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An object is headed towards a spring at a constant velocity, no external force will act upon it, except for any force applied by the spring itself after the collision.

Let's say the spring is of arbitrary tension and arbitrary length.

My question is whether the kinetic energy of object will make a difference in how far the spring is compressed, or whether it is only a matter of momentum?

For example, below, Object1 and Object2 have the same momentum, but Object2 has twice the kinetic energy. Would they compress the spring the same distance, or would Object2 compress it further?

Object1 has a mass of 1kg and a velocity of 1m/s.

Object2 has a mass of .25kg and a velocity of 4m/s.

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    $\begingroup$ Nothing is compressed "by" energy and momentum. Energy and momentum are just qualities that physical objects, which actually do the compressing, can have. $\endgroup$ – knzhou Jun 19 '19 at 14:42
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    $\begingroup$ The compression distance $x$ is determined by $mv^2 = kx^2$. So $x$ depends on both $m$ and $v$. Whether you call that "depending on momentum" or "depending on energy" is kind of up to you. $\endgroup$ – knzhou Jun 19 '19 at 14:43
  • $\begingroup$ @knzhou according to your equation x, in the case of the two example objects, is directly proportional to energy, not momentum $\endgroup$ – TheCatWhisperer Jun 19 '19 at 14:49
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    $\begingroup$ No, $x^2$ is proportional to energy. And meanwhile $x$ is proportional to $p$ if you allow extra factors of $m$. So again I don't know what you mean by whether it's energy or momentum -- it depends on both $m$ and $v$, and you can call that what you want. $\endgroup$ – knzhou Jun 19 '19 at 14:55
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    $\begingroup$ @TheCatWhisperer the conservation of momentum would only become relevant if the spring was not attached to say a wall ... then the centre of mass and would move and blah blah .. thats a different problem probably not the one your referring to $\endgroup$ – More Anonymous Jun 19 '19 at 14:57
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When the object collides with the spring there will be a change in both its kinetic energy and momentum. It is the kinetic energy of the object that determines the compression of the spring. If the objects have the same mass $m$ and the same velocity, they will have the same momentum and kinetic energy.

In terms of the effect of an object on the spring, you can apply the work energy theorem, which states that the net work done on an object equals the change in its kinetic energy. In the case of the following equation applying the theorem the kinetic energy is that before the collision, and is zero at the maximum compression of the spring when the object is brought to a stop. The spring does negative work on the object taking its kinetic energy and storing it as spring potential energy.

$$\int F(x)dx=-\frac{mv_{i}^2}{2}$$

For the spring,

$$\int F(x)dx=\frac{kx^2}{2}=-\frac{mv_{i}^2}{2}$$

Solve for the displacement $x$.

In your example object 2, even though it has the same momentum as object 1, has twice the kinetic energy so it will compress the spring more.

Hope this helps.

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As mentioned by knzhou in the comments, I don't think it's correct to really say that a spring is directly compressed by either momentum or by kinetic energy. As dmckee's comment rightfully points out; springs are compressed by force. The relationships between the applied force, kinetic energy, and momentum, will all depend on the situation you describe.

That said, it can easily be shown that in your example scenario, given the same momentum, we get different values for spring compression. We know that kinetic energy can be determined by $KE = \frac 12 m v^2$ and the energy in the spring can be shown as $PE = \frac 12 k x^2$. Due to conservation of energy, that gives us $$PE = KE$$ $$\frac 12 mv^2 = \frac 12 k x^2$$

In your example, if you solve for $x$, object 1 will compress the spring $x = \sqrt{\frac 1 k}$ and for object 2 the spring will compress $x = \sqrt{\frac 8 k}$. You can see that increasing kinetic energy while leaving momentum the same will increase the compression of the spring. This means that the compression is proportional to energy, not momentum; but does not necessarily say anything about the "cause" of the compression.

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    $\begingroup$ If you let the ball hit a free spring you'll find that momentum is also conserved (and that the energy transfer is never complete--the ball never comes to a stop). The reason that momentum is not conserved but energy is here is that the system is subject to external impulse (from the fixed end of the spring) but not to external work (because they fixed end is fixed). So one absolutely should not say that the spring is compressed by anything except force. $\endgroup$ – dmckee --- ex-moderator kitten Jun 19 '19 at 14:56
  • $\begingroup$ @dmckee Very good point. I added in that springs are compressed by force, since I didn't remotely address that in the answer and absolutely should have. $\endgroup$ – JMac Jun 19 '19 at 15:00
  • $\begingroup$ @JMac Sorry, I was editing my answer and posted it before seeing yours which essentially says the same thing. Not trying to one up on you. Bob $\endgroup$ – Bob D Jun 19 '19 at 15:09

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