- #1

- 13

- 0

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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter GingerKhan
- Start date

- #1

- 13

- 0

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

gneill

Mentor

- 20,945

- 2,886

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

- #3

- 13

- 0

Would it be kinetic energy?

- #4

gneill

Mentor

- 20,945

- 2,886

Would it be kinetic energy?

That's one. There's another.

- #5

- 13

- 0

Electric potential energy?

- #6

gneill

Mentor

- 20,945

- 2,886

Electric potential energy?

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

- #7

- 21

- 0

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

gneill

Mentor

- 20,945

- 2,886

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

- 21

- 0

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

gneill

Mentor

- 20,945

- 2,886

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?

Yes, that looks much better!

- #11

- 21

- 0

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

gneill

Mentor

- 20,945

- 2,886

Glad to be of help!

Share: