8
$\begingroup$

The below homework question just sets the context for my question. I am not asking for a solution to this question

In a collinear collision, a particle with an initial speed $v_0$ strikes a stationary particle of the same mass. if the final total kinetic energy is $50\ \%$ greater than the original kinetic energy, what is the magnitude of the relative velocity between the two particles after the collision?

How can kinetic energy after collision be more than initial kinetic energy in this particular case, where there is no disintegration of matter or something?

$\endgroup$
  • 1
    $\begingroup$ Would it be easier to find the answer for the reverse process? "After a collision, one particle is at rest and the other moving at speed $v_0$. Also, the total kinetic energy was $50\%$ greater before the collision, compared to after the collision. The magnitude of relative velocity between the particles before the collision was ...?" $\endgroup$ – Hagen von Eitzen Nov 9 at 22:20
14
$\begingroup$

I'm not sure why answers here only discuss changing reference frames. You can operate in the same reference frame and still have an increase in kinetic energy. For example, if one object has a compressed spring attached to it that is set to release upon collision. Then the extra energy comes from what was the potential energy.

All you need to do is apply momentum conservation as well as the condition of a $50\%$ increase in kinetic energy. Or use the coefficient of restitution. It is totally possible.

$\endgroup$
6
$\begingroup$

I have written this answer to point out that the increase in kinetic energy has nothing to do with changing the reference frame and answers which hint at this being true are wrong.
Although individual particles might appear to have a different kinetic energy in another reference frame the total kinetic energy of a system under consideration does not change.

In the question there is the following statement which hints that there is “something” which generates kinetic energy.

How can kinetic energy after collision be more than initial kinetic energy in this particular case, where there is no disintegration of matter or something?

A change of frame of reference does not "generate" the extra kinetic energy even though the kinetic energies of each of the particles might change.
There is an illustration of this idea in the answer to Kinetic energy with respect to different reference frames. If one changes the frame of reference the law of conservation of energy is still valid so in every frame of reference there will be the $50\%$ increase in the kinetic energy.

Classically this is a super-elastic collision with the extra kinetic energy coming from the internal energy of the system eg a chemical reaction, a compressed spring releasing its energy etc.

$\endgroup$
  • $\begingroup$ But the question explicitly states that one of the particles is initially stationary, which means that the kinetic energy has to be evaluated with respect to the initial rest frame of that particle, not the centre-of-mass frame. $\endgroup$ – TonyK Nov 9 at 16:06
  • $\begingroup$ What is the relationship between your required relative velocity, $2v_f$ and the choices given in the problem? $\endgroup$ – Bob D Nov 9 at 19:10
  • $\begingroup$ When you say "A change of frame of reference does not "generate" the extra kinetic energy" do you mean a change in kinetic energy is the same in all inertial reference frames? $\endgroup$ – Bob D Nov 9 at 21:27
  • $\begingroup$ @BobD If you were moving along and observing a collision how could the speed that you were travelling at affect the change in kinetic energy during the collision? $\endgroup$ – Farcher Nov 9 at 22:06
  • $\begingroup$ @Farcher My question was not necessarily in the context of a collision. I'm simply asking if you mean by your statement a change in kinetic energy of an object will be the same in all inertial reference frames? $\endgroup$ – Bob D Nov 9 at 22:10
3
$\begingroup$

This is called a super-elastic collision. It is not physically impossible - in a sense, it's just the reverse of an inelastic collision, and by following that reasoning, you can then easily come to understand how it's possible. Here's how it goes.

  • In an inelastic collision, the energy after collision is lower than the energy before. You still have a conservation of energy problem, just in the "other direction". But of course we know that conservation of energy is still valid, even though these kinds of collisions exist in great abundance in our world. So there's got to be an answer.
  • The answer? The "disappearing" energy goes into internal energy in the objects (or to some external, non-kinetic sink, e.g. sound waves.). Mostly, this is in the form of heat, causing a slight elevation of the temperature.

  • Hence, what must a super-elastic collision involve? Logically, by simply turning this around, it means some internal energy must "go out" and become external energy, i.e. kinetic energy. (Or equivalently in mathematical terms, the objects "absorb" negative energy during the collision, which is equivalent to turning the direction of energy transfer around, just as me giving you negative money means you paying me positive money.)

