A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 s to move through the barrel of the gun and acquires a velocity of 100 m/s. Calculate :
A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 s to move through the barrel of the gun and acquires a velocity of 100 m/s. Calculate :
(i) the velocity with which the gun recoils.
(ii) the force exerted on gunman due to recoil of the gun.
Mass of the gun m_{1} = 3 kg
Mass of bullet m_{2} = 30 g = 0.03 kg
Velocity of bullet v_{2} =100 m/s
(i) According to the law of conservation of momentum
m_{1} x v_{1} = m_{2} x v_{2}
3 x v_{1} = 0.03 x 100
Recoil velocity v_{1 }= \(\frac{100 \times 0.03}{3}\) = 1 m/s
(ii) Initial velocity of the gun u = 0 m/s
Final velocity of the gun v = 1 m/s
Time t = 0.003 s
Acceleration a = \(\frac{v  u}{t}\) = \(\frac{1  0}{0.003}\)
= \(\frac{1000}{3}\) m/s^{2}
Force = m x a = 3 x \(\frac{1000}{3}\) = 1000 N

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