Examkin

Question

Difficulty: MediumNewton's Laws of Motion and Linear Momentum

A machine gun fires bullets, each of mass 0.020 kg0.020\text{ kg}, at a speed of 400 m s−1400\text{ m s}^{-1}. If the gun fires 5 bullets per second5\text{ bullets per second}, what is the magnitude of the average recoil force exerted on the gun in newtons?

Answer: 40 N

Answer

The magnitude of the average recoil force exerted on the gun is 40 N40\text{ N}.
According to Newton's second law of motion, force is equal to the rate of change of momentum (F=ΔpΔtF = \frac{\Delta p}{\Delta t}). The total mass of ammunition leaving the gun per second is 5×0.020 kg=0.10 kg s−15 \times 0.020\text{ kg} = 0.10\text{ kg s}^{-1}. Multiplying this mass rate by the velocity (400 m s−1400\text{ m s}^{-1}) gives a rate of momentum change of 40 N40\text{ N}, which corresponds directly to the average recoil force.

Step-by-Step Solution

1
Calculate the mass of bullets fired per unit time (mass flow rate).
ΔmΔt=5 bullets/s×0.020 kg/bullet=0.10 kg s−1\frac{\Delta m}{\Delta t} = 5 \text{ bullets/s} \times 0.020 \text{ kg/bullet} = 0.10 \text{ kg s}^{-1}.
Force is defined as the rate of change of linear momentum, which requires knowing the total mass delivered per second.
2
Apply Newton's second law in terms of momentum change per second.
F=ΔpΔt=(ΔmΔt)v=0.10 kg s−1×400 m s−1=40 NF = \frac{\Delta p}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) v = 0.10 \text{ kg s}^{-1} \times 400 \text{ m s}^{-1} = 40 \text{ N}.
The rate of momentum change of the bullets equals the magnitude of the force exerted on them, which by Newton's third law equals the recoil force on the gun.

Key Concept

Newton's Second Law of Motion and Rate of Change of Linear Momentum
Rate this question

Topics

Newton's Laws of Motion and Linear Momentum

More questions from this topic