# Bullet fired from a gun

1. Oct 2, 2007

### kateman

1. The problem statement, all variables and given/known data
A 500g pistol lies at rest on an essentially frictionless table. It accidentally discharges and shoots a 10g bullet parallel to the table. How far has the bullet moved by the time the gun has recoiled 2.00mm?

2. Relevant equations
0=m1v1+m2v2
m1v1= (m1+m2)Vf

3. The attempt at a solution
well i tried the 0=m1v1+m2v2 formula and tried to substitute the value of v as s/t and then eliminating time, i rearranged it for the displacement value of the bullet and got a weird value of -.1 when the answer is 9.8cm
i know now that this is the wrong way to approach it when i consider the conservation of momentum but i have no other ideas on how to do this or how to start it!

if anyone does it would be most appreciated!

2. Oct 2, 2007

### Staff: Mentor

There's nothing at all wrong with your approach, or even your answer, depending upon the units you are using. Realize that if the gun goes moves one way (+ say), then the bullet will move in the opposite direction (call it -).

0.1 m = 10 cm. If you subtract the 2 mm (0.2 cm) that the gun moved you'll get 9.8 cm. (But I see no reason why you would subtract the gun's movement from the bullet's, so I disagree with that answer.)

3. Oct 2, 2007

### Astronuc

Staff Emeritus
Well, the force pushing the bullet is pushing back on the gun.

So F=ma

then both bullet and gun are starting at rest, and force applied over distance (F*d) is energy, and the energy is kinetic.

Assume constant acceleration. The lighter mass accelerates faster.

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html

4. Oct 2, 2007

### kateman

ahh yeah converting back to cm might help :P

hmm thats very interesting (what you found), but i agree, it does seem to be just a random coincidence

yup, its all good now, thanks to both of you

Last edited: Oct 2, 2007
5. Oct 2, 2007

### Staff: Mentor

Also realize that the (perfectly valid) approach that you took is equivalent to noting that the center of mass of "bullet + gun" does not change. If the gun moves 2mm to the left, then the bullet must move 10 cm to the right to maintain the center of mass.

6. Oct 2, 2007

### kateman

yeah i do now thanks