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An ideal spring is attached to a wall, and the other end is attached to a mass $m$. The spring is initially compressed a distance $x$. After it is released, the mass collides with another mass $2m$ at a distance $x/2$ to the right of the spring equilibrium. The collision is inelastic and they slide together. How far will they slide before coming to a momentary stop?

My work

Since the collision is inelastic then mechanical energy is not conserved. And since the two-mass system has an external force, momentum is not conserved. So one has to use only mechanical principles to solve this.

We can compute the velocity of the mass at the moment of impact using simple harmonic motion, $r(t) = -x\cos\left( t\sqrt{k/m}\right)$ so that $x/2 = -x\cos \left(t^{*}\sqrt{k/m}\right)$. With this I could do an awkward solution for time and maybe continue from there.

At this point, though, I'm not perfectly clear on how to move forward. Do I just assume that momentum is conserved for a very, very short time period during the collision and use that to find the velocity of the combined mass system, then put that into another simple harmonic motion problem?

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  • $\begingroup$ I would probably do it as a relation of spring energy to KE. But where is that formula for $r(t)$ coming from? Shouldn't that function use $\sqrt{\frac{k}{m}}$ instead of $\frac{k}{m}$? $\endgroup$
    – BowlOfRed
    Sep 7, 2014 at 18:31
  • $\begingroup$ @BowlOfRed You're right, I was solving the Differential Equation in my head quickly before running out the door, so clearly that didn't go well. I'm fixing it now. $\endgroup$
    – Addem
    Sep 7, 2014 at 19:55

2 Answers 2

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Remember that an inelastic collision means energy is lost but not momentum. Momentum is always conserved in collisions. So you have to figure out what kind of velocity the two joint masses have and use your knowledge of springs to compute the halt at the given position, velocity and mass.

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  • $\begingroup$ But isn't momentum not conserved for a different reason, namely the external (spring) force? $\endgroup$
    – Addem
    Sep 7, 2014 at 19:52
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    $\begingroup$ Yes, but it does not change in the collision itself. The momentum of the two bodies right before the collision and right afterwards will be the same. The spring continuously steals and gives the momentum to the body and transfers it to/from the wall/Earth - the effects of a momentum transfer from a small body on a spring to the Earth is neglected. $\endgroup$
    – Void
    Sep 7, 2014 at 20:07
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    $\begingroup$ So the answer to your added question is yes. You assume the collision instantaneous, find out the velocity and proceed. $\endgroup$
    – Void
    Sep 7, 2014 at 20:10
  • $\begingroup$ OK, got it. And just to be clear: If the collision were not instantaneous, then you wouldn't be able to use conservation of momentum at that moment--but in that case, the collision is somewhat elastic. And in a perfectly elastic collision, the conservation of energy would still hold. $\endgroup$
    – Addem
    Sep 7, 2014 at 20:14
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I would proceed as follows:

Assuming that the stored energy in the spring, compressed/extended a distance $d$ from equilibrium, is given by$$E_{stored}=kd^2$$

  1. Find stored energy, kinetic energy, and thus total energy at the moment of release;

  2. Find stored energy, and thus kinetic energy, just before the collision

  3. Step aside for a moment, and use conservation of momentum to find out what happens to the kinetic energy when a mass $m$ collides inelastically with a stationary mass of $2m$;

  4. Combine stored energy and reduced kinetic energy to find new total energy, just after the collision;

  5. Find extension/compression needed to store all this new total energy...

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  • $\begingroup$ In a spring, isn't stored energy--which I assume is the same as potential energy--equal to $\frac{1}{2}kd^{2}$? $\endgroup$
    – Addem
    Sep 7, 2014 at 20:05
  • $\begingroup$ And so basically, at step 3, you're saying to assume conservation of momentum over a very short time during the collision, right? In general the momentum is not conserved when there's a spring force, if I'm not mistaken. $\endgroup$
    – Addem
    Sep 7, 2014 at 20:06
  • $\begingroup$ I just defined a new $k$ for this problem; I didn't plan to use the spring force anywhere, and $k$ just cancels out in the end, anyway. $\endgroup$
    – DJohnM
    Sep 7, 2014 at 20:10

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