How Does Magnetic Repulsion Affect Puck Velocities in Elastic Collisions?

[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

 

[Doc Al]

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

 

[pinkyjoshi65]

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

 

[Doc Al]

velocities become the same..?

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

 

[pinkyjoshi65]

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

 

[pinkyjoshi65]

if that's the case, then M1v= (M1+M2)V
hence V= M1v/M1+M2

 

[Doc Al]

Exactly.

 

[pinkyjoshi65]

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

 

[pinkyjoshi65]

and also this is all 1-dimensional yes?

 

[Doc Al]

ahh k..so for part c its quite staight foreward. we have to use0.5(M1+M2)V2

Right.

For part d potential energy will be max when the h is max. How do we find the h..?

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

and also this is all 1-dimensional yes?

Yes. (It's a "head-on" collision.)

 

[pinkyjoshi65]

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

 

[Doc Al]

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

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

 

[pinkyjoshi65]

initial total energy is M1v1^2+ M2v2^2

 

[Doc Al]

initial total energy is M1v1^2+ M2v2^2

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

 

[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

 

[Doc Al]

Now you've got it.