Answered

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

Verified Answer :

TaylorBayley

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.

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:

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

Number 2

Obs: approximate results.

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

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