# Homework Help: Minimum separation between pucks

1. Nov 20, 2011

### GingerKhan

Two frictionless pucks are placed on a level surface at an initial distance of 20.0 m. The mass of puck 1 is 0.80 kg and it has a charge of +3.0 X 10^-4 C., while puck 2 is 0.40 kg with a charge of +3.0 X 10^-4 C. the initial velocity of puck 1 is 12.0 m/s (E) and the initial velocity of puck 2 is 8.0 m/s (W). Find the minimum separation of the two pucks.

I've got no idea where to start. I suppose the total electric force has to cancel out the net force of the two pucks?

2. Nov 20, 2011

### Staff: Mentor

Think in terms of conservation of energy. What forms of energy are at play here?

3. Nov 20, 2011

### GingerKhan

Would it be kinetic energy?

4. Nov 20, 2011

### Staff: Mentor

That's one. There's another.

5. Nov 20, 2011

### GingerKhan

Electric potential energy?

6. Nov 20, 2011

### Staff: Mentor

Yup. There will be an exchange between kinetic energy and electric potential energy as the pucks approach each other.

7. Dec 10, 2011

### Seinfeld4

I believe I might have done it correctly, but the answer I got doesn't seem too realistic, so I would definitely appreciate it if someone could check my work!

First, I found the total energy of the system to be 110.9 J by finding the sum of the kinetic energy and electrical potential energy at a distance of 20.0 m.

Then, I used conservation momentum since the velocity of the two masses will be the same when they're at minimum separation.

v = ((0.8)(12) + (0.4)(8.0)) / (0.4 + 0.8)
v= 10.67 m/s

Then, to find the distance at minimum separation, I set the sum of the new kinetic energy (with v = 10.67 m/s for both masses) and the electrical potential energy equal to 110.9 J.

This yields a distance of 19.0 m or thereabouts, which is only 1 m away from where they started, which leads me to believe that this isn't a realistic answer. Also, I don't understand why the velocities of the two masses are the same at the point of minimum separation. My book simply stated this fact without elaborating. I don't feel too confident about this...

8. Dec 10, 2011

### Staff: Mentor

The velocity v that you calculated should be the velocity of the center-of-momentum frame of reference. As such you need to be sure not to drop the signs of the velocities when calculating it!

Note that you could, if you wish, move all your calculations to that new frame of reference by adding v to each of the original velocities. What's the new KE there? In that frame the pucks will have zero velocity when they have minimum separation, making KE calculation for that moment easy!

The center of momentum frame is useful for collision problems because at the instant the collision is taking place the two objects are at the origin of that frame and motionless for an instant. In this case where the collision is "soft" (the charges prevent physical contact) they will be motionless with some separation between them.

Any object with zero velocity in the center-of-momentum frame will have, in the original frame of reference, a velocity equal to that of the center-of-momentum frame's.

9. Dec 10, 2011

### Seinfeld4

That's very interesting, I didn't realize that. So, if I assign a negative value to one of the initial velocities, then the magnitude of the conservation of momentum calculation becomes 5.33 m/s instead of 10.67 m/s.

With this new velocity value, the distance becomes 8.63 m instead of 19.0 m, which seems more realistic, I think...

Is that better?

10. Dec 10, 2011

### Staff: Mentor

Yes, that looks much better!

11. Dec 10, 2011

### Seinfeld4

Thanks a lot! In retrospect, it was a foolish mistake.

If the velocity had been 10.67 m like I had originally thought, then the momentum would not have been conserved at all. In fact, the momentum at the point of minimum separation would have been twice the value that the system started with! Rookie mistake.

Thanks again.

12. Dec 10, 2011