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I think this problem is much more difficult than what I've learned so far.


B) is the problem I'm having a hard time with. I think it is much more difficult to consider because as the red object begins compressing the spring the blue object will begin to move as well.

It also seems apparent that there must be some kind of relationship of how much force is instantaneously transferred to the blue object as it is compressed by the red object.

I'm really not sure how to do this to be honest. I haven't learned how to make work with springs besides energy equations. I've tried the spring constant (N/m) x the incoming speed (m/s) to determine how many newton's per second are being exerted during the collision, but I haven't figured out where to go from there.

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closed as off-topic by John Rennie, centralcharge, Emilio Pisanty, Kyle Kanos, Brandon Enright Jan 20 '14 at 19:37

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Use conservation of energy and the fact that at max compression the relative velocity of the masses must be zero. – Sandesh Kalantre Jan 20 '14 at 4:30
I've tried finding the kinetic energy of the red mass and finding the max compression of the spring, but then I realized that it wouldn't be accurate since the surface is frictionless and the blue mass will move as the red mass compresses the spring and therefore the spring will be compressed less than by just applying a simple kinetic energy to spring compression series of equations. Correct me if I'm wrong though, that's just my intuition. – Klik Jan 20 '14 at 4:46
Why don't you look at the whole thing as a system? (A + spring + B)? Then you don't need to worry about internal forces, and you can safely assume that the total energy of the system is constant. So whatever energy was possessed by A is now stored as potential energy of the spring + KE of both bodies *moving at the same speed* . Can you say why? – mikhailcazi Jan 20 '14 at 6:09
How do you look at the whole thing as a system? I'm unsure by what you mean exactly. In Ross' answer he mentioned something about cm frame and I'm currently looking into that as a solution. Would you be suggesting the same thing? – Klik Jan 20 '14 at 6:17
up vote 1 down vote accepted

The potential energy in the spring is $\frac 12 kx^2$, where $x$ is the compression of the spring. That changes the collision dynamics, but doesn't matter for part a. The same equations you used to balance energy and momentum without the spring apply, because well after the collision the spring is extended. For part b, transform to the cm frame. At minimum separation both masses are stationary in this frame and all the energy is stored in the spring.

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This is what I was thinking. I mean originally I approached the problem as a kinetic energy to spring energy problem ($1/2m*v^2=1/2k*x^2$), but then I began to think about it more and it seems more difficult that I had originally anticipated. Please let me know if this sounds wrong. Instant one, the red mass connects with the spring. Instant two the spring compresses between the masses. Instant three the red mass continues compressing the spring, the spring begins acting and accelerating the blue mass. And so on... It seems to me that the spring will not hold all of the energy in one instant. – Klik Jan 20 '14 at 5:55
In the cm frame, the spring accelerates both masses any time it is compressed. Since the total momentum in that frame is zero, at the moment one mass turns around (and has zero velocity) the other one must as well. At that moment there is no KE, so it must all be PE. – Ross Millikan Jan 20 '14 at 5:58
Great, thanks for the answer. – Klik Jan 20 '14 at 6:47

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