# Homework Help: Problem Concerning Inelastic collisions.

1. Jul 24, 2015

### HermitOfThebes

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

http://www.pdx.edu/physics/sites/www.pdx.edu.physics/files/Nov-19th_version_Comprehensive Exam I_Nov-2006.pdf

2. Relevant equations

3. The attempt at a solution
This problem is driving me nuts. Its the problem number 1 in the afternoon session exam provided in the link above. It's about conservation of linear momentum. I have analysed the first situation and got the equation V2=V(1-F/2)
*given that (1-f)^1/2 = 1-f/2 as f << 1. This is given in the Feynman's exercise book.
I can't for the life of me analyse the second situation. Could someone please at least tell me what the equation for the conservation of Kinetic Energy be?

Last edited: Jul 24, 2015
2. Jul 24, 2015

### Nathanael

Your equation is meaningless unless you state what the variables are. F is the applied force, but what are "V" and "V2"? The velocities of the respective masses? If so then it's dimensionally wrong, and why are you even considering the velocities? Please show your work you want to get help.

For problem 2, also try to show your work. When the block first transitions from the ground to the triangle there is an inelastic collision.
(Unless you meant part (b) of problem 1?)

3. Jul 24, 2015

### HermitOfThebes

Nathanael I am asking about the other problem, not this one. I am talking about problem 1 in mechanics in the afternoon exam, if you scroll down you will find it. It's about two gliders colliding.
F is defined as the fraction of lost K.E.. I would really appreciate it if you could help me with this problem as its killing me.
V2 is rebound velocity for both masses in Case 1, and V is the velocity by which they approach each other.

4. Jul 24, 2015

### Nathanael

Okay. Your equation is good then. For the second case, if you switch to a certain reference frame then it is the same as case 1. So you can use your equation to calculate the speeds in that reference frame and then transform back to the original reference frame.

5. Jul 24, 2015

### HermitOfThebes

I did exactly that but I keep getting vf/2 not vf/4 :S

6. Jul 24, 2015

### Nathanael

I haven't done it but I'd like to see your work.

Edit:
Okay now I've done it and I got vf/2 as well.

7. Jul 24, 2015

### HermitOfThebes

I switched to a frame of reference where the speed of both boxes becomes v/2 instead of one having v and the other being at rest.
Now if I substitute into the equation(V-rebound=V-approach*(1-f/2)) I get V-rebound=V/2(1-f/2)
How do I revert back to the original frame of reference?

8. Jul 24, 2015

### Nathanael

Well to get into that reference frame, you had to start moving +v/2 with respect to the original reference frame. So now to get back to the first frame you must move -v/2 relative to this second frame.

I solved the problem using their hint as well, and I still get vf/2

Edit: No I don't! I get vf/4 both times! Silly me I was using V2=V(1+f)

9. Jul 24, 2015

### HermitOfThebes

yea I did that too, still :S Even in Feynman's exercise book it says it's vf/4

10. Jul 24, 2015

### Nathanael

No no I get vf/4 with both methods. I made a mistake of using V2=V(1-f) instead of V2=V(1-f/2) by accident.

You get this too, check your equation.

[edited to fix typo]

Last edited: Jul 24, 2015
11. Jul 24, 2015

### HermitOfThebes

12. Jul 24, 2015

### Nathanael

You already told me this equation: Vrebound=V/2(1-f/2) ... You pretty much had it, but your equation has a small mistake: it is off by a factor of -1, because you see if you imagine which block we are interested in, it is the one which is traveling backwards in this new reference frame.

To get into this new reference frame we just subtracted v/2 from all the velocities right? The first block went from v to v/2 and the second block went from 0 to -v/2...
Well now to get back to the old reference frame we just add v/2 to all velocities. Doing so gives you vf/4

13. Jul 24, 2015

### HermitOfThebes

Oh. That bloody -1 :D thank you very much :D I wasted alot of time trying all kinds of crazy things, should have paid more attention. Thanks :D

14. Jul 24, 2015

### Nathanael

No problem.

Ditto about the paying attention haha