Collisions


by pinkyjoshi65
Tags: collisions
pinkyjoshi65
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#1
Oct26-07, 02:13 PM
P: 263
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
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Doc Al
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#2
Oct26-07, 02:35 PM
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Hint: When the pucks are as close to each other as they get, what's the relationship of their velocities?
pinkyjoshi65
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#3
Oct26-07, 02:37 PM
P: 263
uhh..i am not sure abt this..velocities become the same..?

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#4
Oct26-07, 02:39 PM
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Collisions


Quote Quote by pinkyjoshi65 View Post
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
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#5
Oct26-07, 02:40 PM
P: 263
so you are saying that the collision becomes the case of a perfectly inelastic collision...?
pinkyjoshi65
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#6
Oct26-07, 02:45 PM
P: 263
if thats the case, then M1v= (M1+M2)V
hence V= M1v/M1+M2
Doc Al
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#7
Oct26-07, 03:00 PM
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Exactly.
pinkyjoshi65
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#8
Oct26-07, 03:03 PM
P: 263
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..?
pinkyjoshi65
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#9
Oct26-07, 03:10 PM
P: 263
and also this is all 1-dimensional yes?
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#10
Oct26-07, 03:20 PM
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Quote Quote by pinkyjoshi65 View Post
ahh k..so for part c its quite staight foreward. we have to use0.5(M1+M2)V2
Right.
For part d potential enegy 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.

Quote Quote by pinkyjoshi65 View Post
and also this is all 1-dimensional yes?
Yes. (It's a "head-on" collision.)
pinkyjoshi65
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#11
Oct26-07, 03:23 PM
P: 263
so then the potential energy will be equal to the kE in part C)..?
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#12
Oct26-07, 03:35 PM
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Quote Quote by pinkyjoshi65 View Post
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
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#13
Oct26-07, 03:37 PM
P: 263
initial total energy is M1v1^2+ M2v2^2
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#14
Oct26-07, 03:42 PM
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Quote Quote by pinkyjoshi65 View Post
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
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#15
Oct26-07, 03:45 PM
P: 263
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
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#16
Oct26-07, 03:47 PM
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Now you've got it.


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