# Recoil Velocity

1. Dec 6, 2008

### Woopy

1. The problem statement, all variables and given/known data
http://hypertextbook.com/physics/mechanics/momentum-energy/problems.shtml [Broken]

It's numerical #1, filling out the table, the part I am stuck on is recoil velocity. I have figured out the bullet momentum and bullet energy, however.

2. Relevant equations
(m1+m2)v=m1v1'+m2v2'

3. The attempt at a solution
The part that I don't understand is what is m2? M1 would be the mass of the bullet, but is the mass of the gun suppose to be m2?

Last edited by a moderator: May 3, 2017
2. Dec 6, 2008

Most likely.

3. Dec 6, 2008

### Woopy

Theres 3 different velocities, and I don't see how there can be. V is recoil velocity what im solving for, v1 would be the muzzle velocity, and v2 as the velocity of the gun (0 m/s?)

4. Dec 6, 2008

### turin

You have your velocity assignments backwards. Treat this like an "inverse inelastic collision". That means that you begin with the two objects combined (like you have in your equation). So, if v is the velocity of the combined object (bullet inside gun before it is fired), then what do you think should be v? Hint: it isn't recoil.

5. Dec 6, 2008

### Woopy

(.0097kg + 4.4kg)(0 m/s) = (.0097kg)(890m/s)+ (4.4kg)(v2') ???

-4.4kg(v2') = 8.633 kgm/s
v2' = -1.96 m/s (since its a recoil, it'd go backwards so negative.

6. Dec 6, 2008

### LowlyPion

Your table has three different bullets fired from 3 different guns with different barrel lengths and total masses.

Filling out the table is really observing Newton's Third Law of Action/Reaction for each bullet/gun combination.

Whatever momentum the bullet leaves the gun with will be equal to the momentum of the gun in the opposite direction.

MVbullet = - MVgun

7. Dec 6, 2008

### turin

I didn't check your numerical calculation this time, but your approach looks good.