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(C) Given:Mass of bullet,$m = 4 \, g = 4 	imes 10^{-3} \, kg$Mass of gun,$M = 4 \, kg$Velocity of bullet relative to ground,$v_b = 50 \, ms^{-1}$Let

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$0.2 \, kg \, ms^{-1}, 0.05 \, ms^{-1}$

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(C) Given:Mass of bullet,$m = 4 \, g = 4 	imes 10^{-3} \, kg$Mass of gun,$M = 4 \, kg$Velocity of bullet relative to ground,$v_b = 50 \, ms^{-1}$Let $V$ be the recoil velocity of the gun.According to the law of conservation of linear momentum,the initial momentum of the system is zero.$M V + m v_b = 0$$4 	imes V + (4 	imes 10^{-3}) 	imes 50 = 0$$4 V = -0.2$$V = -0.05 \, ms^{-1}$The magnitude of the recoil velocity is $0.05 \, ms^{-1}$.Impulse imparted to the gun is equal to the change in momentum of the gun:$J = |M \Delta V| = |4 	imes (-0.05) - 0| = 0.2 \, kg \, ms^{-1}$.

$A$ 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 the velocity of recoil of the gun are:

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