A rifle of mass 3kg fires a bullet of mass 0.03kg. The bullet leaves the barrel of the rifle at a velocity of 100m/s. If the bullet takes 0.003s to move through its barrel, calculate the force experienced by the rifle due to its recoil.

A rifle of mass 3kg fires a bullet of mass 0.03kg. The bullet leaves the barrel of the rifle at a velocity of 100m/s. If the bullet takes 0.003s to move through its barrel, calculate the force experienced by the rifle due to its recoil.  

Solution:

Mass of the rifle, m₂ = 3 kg

Mass of the Bullet, m₁ = 0.03 kg

The velocity of the bullet, v₁ = 100 m/s

Time taken by the bullet to move through the barrel, t = 0.003 s

So,

Initial velocity of rifle, u₂ = 0 m/s.

Initial velocity of bullet, u₁ = 0 m/s.

Final velocity of bullet, v₁ = 100 m/s.

Final velocity of rifle = v₂ m/s.

Now, we know from the law of conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

So,

(0.03 x 0) + (3 x 0) = (0.03 x 100) + (3 x v₂)

0 = 3 + 3.v₂

v₂ = -1 m/s

Therefore, acceleration of the rifle is given by,

a = (v - u)/t = -1 - 0/0.003 = -333.33 m/s^2

So,

The Force experienced by rifle, therefore, would be given by

F = m.a

F = 3 x (-333.33)

F ≈ 1000 Newtons

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Updated on: 10-Oct-2022

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