# Inelastic collision

1. May 13, 2007

### Aikenfan

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 13, 2007

### Dick

Conservation of momentum?

3. May 13, 2007

### Aikenfan

I saw these formulas but i dont exactly know what to use
V2 = mA * V1 ⁄ (mA + mB)
mAVA1 + mB1 = (mA + mB)V2

4. May 13, 2007

### Dick

Your best bet is not to memorize derived formulas. The momentum before the collision is mA*vA+mB*vB. The momentum after is (mA+mB)*vC where vC is the velocity of the wreckage. And they are equal. Remember that velocity has a sign depending on the direction of travel.

5. May 13, 2007

### Aikenfan

That makes a lot more sense...when i googled inelastic collision, everything seemed really complicated

6. May 13, 2007

### Aikenfan

I got 35.2 m/s
that seems right
where did you get the formula...it was so much simpler that what i tried to do before

7. May 13, 2007

### Dick

I didn't use a formula. I just remembered that momentum is equal to mass*velocity - and that the sum of the momenta before the collision is equal to the sum of the momenta after. Unfortunately, your answer is not correct. Remember that the collision is head-on. One of the velocities needs to be negative. They aren't both headed in the same direction!

8. May 14, 2007

### Aikenfan

I am going to post the questions and all of my work...these answers
do not have to be exact, we can make assumptions and estimations about anything in the problem

9. May 14, 2007

### Aikenfan

10. May 14, 2007

### hage567

You lost me at the very last step you did.

This line looks OK: 14030943.33 = (322591.16 + 104930.7)Vc

But I don't understand what you did after that. I don't get the same answer as you did.

11. May 14, 2007

### Aikenfan

12. May 14, 2007

### hage567

That's better!

13. May 14, 2007

### Aikenfan

Thank you for taking the time to read my work, i know it was a lot

14. May 14, 2007

### hage567

No problem. There is one thing on the first page near the bottom that I noticed; you have the momentum terms as m+v. I'm guessing you just copied it wrong or something because you used it correctly later on.

15. May 14, 2007

### Aikenfan

Thanks! I just corrected that :-)