Question

A 3.1-kilogram gun initially at rest is free to

move. When a 0.015-kilogram bullet leaves the
gun with a speed of 500. meters per second, what
is the speed of the gun?
(1) 0.0 m/s (3) 7.5 m/s
(2) 2.4 m/s (4) 500. m/s

Answer 1

Let us consider the gun with the index 1 and bullet with the index 2. Using the Momentum consevation Equation, we have:


`\Delta Q= 0 \\ m_(1)*v_(1)-m_(2)*v_(2)=0 \\ m_(1)*v_(1)=m_(2)*v_(2)`

Entering the unknown, comes:


`m_(1)*v_(1)=m_(2)*v_(2) \\ 3.1*v_(1)=0.015*500 \\ v_(1)= (7.5)/(3.1) \\ \boxed {v_(1)=2.4m/s}`

Number 2

Obs: approximate results.

If you notice any mistake in my english, please let me know, because i am not native.

Answer 2

The conservation of momentum states that the total momentum in a system is constant if there is no external force acting on the system. The total momentum in the gun bullet system is 0 so it must stay that way.

The momentum of the bullet is mv = 0.015*500=7.5
The momentum of the gun must be the same to keep the total momentum of the system equal to zero, so we know that p = 7.5 for the gun.
Substituting this in we get:

7.5=3.1x
x=7.5/3.1
x=2.42

So the speed of the gun is 2.4m/s.