Back to Directory
NEET PHYSICSEasy

A machine gun is mounted on a 2000 kg car on a horizontal frictionless surface. At some instant, the gun fires bullets of mass 10 g with a velocity of 500 m/s with respect to the car. The number of bullets fired per second is ten. The average thrust on the system is:

A

550 N

B

50 N

C

250 N

D

250 dyne

Step-by-Step Solution

  1. Principle: The average thrust (force) exerted by the machine gun is equal to the rate of change of momentum of the bullets fired, in accordance with Newton's Second Law of Motion (F=dpdtF = \frac{dp}{dt}) [Source 65].
  2. Given Data:
  • Mass of each bullet, m=10 g=0.01 kgm = 10 \text{ g} = 0.01 \text{ kg}.
  • Velocity of bullets relative to the car, v=500 m/sv = 500 \text{ m/s}.
  • Number of bullets fired per second, n=10n = 10.
  1. Calculation: The rate of change of momentum is the product of the mass flow rate and velocity. Fthrust=Total masstime×velocity=(n×m)×vF_{thrust} = \frac{\text{Total mass}}{\text{time}} \times \text{velocity} = (n \times m) \times v Fthrust=(10×0.01 kg/s)×500 m/sF_{thrust} = (10 \times 0.01 \text{ kg/s}) \times 500 \text{ m/s} Fthrust=0.1×500F_{thrust} = 0.1 \times 500 Fthrust=50 NF_{thrust} = 50 \text{ N} (Note: The mass of the car is not required to calculate the thrust force).
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started