A machine gun is mounted on a 2000 kg car on a horizontal frictionless surface. At some instant, the gun fires

Question

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:

Accepted Answer

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 ($F = \frac{dp}{dt}$) [Source 65].
2. **Given Data:**
   - Mass of each bullet, $m = 10 	ext{ g} = 0.01 	ext{ kg}$.
   - Velocity of bullets relative to the car, $v = 500 	ext{ m/s}$.
   - Number of bullets fired per second, $n = 10$.
3. **Calculation:**
   The rate of change of momentum is the product of the mass flow rate and velocity.
   $$ F_{thrust} = \frac{	ext{Total mass}}{	ext{time}} 	imes 	ext{velocity} = (n 	imes m) 	imes v $$
   $$ F_{thrust} = (10 	imes 0.01 	ext{ kg/s}) 	imes 500 	ext{ m/s} $$
   $$ F_{thrust} = 0.1 	imes 500 $$
   $$ F_{thrust} = 50 	ext{ N} $$
   (Note: The mass of the car is not required to calculate the thrust force).

Step-by-Step Solution

Practice All PHYSICS Questions