Question

A $60g$ bullet fired from a $5kg$ gun leaves with a speed of \[500m/s\] . Find the speed (velocity) with which the gun recoils (jerks backward).

Answer 1

Hint: Newton’s third law states that every action has an equal and opposite reaction. When a bullet is fired it will gain some momentum, to conserve the momentum of the system, the gun will move in the direction opposite to the bullet, this is called the recoil of the gun.

Complete step by step solution:
Let us first write the information given in the question.
Mass of bullet ${m_b} = 60g = 0.06kg$, mass of gun ${m_g} = 5kg$, speed of bullet = \[500m/s\], we have to find the recoiling speed of the gun.
According to the law of conservation of momentum, the initial momentum of a system is equal to the final momentum of the system.
We can write momentum conservation for the bullet-gun system as below.
${m_g}{v_g} = {m_b}{v_b}$
Let us substitute the values in the above formula.
$5 \times {v_g} = 0.06 \times 500$
Let us simplify this.
${v_g} = \dfrac{{30}}{5} = 6m/s$
Hence, the recoiling speed of the gun is \[6m/s\] .

Note:
Conservation of momentum is valid for all physics problems.
For the gun-bullet system, the initial momentum is zero. It means the final momentum should also be zero. For this to happen, the gun has to acquire the same momentum as the bullet in the opposite direction. s
Momentum is a quantity that measures the heaviness of an object.
When the same force is acting on two objects, the lighter object will gain momentum greater than the heavier object. The force required to move the heavy object will be greater than the lighter object.