A machine gun of mass $$10$$ kg fires $$20$$ g bullets at the rate of $$180$$ bullets per minute with a speed of $$100$$ m s$$^{-1}$$ each. The recoil velocity of the gun is:

Solution & Explanation

1. Identify the Given System Parameters

We are dealing with a recoil problem based on the Law of Conservation of Linear Momentum. Let us write down the given values from the system:

• Mass of the machine gun ($$M$$):

  $$M = 10 \,\, \text{kg}$$

• Mass of each bullet ($$m$$): Converting grams to kilograms:

  $$m = 20 \,\, \text{g} = \frac{20}{1000} \,\, \text{kg} = 0.02 \,\, \text{kg}$$

• Muzzle velocity of each bullet ($$v$$):

  $$v = 100 \,\, \text{m s}^{-1}$$

• Firing rate ($$n$$): Number of bullets fired per second:

  $$n = \frac{180 \,\, \text{bullets}}{60 \,\, \text{seconds}} = 3 \,\, \text{bullets/s}$$

2. Apply the Principle of Conservation of Linear Momentum

Initially, both the machine gun and the bullets are at rest, making the total initial momentum of the system equal to zero ($$P_{\text{initial}} = 0$$).

According to Newton's Third Law, the forward momentum imparted to the bullets every second must be perfectly balanced by the backward momentum gained by the gun every second to keep the total net momentum conserved ($$P_{\text{final}} = 0$$):

$$\text{Momentum gained by the gun per second} = \text{Momentum of bullets fired per second}$$

$$M \cdot V_{\text{recoil}} = n \cdot (m \cdot v)$$

Where $$V_{\text{recoil}}$$ is the backward recoil velocity of the machine gun.

3. Substitute and Solve for Recoil Velocity ($$V_{\text{recoil}}$$)

Isolating the recoil velocity variable from the momentum balance equation:

$$V_{\text{recoil}} = \frac{n \cdot m \cdot v}{M}$$

Substituting our known structural values into the expression:

$$V_{\text{recoil}} = \frac{3 \cdot 0.02 \cdot 100}{10}$$

$$V_{\text{recoil}} = \frac{3 \cdot 2}{10}$$

$$V_{\text{recoil}} = \frac{6}{10} = 0.6 \,\, \text{m s}^{-1}$$

Correct Option Key: Option D (0.6 m s-1)