# Impact problem

1. Nov 14, 2006

### Alec

Hi,
two wagons with elastic bumpers colide without any friction. Wagon A has the mass 0.1 kg while wagon B has 0.4 kg. After the collision wagon A has the speed 0.5 meters / second and wagon B lies still. Taken in consideration that the collision was elastic, what velocity does each wagon have before the collision?

I've came to the conclusion that Wk before = Wk after since it's elastic.
And that you can calculate the Wk after since you have all the information, yet this leaves me with two unknown constants, V1 and V2.

2. Nov 14, 2006

### OlderDan

What else besides the energy has to be conserved?

3. Nov 14, 2006

### Alec

The momentum: P before = P after.
But still, I have V1 and V2 unknown.

4. Nov 14, 2006

### physics girl phd

You have only two unknowns, right? (V1 and V2).

you have decided on two equations:

Conservation of energy (NOT work!)
Conservation of momentum.

With two equations and two unknowns, you can solve the set of equations

5. Nov 14, 2006

### Alec

Ok, how?
I seriously have no idea.
Could it be that Wk = p^2/2m ?

6. Nov 14, 2006

### physics girl phd

ENERGY is p^2/2m (I'm a bit disturbed by your use of the variable Wk -- is that how your book denotes it? I'd say the kinetic energy KE or E_k is p^2/2m).

Calculate the total energy of the two objects before the collision.
Write an expression for the total energy after the collision (using variables for the unknowns).
Set these two equal to each other.
THAT means that energy is conserved -- since energy before the collision equals the energy after the collision -- which is the case in an elastic collision (no energy is used in heating or deforming the objects.

Do the same for the momentum.

Then you have two equations to solve for the unknowns. (look up how to solve a set of equations).

7. Nov 14, 2006

### Alec

Sorry, yeah my book says Energy is Wk, although Ek is much more logical in English (I'm european if that explains it).

8. Nov 14, 2006

### Alec

Well I've tried figuring it out but I have no idea how to solve for two unknown s. Physics is really not my cup of tea.

9. Nov 14, 2006

### Hootenanny

Staff Emeritus
Perhaps you could show us what you have written?

10. Nov 14, 2006

### Alec

yes, sorry.
I wrote:
Ek = (0.1^2 * V(1)^2) / 0.2 + (0.4^2 * V(2)^2 / 0.8) = 0.0125.
Momentum = V1 * 0.1 + V2 * 0.4 = 0.0125.

This was before the collision.
After the collision it's the same but V(2) = 0 and V(1) = 0.5. So:
Ek = (0.1^2 * 0.5^2) / 0.2 + (0.4^2 * 0^2 / 0.8) = 0.0125.
Momentum = 0.5 * 0.1 + 0 * 0.4 = 0.0125.

11. Nov 15, 2006

### OlderDan

See the annotations in the quote