Question

A car with a gun mounted on it is kept on a horizontal frictionless surface. Total mass of the car, gun and shell is 50 kg. Mass of each shell is 1 kg. If the shell is fired horizontally with relative velocity 100 m/sec with respect to the gun. What is the recoil speed of the car after a second shot at the nearest integer?

Answer 1

Hint : In this solution, we will use the law of conservation of momentum that tells us that in absence of external force, the net momentum of will be zero. When the shell is fired from the gun mounted on the car, the car will be pushed backwards.

Formula used: In this solution, we will use the following formula
 $ P = mv $ where $ P $ is the momentum of an object of mass $ m $ travelling with velocity $ v $ .

Complete step by step answer:
We’ve been told that a car with a gun is firing shells with relative velocity $ 100\,m/s $ is moving with respect to the car and we want to find the recoil velocity of the car.
When the shell is fired from the gun, it will gain momentum in the forward direction. Now, since the net external force on the system of the car, the gun and the shell are zero, according to the law of conservation of momentum the net momentum of the system must be zero.
The recoil of the shell is experienced by the net mass of the car and the gun. Let us denote the net mass of the car and the gun system as $ {m_{gc}} $ and its momentum as $ {P_{gc}} $ .
Since the net momentum of the system must be zero, the momentum of the gun-car system must be equal to the momentum of the bullet. So after the first shot
 $ {P_{gc}} = {P_{{\text{bullet}}}} $
So,
 $ {m_{gc}}{v_{gc}} = {m_{{\text{bullet}}}}{v_{{\text{bullet}}}} $
Substituting the values of $ {m_{{\text{bullet}}}} = 1\,kg $ , $ {m_{gc}} = 50 - 1 = 49kg $ , and $ {v_{{\text{bullet}}}} = 1\,00m/s $ , we get
 $ 49{v_{gc}} = 1 \times 100 $
Which gives us
 $ {v_{gc}} = 2.04\,m/s $
Similarly, when the second shell is fired, the mass of the car gun system will be $ 48\,kg $ so the recoil of the car after the second shot will be:
 $ 48{v_{gc}} = 1 \times 100 $
 $ {v_{gc}} = 2.08m/s $
The net velocity of the car ginsystem after the second shot will actually be the sum of the recoil velocity of the car after the first and the second shot which is $ 2.04 + 2.08 = 4.12\,m/s $ .
The closest integer velocity of the car will be $ 4\,m/s $ .

Note:
Here we have assumed that the car does not slow down between shots. So, it will maintain its velocity without slowing down due to friction. After the first shell is fired, the car will still be moving with a constant velocity and not accelerating so the net force on the system will be zero.