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Force and Laws of Motion
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01/11/2021 7:56 pm
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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.
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01/11/2021 8:05 pm
Mass of the gun m1 = 3 kg
Mass of bullet m2 = 30 g = 0.03 kg
Velocity of bullet v2 =100 m/s
(i) According to the law of conservation of momentum
m1 x v1 = m2 x v2
3 x v1 = 0.03 x 100
Recoil velocity v1 = \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/s2
Force = m x a = 3 x \frac{1000}{3} = 1000 N