a gun of mass 3 kg fires a bullet of mass 30g. The bullet takes 0.003s to move the barrel of the gun and acquires a velocity of 100m/s. Calculate the force exerted on the gunman due to recoil of the gun.

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Solution

$The acceleration of the bullet is a = \frac{v - u}{t} = \frac{100 - 0}{0.003} = \frac{10^{5}}{3} {ms}^{- 2} So force is f = m a = 0.03 \times \frac{10^{5}}{3} = 1000 N$

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a gun of mass 3 kg fires a bullet of mass 30g. The bullet takes 0.003s to move the barrel of the gun and acquires a velocity of 100m/s. Calculate the force exerted on the gunman due to recoil of the gun.

The acceleration of the bullet is a=v-ut=100-00.003=1053 ms-2So force is f=ma=0.03×1053=1000 N

$The acceleration of the bullet is a = \frac{v - u}{t} = \frac{100 - 0}{0.003} = \frac{10^{5}}{3} {ms}^{- 2} So force is f = m a = 0.03 \times \frac{10^{5}}{3} = 1000 N$