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

JEE Mains - Recoil Velocity of a Machine Gun

Question: A machine gun of mass \(10\,\text{kg}\) fires \(20\,\text{g}\) bullets at the rate of \(180\) bullets per minute with a speed of \(100\,\text{m/s}\) each. Determine the recoil velocity of the gun.
[30-Jan-2023 Shift 2]

Select the correct answer:

(A) \(0.06\,\text{m/s}\)

(B) \(0.6\,\text{m/s}\)

(C) \(1.2\,\text{m/s}\)

(D) \(6\,\text{m/s}\)

Solution

Step 1: Convert the bullet mass into kilograms:
\(20\,\text{g} = 0.02\,\text{kg}\).

Step 2: Determine the momentum of one bullet:
\[ p_{\text{bullet}} = m_{\text{bullet}} \times v_{\text{bullet}} = 0.02 \times 100 = 2\,\text{kg·m/s}. \]

Step 3: Calculate the rate of fire in bullets per second:
\[ \text{Rate} = \frac{180\,\text{bullets}}{60\,\text{s}} = 3\,\text{bullets/s}. \]

Step 4: Determine the total momentum imparted per second (impulse per second):
\[ \Delta p/\Delta t = 3 \times 2 = 6\,\text{kg·m/s}^2. \]

Step 5: Apply conservation of momentum (or impulse-momentum theorem) to the machine gun of mass \(10\,\text{kg}\):
\[ v_{\text{recoil}} = \frac{\text{Total momentum per second}}{\text{Mass of the gun}} = \frac{6}{10} = 0.6\,\text{m/s}. \]

Hence, the recoil velocity of the machine gun is \(0.6\,\text{m/s}\).