Question

A gunman with 3kg gun fires a bullet of 30gm .It takes .003 second to come out and acqires a velocity of 60m/s .Calculate the recoil velocity of gun .

Answer 1

Heya !!

Here,

• Mass of bullet, m₁ = 30 g
• Final Velocity of bullet, v₁ = 60 ms⁻¹
• Mass of gun, m₂ = 3 kg = 3000 g
• Recoil Velocity of Gun, v₂ = ?
• Initial Velocity of bullet, u₁ = Initial Velocity Of gun, u₂ = 0 ms⁻¹

According to the Law of Conservation of Linear Momentum,

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} \\ \\</p> <p>\implies (30)(0) + (3000)(0) = (30)(60) + (3000)(v_{2}) \\ \\</p> <p>\implies 0 + 0 = 1800 + (3000)(v_{2}) \\ \\</p> <p>\implies 0 = 1800 + (3000)(v_{2}) \\ \\</p> <p>\implies (3000)(v_{2}) = -1800 \\ \\</p> <p>\implies v_{2} = - \dfrac{1800}{3000} \\ \\</p> <p>\implies \boxed{v_{2} = 0.6 ms^{-1}}$

Therefore, recoil velocity of the gun will be 0.6 ms⁻¹ .

Hope it helps !!