Now, of course, because of the second law of thermodynamics (the law of increasing entropy), it is "exceedingly unlikely" (in a sense that can be made very precise) for the energy source here to be the same kind of energy - i.e. internal thermal energy - where the energy goes following an inelastic collision. Instead, for it to happen within the lifetime of any being within the observable universe, some sort of low-entropy energy must be present within the objects prior to collision that is ripe for conversion to mechanical energy.

One example may be to imagine the balls' surfaces coated with a thin layer of mild explosive for which the initial impact force is just enough to trigger the chemical reactions that release the explosive energy, while also not being powerful enough to outright shatter the balls so that the analysis is still meaningful, and also that its contribution to the masses of the balls is negligible, so that when it cooks off the masses can be treated as the same after collision. But there's many other ways to arrange it, too: it doesn't matter how, just as long as the relevant conversion of the internal energy is possible.

Or, in short: the objects are "powered" in some fashion.


Of course, this is just the practical way. If we want to seriously entertain the profound way, then we can return to my comment regarding the second law of thermodynamics, and imagine the exact time-reversal, even all the way down to the microscopic level, of ordinary balls undergoing inelastic collision. While we can't realistically arrange for this (see "extremely unlikely"), it is not prohibited by the laws of physics of our world as far as we understand them, because they are time-symmetric, so the reverse evolution will be generated by the same laws as the forward one. In this case, when the balls come back together, the thermal jiggling of their atoms will all happen to conspire by sheer happenstance "just right" so as to give them a "kick" from that internal thermal energy at that point of collision, that is enough to send them out with greater kinetic energy than when they came in (and cooling them down proportionately by the needed amount). Again, it's still conversion of an internal reservoir ... just now the one that "isn't supposed to" happen.

$\endgroup$
0
$\begingroup$

Macroscopic kinetic energy is reference frame dependent. For example a car with velocity $v$ with respect to the reference frame of the road has kinetic energy of $\frac{mv^2}{2}$ with respect to a person standing on the road, but it has zero kinetic energy in the reference frame of the driver. Note that the problem asks for the RELATIVE final velocity between the particles.

Hope this helps

$\endgroup$
  • $\begingroup$ But relative velocity isn't frame dependent. No matter what frame you are in, you will still see the two objects have the same relative velocity. So really you think the question is asking "in what reference frame will you observe a $50\%$ increase in total kinetic energy"? And then are we assuming the collision is elastic? $\endgroup$ – Aaron Stevens Nov 10 at 13:21
  • $\begingroup$ Which at that point there isn't a need for the collision in the first place. $\endgroup$ – Aaron Stevens Nov 10 at 13:30
  • $\begingroup$ @AaronStevens I'm sorry, but where did I say relative velocity is frame dependent? Although it might not apply to this collision example, I'm saying in general kinetic energy, and therefore change in kinetic energy, is frame dependent. Do you not agree? $\endgroup$ – Bob D Nov 10 at 14:51
  • $\begingroup$ I agree that kinetic energy of frame dependent. I don't think this problem has anything to do with that though. $\endgroup$ – Aaron Stevens Nov 10 at 15:13
  • $\begingroup$ @AaronStevens That may be, but I'm still trying to sort it out. E.g., is the 50% increase in KE seen from the original rest frame of the second particle, or is it as seen from the rest frame of the second particle. Seems it can't be the latter because if the relative velocity after the collision is root 2 x $v_o$ then the increase as seen from the second particle would be 100% $\endgroup$ – Bob D Nov 10 at 15:28
-1
$\begingroup$

Yes final kinetic energy may be greater/less than the initial if you view from a different frame of reference as it is a frame dependent physical quantity.In this problem it is clearly mentioned that it is asking for relative velocity between particles.

$\endgroup$
  • 1
    $\begingroup$ For what it's worth, I'm still not convinced that your answer and mine are not relevant, at least in the case where there is no initial stored energy in one of the particles that is released upon the collision. See my comments to Aaron Stevens and Farcher. Although we are getting pounded on for the moment, hang in there. $\endgroup$ – Bob D Nov 9 at 22:41

protected by AccidentalFourierTransform Nov 11 at 2:16

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.