Bullet and Can Collision: Determining Bullet Speed After Impact

[softball1394]

Homework Statement

A 7g bullet moving horizontally at 200 m/s strikes and passes through a 150 g tin can sitting on a post. Just after impact, the can has a horizontal speed of 180 cm/s. What the bullet's speed after leaving the can? ANSWER = 161 m/s

Homework Equations

7g = .007 kg
150 g = .15 kg
180 cm/s = 1.8 m/s

The Attempt at a Solution

momentum = m1 x v1 + m2 x v2
=.007kg (200 m/s) + .15kg (0)
=1.4 kg m/s

conversation of momentum--
1.4 kg m/s = 1.4 kg m/s
1.4 = (m1+m2) v
1.4 = (.007 + .15) v
v= 8.92

obviously very wrong :P

 

[ehild]

Think, what happens with the bullet after it hits the can? The wall of a can is very thin...

ehild

 

[softball1394]

Well it slows down, but only slightly.

But I can't figure out how to show that in a formula...

 

[softball1394]

...anyone there?

 

[ehild]

How many times hits the bullet the wall of the can when it flies through it?

ehild

 

[softball1394]

Twice.

 

[ehild]

So, there are two impacts, and neither of them can be considered inelastic collision as the bullet leaves the can: the bullet and can have different velocities. Neither are they elastic collisions. Some part of the kinetic energy of the bullet is lost while it goes through the wall of the can. I am not sure if it can be assumed that this loss of energy is the same at both impacts or the force exerted by the walls are the same in both cases. Haven't you got any hint from your teacher?

ehild