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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 :

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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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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

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