# Conservation of Momentum/Energy related problem

1. Oct 17, 2008

### maniacp08

A bullet of mass m is fired vertically from below into a thin horizontal sheet of plywood of mass M that is initially at rest, supported by a thin sheet of paper. The bullet punches through the plywood, which rises to a height H, above the paper before falling back down. The bullet continues rising to a height h, above the paper.

a)Express the upward velocity of the bullet and the plywood immediately after the bullet exits the plywood in terms of h and H.
b)What is the speed of the bullet?
c)What is the mechanical energy of the system before and after the inelastic collision?

Relevant equations:
Momentum = M * V
Conservation of energy = Ui + Ki = Uf + Kf

What should I do first?
The bullet makes an inelastic collision first with the plywood so the KE is not conserved and it continues to go past the plywood.

For part A:
Im not too sure on how to express the upward velocity in terms of h and H, actually Im just confused at the question.

For Part B:
Im stuck here too...

For Part C:
Do I do Ui + Ki = Uf + Kf where initial is before the collision and final is after?

2. Oct 17, 2008

### Staff: Mentor

That's true. But what is conserved?

If you throw a ball straight up with speed v, how high does it rise? Same problem, only you are given the height and need to find the speed v.

What's conserved?

Energy is not conserved. But you can calculate the KE since you'll have the speeds.

3. Oct 17, 2008

### maniacp08

Momentum is always conserved.

so mass of bullet * V(unknown) is the momentum.

So I should use the conservation of energy equation...
Change in KE + change in PE = 0
Ki would be 1/2 mv^2
kf would be 0
Ui would be 0
Uf would be mgh or mgH?

Im not sure if this is the right approach...please guide.

4. Oct 17, 2008

### Staff: Mentor

You'll need that for part B.

You could certainly use this approach (or kinematics) to solve for the speeds in part A.

5. Oct 17, 2008

### maniacp08

Shouldn't the momentum of the bullet be
(m+M) * V?

Change in KE + change in PE = 0
So I have -Ki + Uf = 0
-1/2(m+M)V^2 + _____ is the Uf when the bullet breaks free from the plywood or Uf is before the bullet breaks free from the plywood while still stuck together? (m+M)gH or mgh?
Is this answer to Part A?

Then I solve for V.
and plug this in to the momentum equation?

Since I have V, so I calculate the KE initial and KE final and the difference is dissipated from the collision?

6. Oct 17, 2008

### Staff: Mentor

The momentum of anything is mass*velocity.

Part A has to do with speeds immediately after the collision. Treat bullet and board separately. There's no reason to use "m+M". (This is a partially inelastic collision--the masses don't stick together.)

Once you've found the speeds of bullet and board after the collision, use momentum conservation to find the speed of the bullet before the collision.

Yes. Once you have all the speeds, just calculate the KE before and after.

7. Oct 17, 2008

### maniacp08

So I should do the conservation of energy on bullet and board separately.

For the bullet:
Ui = 0
Uf = mgh
Ki = 1/2 mVi^2
Kf = 0 or should be 1/2 m Vf^2?

For plywood
Ui = 0
Uf = MgH
Ki = 0
Kf = 1/2M Vf^2

Solve for Vf for the plywood?

8. Oct 18, 2008

### Staff: Mentor

Yes, in order to solve for the speeds of bullet and board after the collision. When doing that calculation, the final speed of each is zero.

You are finding the speed of the bullet (after the collision). The final speed, when it's risen to height h, is zero.

No. Again you are solving for the speed of the plywood immediately after the collision, so the final speed is zero.

Note that here you are not really analyzing the collision at all. (That comes later.) You are analyzing the speeds after the collision. Using the height they each rise to, you can calculate the speeds they must have had right after the collision.

9. Oct 18, 2008

### maniacp08

So for the bullet after the collision is
Ui = 0 and Kf = 0
-1/2 m * Vi^2 = mgh
Vi = root of 2gh

The plywood after the collision is
Ui = 0 and Kf = 0
-1/2 M * Vi^2 = MgH
Vi = root of 2gH

So now I should use conservation of momentum to find the speed before the collision?
p = mass * velocity

momentum for the bullet is
p = m * root of 2gh?

momentum for the plywood is
p = m * root of 2gH?

For Part C:
1/2 m * Vi(bullet)^2 + 1/2 M * Vi(plywood)^2 = KE initial
Using the V from part A?
1/2 m * Vf(bullet)^2 + 1/2 M * Vf(plywood)^2 = KE final
Using the V from Part B?

10. Oct 18, 2008

### Staff: Mentor

Good.

Set the initial momentum of the bullet equal to the final momentum of bullet and board. Use that to solve for the speed of the bullet before the collision.

The speeds from part A will allow you to calculate the KE after the collision; The speed from part B will allow you to calculate the KE before the collision.

11. Oct 18, 2008

### maniacp08

m*Vi(bullet) = (m+M)*Vf
plug in Vi as root of 2gh
and solve for Vf?

This Vf will go into the Vf for bullet and plywood for Part C
1/2 m * Vf(bullet)^2 + 1/2 M * Vf(plywood)^2 = KE final
correct?

If this is correct, Thanks again Doc for helping! =]

12. Oct 18, 2008

### Staff: Mentor

No. As I said earlier, the bullet and board do not stick together so there's no reason to use "m+M". Bullet and board have different speeds after the collision--you found those speeds in part A.
When analyzing the collision (in part B) you are solving for the initial speed (just before the collision). You already have the final speeds (just after the collision)--they were calculated in part A.

13. Oct 18, 2008

### maniacp08

oooh, and Doh! I feel dumb.
m*Vi(bullet) = m*root of 2gh + M*root of 2gH
Solve for Vi(bullet)
That's for Part B

And For part C is easy since I know the the speed of bullet before and after collision.
1/2 m * Vi(bullet)^2 + 1/2 M * Vi(plywood)^2 = KE initial
Vi(plywood) = 0
Vi(bullet) from part B

1/2 m * Vf(bullet)^2 + 1/2 M * Vf(plywood)^2 = KE final
Vf(bullet) = root of 2gh and Vf(plywood) = root of 2gH