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Collisions homework problemby pinkyjoshi65
Tags: collisions 
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#1
Oct2607, 02:13 PM

P: 263

A 1.0 kg magnetized air puck moving across a level table at 0.24 m/s approaches headon 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 (1dimensional) For B) I am not sure how to find th miniimum distance 


#2
Oct2607, 02:35 PM

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P: 41,477

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



#3
Oct2607, 02:37 PM

P: 263

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



#4
Oct2607, 02:39 PM

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P: 41,477

Collisions homework problem



#5
Oct2607, 02:40 PM

P: 263

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



#6
Oct2607, 02:45 PM

P: 263

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


#8
Oct2607, 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..? 


#9
Oct2607, 03:10 PM

P: 263

and also this is all 1dimensional yes?



#10
Oct2607, 03:20 PM

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P: 41,477




#11
Oct2607, 03:23 PM

P: 263

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



#12
Oct2607, 03:35 PM

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P: 41,477




#13
Oct2607, 03:37 PM

P: 263

initial total energy is M1v1^2+ M2v2^2



#14
Oct2607, 03:42 PM

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P: 41,477




#15
Oct2607, 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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