# Collisions homework problem

1. Oct 26, 2007

### pinkyjoshi65

A 1.0 kg magnetized air puck moving across a level table at 0.24 m/s approaches head-on a stationary, similarly magnetized air puck of mass 0.50 kg. If the "magnetic collision" is repulsive and perfectly elastic, determine:
(a) the velocity of each puck after the collision
(b) the velocity of both pucks at minimum separation
(c) the total kinetic energy at minimum separation
(d) the maximum potential energy stored in the magnetic force field during the collision

For A) i simply used the conservation of momentum and KE and found the final velocities (1-dimensional)

For B) I am not sure how to find th miniimum distance

2. Oct 26, 2007

### Staff: Mentor

Hint: When the pucks are as close to each other as they get, what's the relationship of their velocities?

3. Oct 26, 2007

### pinkyjoshi65

uhh..i am not sure abt this..velocities become the same..?

4. Oct 26, 2007

### Staff: Mentor

Exactly! (If the velocities weren't the same, then they'd keep getting closer.) So figure out what that velocity must be.

5. Oct 26, 2007

### pinkyjoshi65

so you are saying that the collision becomes the case of a perfectly inelastic collision...?

6. Oct 26, 2007

### pinkyjoshi65

if thats the case, then M1v= (M1+M2)V
hence V= M1v/M1+M2

7. Oct 26, 2007

### Staff: Mentor

Exactly.

8. Oct 26, 2007

### pinkyjoshi65

ahh k..so for part c its quite staight foreward. we have to use0.5(M1+M2)V2
For part d potential enegy will be max when the h is max. How do we find the h..?

9. Oct 26, 2007

### pinkyjoshi65

and also this is all 1-dimensional yes?

10. Oct 26, 2007

### Staff: Mentor

Right.
What do you mean by "h is max"? Just use the fact that total energy is constant.

11. Oct 26, 2007

### pinkyjoshi65

so then the potential energy will be equal to the kE in part C)..?

12. Oct 26, 2007

### Staff: Mentor

No, but the sum of PE + KE must be constant. (What's the initial total energy?)

13. Oct 26, 2007

### pinkyjoshi65

initial total energy is M1v1^2+ M2v2^2

14. Oct 26, 2007

### Staff: Mentor

Not exactly. Initially, only one puck is moving. (And that's not the correct formula for KE!)

15. Oct 26, 2007

### pinkyjoshi65

yes so M2v2^2 will 0..i know that hence the initial total energy will be 0.5*M1v1^2= 0.0288J. so then we can subtract the KE (from part c)frm this energy to find the PE

16. Oct 26, 2007

### Staff: Mentor

Now you've got it.