A bullet of 4 g mass is fired from a gun of mass 4 kg. If the bullet moves with the muzzle speed of 50 ms$$^1$$, the impulse imparted to the gun and velocity of recoil of gun are

We apply the law of conservation of momentum. Initially, both the bullet and gun are at rest, so the total momentum is zero.

The bullet has mass $$m = 4 \text{ g} = 0.004 \text{ kg}$$ and muzzle speed $$v = 50 \text{ ms}^{-1}$$.

The impulse imparted to the gun equals the change in momentum of the bullet (Newton's third law). The magnitude of impulse is: $$J = mv = 0.004 \times 50 = 0.2 \text{ kg m s}^{-1}$$

By conservation of momentum, the gun (mass $$M = 4 \text{ kg}$$) recoils with velocity $$V$$ such that: $$MV = mv$$ $$V = \frac{mv}{M} = \frac{0.004 \times 50}{4} = \frac{0.2}{4} = 0.05 \text{ m s}^{-1}$$

Therefore, the impulse imparted to the gun is $$0.2 \text{ kg m s}^{-1}$$ and the velocity of recoil is $$0.05 \text{ m s}^{-1}$$.