial velocity of the gun, $u_{g} = 0 m / s$ (The gun is at rest)
Initial velocity of the projectile, $u_{p} = 0 m / s$ (The gun has not been fired)
Velocity of the projectile after getting fired, $v_{p} = 500 m / s$
Velocity of the gun $v_{g} = - 12.5 m / s$ (Direction of the velocity of the gun is opposite to the projectile).
Concept to be used
We will use the concept of conservation of momentum.
Momentum before firing the gun = Momentum after firing the gun
Formula for momentum is given by: $p = m v$ .
Calculation of the mass of the gun
Initial momentum of a system = Final momentum of a system. (According to the law of conservation of momentum).
Initial momentum of gun + Initial momentum of projectile = Final momentum of gun + Final momentum of the projectile.
$m_{g} u_{g} + m_{p} u_{p} = m_{g} v_{g} + m_{p} v_{p} .$
$m_{g} \times 0 + 200 \times 0 = m_{g} \times (- 12.5) + 200 \times 500 .$
$0 = m_{g} + (- 12.5) + 100000 .$
$12.5 m_{g} = 100000 k g .$
$m_{g} = \frac{100000}{12.5} = 8000 k g .$
Hence, the mass of the gun is $8000 k g$ .

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