A bullet leaves the barrel of a gun with a kinetic energy of 90 J. The barrel of the gun is 50 cm long. The gun has a mass of 4 kg, the bullet 10 g. (a) Find the bullets final velocity. (b) Find the bullets final momentum. (c) Find the momentum of the recoiling gun. (d) Find the kinetic energy of the recoiling gun.
KE = 1/2mv^2
p = mv
The Attempt at a Solution
(a) Since I know the kinetic energy of the bullet as it leaves the barrel, I attempted to find the velocity with this information.
90 J = (1/2)(.01 kg)v^2
which gives v = 134.16 m/s
(b) To find momentum of the bullet I plugged this value for v into p= mv, this gives the result p = 1.3416 kg m/s.
for (c) I used conservation of momentum, since the initial momentum of both the bullet and gun are zero.
0 = (1.3416 kg m/s) + (4 kg)v
solving this give v = -.3354 m/s
(d) KE =(1/2)mv^2 so the kinetic energy of the gun is KE = (1/2)(4 kg)(-.3354 m/s)^2 = .225 J
I'm not entirely sure if I solved this correctly. My main problem is that it gives the length of the barrel of the gun and I did not use this in the solution. Is this unnecessary information? or did I do this incorrectly? I thought that the 50 cm could be used to find the work done on the gun, but 50 cm is how far the bullet moved, not how far the gun moved.