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I am a biologist developing an interest in physics. I am struggling with the implications of changing reference frames on momentum, mechanical energy and work done calculations. I invented the situation in the diagrams below to try and work stuff out but the outcomes do not fit with what I believe are the correct physical principles. Can anyone assist me?

The diagrams show the situation. Starting in the ground reference frame I have first calculated velocity after collision via conservation of momentum and mechanical energy. I hope this is correct.

Velocities calculated after elastic collision in ground reference frame

I believe that while the momentum and mechanical energy values will differ for different reference frames, the values should be conserved within a reference frame before and after an elastic collision. I have calculated momentum and kinetic energy values for the 2kg ball reference frame and momentum does not appear conserved in this scenario. I cannot figure out what my error is.

Momentum does not appear to be conserved in the 2kg ball reference frame

A final question. Imagine we have another ball to the left which is not involved in the collision and is stationary. In the ground reference frame the momentum and mechanical energy are both conserved. In the reference frame of the 2kg ball, neither momentum nor mechanical energy are conserved. Is this something wrong with the way I have defined the system as I could add any number of stationary objects fixed to the Earth which in the ground reference frame would have no momentum or mechanical energy but in the reference frame of the ball would have momentum and kinetic energy meaning the conservation principles do not work. In general I am unclear on the concept of system in mechanics. I have this working notion: In mechanics a system consists of an object or group of objects whose motion is being described.

Neither momentum or KE conserved in the 2kg ball reference frame when a third ball, not involved in the collision, is added to the system

The velocities in the 2kg reference frame examples should be correct based on the original velocities I stipulated in the ground reference frame. Any help would be much appreciated.

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3 Answers 3

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I didn't redo your calculations and I assume that they are correct, which actually doesn't play any role in what I'll describe now. Notice that in the second scenario the 2kg ball will inevitably start to move. By keeping it still you change the reference frame one more time, which invalidates the use of conservation laws.

You cannot use the conservation of energy or momentum in two reference frames$^*$. It is pretty straight forward to see why this must be the case. Assume there is a ball with mass $m$ and from the reference frame of the stationary observer it is moving with velocity $v$. Clearly the ball has a kinetic energy $mv^2/2$ and a momentum $mv$. However from the reference frame of the ball, it is not moving therefore its momentum and energy is zero. If you were to use the conservation of energy or momentum, you would see that it is violated. Thus you conclude that you cannot use the conservation laws for two different reference frames.

I am not sure I understand what your last question is.

$^*$I am assuming that you don't know about momentum-energy four vector, which combines these two and is in fact conserved.

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  • $\begingroup$ Thank you gonenc, most appreciated. As I have alluded below, I cannot see how I have changed reference frame. My (probably erroneous) interpretation of what I have done is in the comment under the next answer. If you could help straighten my understanding it would be much appreciated. And you were quite right in assuming I don't know about the momentum-energy four vector! All in good time. $\endgroup$
    – James Kempton
    Commented Jun 13, 2015 at 21:28
  • $\begingroup$ Notice that after the 4kg ball hits the 2kg ball it should start moving as a consequence of conservation of momentum. In your after picture you have drawn the 2kg ball stationary, which means that you have change the reference frame. $\endgroup$
    – Gonenc
    Commented Jun 13, 2015 at 21:30
  • $\begingroup$ I think I see your confusion. Don't think of reference frames as if they were to belong an object. Think of them as an observer of the experiment. We say that we are in the reference frame of an object if the observer moves with the object with exactly the same velocity. Take this example the observer moves with 5m/s to the right so that the 2kg ball appears to be stationary. After the collision the observer keeps moving, while the 2kg ball changes its direction. If you want to keep the 2kg ball stationary, you need to change the observer, which is a change of the reference frame. $\endgroup$
    – Gonenc
    Commented Jun 13, 2015 at 21:48
  • $\begingroup$ Yes it works! Thank you gonenc for your explanation, you have made me a happy man. It takes one to understand a concept, it takes another to understand the mind of a fool and make them understand that concept - my appreciation indeed. A very positive first use of this website - i'll be back and you have been warned. $\endgroup$
    – James Kempton
    Commented Jun 13, 2015 at 21:56
  • $\begingroup$ If this answer helped you solve the problem consider accepting it as your answer by clicking the check mark on the left hand side. See this meta post to learn more about accepting an answer. $\endgroup$
    – Gonenc
    Commented Jun 13, 2015 at 22:05
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Great question. I didn't check all the math, but I see a problem. In the first graphic, you described a collision process in a single reference frame. Your picture is lovely. In the second graphic there is a problem. In the rest frame of the 2kg ball, the 2kg ball sees a 4kg ball approaching it. During the collision, momentum is transferred from the 4kg ball to the 2kg ball. Therefore, after the collision the 2kg ball is no longer at rest. Therefore, in the second graphic the "after diagram" is incorrect. Essentially, in the second graphic you have depicted difference reference frames before and after the collision. Therefore, you shouldn't expect to get the same total momenta for your "before" and "after" calculations in the second graphic.

Good question in the third graphic. Since the "external bodies" like the Earth play no part in the collision (to an extremely reasonable approximation), their momenta and energies do not change during the collision, so we need not consider them when checking that momentum is conserved in a collision.

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  • $\begingroup$ Most appreciated. A remaining problem-I cannot see how I have switched reference frames(RF). The collision in the second diagram is supposed to be the same as in the first, this time from the 2kg ball perspective. I thought that when determining velocities in an RF, you consider an observer stationary with respect to the RF and describe the velocities relative to that. For the 'after' of the collision, even though the 4kg ball has imparted momentum to the 2kg ball, isn't it seen from the perspective of the 2kg ball that the 4kg ball has reversed its direction and kept the same speed? $\endgroup$
    – James Kempton
    Commented Jun 13, 2015 at 21:22
  • $\begingroup$ @JamesKempton Inertial reference frames can be moved, you can't slow one down or turn one. So if you took the inerital reference frame where the 2kg bank was originally at rest, then the frame keeps moving that way forever regardless of whether or when or how the ball moves later. So when you say "the frame of the 2kg ball" you can pick the frame where it was originally at rest or the frame where it ends up at rest but they are different frames since they move relative to each other. $\endgroup$
    – Timaeus
    Commented Jun 14, 2015 at 3:01
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There is a mistake in your calculation . In 2nd photo, remember , $V(1)$ is for $4\,\mathrm{ kg}$ mass And, $V(2)$ is for $2\,\mathrm{ kg}$ mass . But in 2nd photo, you had used $V(1)=0$ and $m(2) =2$ but it should be $m(1)=4$ and $V(1)=0$ so, $m(1)+m(2)=0+2\cdot 15=30$ which is correct .

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