What Is the Average Force Exerted by a Recoiling Shotgun?

[Emily_20]

Homework Statement

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?

Homework Equations

p = m • v[/B]
vf=vi + at
F=ma

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.

 

[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

 

[Emily_20]

Thanks!