Speed of object after collision related to spring

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The discussion focuses on calculating the speed of object A after a collision involving a spring. The conservation of energy equation is initially applied, equating the kinetic energy of object B before the collision to the elastic potential energy of the spring and the kinetic energy of object A. The participants confirm that momentum conservation applies since the system is closed, leading to the conclusion that both objects A and B will move with the same speed at maximum spring compression. The derived speed of object A is determined to be half of the initial speed of object B. The conversation highlights the successful application of both conservation principles to solve the problem.
songoku
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Homework Statement


As shown in the figure below, object A (mass: m) is at rest on a smooth, horizontal surface, and a lightweight spring that compress / stretches in the horizontal direction is attached to it. Object B, which has the same mass m, approaches A from the left side with speed v and collides with the spring. The spring compresses and A begins to move. A, B, and the spring travel in a straight line. What is the speed of A when the spring is at maximum compression?
qwe_zps299uruf2.png


Homework Equations


conservation of energy
momentum?

The Attempt at a Solution


I tried using conservation of energy:

Kinetic energy of B before hitting the spring = elastic potential energy of spring at maximum compression + kinetic energy of A

But then I stuck because I don't know how to change the spring constant and maximum compression to other variables
 
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songoku said:
But then I stuck
so consider
songoku said:
momentum?
Do the conditions for momentum conservation exist here?
 
haruspex said:
Do the conditions for momentum conservation exist here?

Yes, taking both boxes and spring as a closed system.

At maximum compression, object A and B will move with same speed?

Using conservation of momentum:
mAuA + mBuB = (mA+mB) V
V = 1/2 v

Is this correct?
 
songoku said:
Yes, taking both boxes and spring as a closed system.

At maximum compression, object A and B will move with same speed?

Using conservation of momentum:
mAuA + mBuB = (mA+mB) V
V = 1/2 v

Is this correct?
That's it.
 
Thanks a lot for the help
 
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