# A bullet is fired from a gun

1. Oct 18, 2014

### Emily_20

1. The problem statement, all variables and given/known data
A 0.031 kg bullet is fired from a shotgun of mass 3.63kg at 420 m/s. If the collision of the recoiling gun with your shoulder takes 0.013s, what is the average force the gun exerts against your shoulder?

2. Relevant equations
p = m • v

vf=vi + at
F=ma
3. The attempt at a solution
I calculated the velocity of the gun from the momentum equation : 0+0=mbf*vbf-mgf*gf
which is 3.587 m/s
Then I used the kinematics equation to solve for the acceleration: 3.587 m/s=0 + a (0.013 s) which is 275.9 m/s^2
Then I used the force equation to find the force: (275.9 m/s/s) * (3.63 kg) = 1002 N. Is it right? I am worried about the acceleration because the number is high. Thanks for help.

2. Oct 18, 2014

### NTW

I believe you are right. The problem can be solved taking another way: the variation of momentum mv is equal to the impulse Ft

The variation of momentum of the bullet is 13,02 m*s, and is the same as the variation of momentum of the gun.

Thus, 13,02 = Ft

As it's stated that the time is 0,013 s, if you solve for F, you get F = 13,02/0,013 = 1002 N

3. Oct 18, 2014

### Emily_20

Thanks!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted