A bullet of ' 4 g ' mass is fired from a gun of mass 4 kg . If the bullet moves with the muzzle speed of $50 \mathrm{~ms}^{-1}$, the impulse imparted to the gun velocity of recoil of gun are :

(B) $0.2 \mathrm{~kg} \mathrm{~ms}^{-1}, 0.05 \mathrm{~ms}^{-1}$

(A) $0.4 \mathrm{~kg} \mathrm{~ms}^{-1}, 0.1 \mathrm{~ms}^{-1}$

(C) $0.4 \mathrm{~kg} \mathrm{~ms}^{-1}, 0.05 \mathrm{~ms}^{-1}$

(D) $0.2 \mathrm{~kg} \mathrm{~ms}^{-1}, 0.1 \mathrm{~ms}^{-1}$

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Conservation of Momentum

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 velocity of recoil of gun are :

(A) $0.4 kg ms^{- 1}, 0.1 ms^{- 1}$
(B) $0.2 kg ms^{- 1}, 0.05 ms^{- 1}$ ✓ Correct
(C) $0.4 kg ms^{- 1}, 0.05 ms^{- 1}$
(D) $0.2 kg ms^{- 1}, 0.1 ms^{- 1}$

Answer: (B)

Solution

Impulse on gun = momentum of bullet =

m_{bullet} \times v_{bullet}

I = \frac{4}{1000} \times 50 = 0.2

kg·m/s

By conservation of momentum:

m_{bullet} v_{bullet} = m_{gun} v_{gun}

\frac{4}{1000} \times 50 = 4 \times v_{gun}

v_{gun} = \frac{0.2}{4} = 0.05

m/s

Therefore, impulse = 0.2 kg·m/s and recoil velocity = 0.05 m/s

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option

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