# Kinetic energy coservation in different frames (non relativistic)

A mass $m$ is thrown horizontally on a ice rink with speed $v$. From the point of view of the thrower, the initial kinetic energy is $\frac12 mv^2$. Finally the mass stops a certain distance away thanks to the ice friction. He concludes that the energy dissipated by action of the friction is $\frac12mv^2$.

The same thing is observed from another observer moving opposite to the motion of the mass with speed $v$. The second observer concludes that initially the mass had the kinetic energy $\frac12m(2v)^2 = 2mv^2$ and finally it had the kinetic energy $\frac12mv^2$. Therefore $she$ makes the conclusion that the energy dissipated is $\frac32mv^2$.

Which conclusion is correct? I can assure you at least one is indeed correct.

Can explain the origin, nature, and the size of the difference?

• Similiar question here: physics.stackexchange.com/q/51220 – Tomi Mar 9 '17 at 0:36
• Thank you for your suggestion but I could not find anything absolutely clarifying the difference. – Jason Mars Mar 9 '17 at 0:49
• – sammy gerbil Mar 9 '17 at 19:40

So due to Newton's third law there is an equal and opposite force on the ground. Treat this as an object with very large mass $M$, your hockey puck has mass $m$. The puck travels over the ice with speed $v$, the observer moves over the ice in the opposite direction with speed $u$. To this observer the puck seems to have speed $v + u$ and the rink seems to have speed $u$, so the total momentum is $M u + m u + m v.$ The total kinetic energy at the beginning is $$K_0 = \frac 12 M u^2 + \frac12 m (u+v)^2.$$After the slowdown the two have the same velocity $u + \frac{m}{M+m}v$ to conserve momentum, and the total kinetic energy is now $$K_1 = \frac12 (M+m) \left(u + \frac{m}{M+m} v\right)^2.$$ Expanding out we see $$K_1 = \frac12 (M+m) u^2 + m u v + \frac 12 \frac {m^2}{M+m} v^2.$$ On the flip-side, $$K_0 = \frac12 (M + m) u^2 + m u v + \frac 12 m v^2.$$ The difference, $$\Delta K = K_1 - K_0 = -\frac12~ v^2 ~\frac{Mm}{M+m},$$ is therefore totally independent of $u$.

In general, the kinetic energy of a bunch of particles under a velocity addition is given by $$K' = \sum_i \frac12 m_i (\vec u + \vec v_i)^2,$$ or after expansion, $$K' = K + \vec u \cdot \vec P + \frac12 M u^2$$where $M = \sum_i m_i$ and $\vec P = \sum_i m_i v_i.$ Therefore in general, subtracting value 1 from value 0, we find $$\Delta K' = \Delta K + \vec u \cdot \Delta \vec P,$$ and any interaction of these particles which conserves momentum $(\Delta \vec P = 0)$ will cause all observers to agree on changes in kinetic energy, even though they won't agree in general on the specific values of that kinetic energy.

• Thank you so much for this beautiful explanation. However, I would like to comment that if you consider higher speeds, how is that $CHANGED$ by the same amount (the kinetic energy)? – Jason Mars Mar 9 '17 at 16:46
• @JasonMars could you explain your question a bit further? I don't understand what you're asking but I think I might have already answered it because $u$ is totally allowed to be whatever speed you desire; it can be $u=0$ or $u=v$ or $u=1000v$ or whatever you'd like. – CR Drost Mar 9 '17 at 20:38
• I am so sorry sir! That's it! I could not recognize in the midst of all those equations. So could you kindly comment any inconsistency in the original question itself- I mean do we have I treat this scenario exactly as has been described by your model? – Jason Mars Mar 9 '17 at 21:31
• I mean your question isn't inconsistent or anything, it's a valid question that occurs to a lot of students at this level of physics. It even happened to the Mythbusters; there was an episode where they claimed that if two cars smash into each other at 100mph, there would be more carnage than one car smashing into a wall at 100mph, since in the one reference frame you'd start with twice the speed and have four times the energy. (They missed that half the energy would then be present at the end, the other factor of 2 is distributed equally among the cars.) – CR Drost Mar 9 '17 at 21:40
• Perfect , that makes sense. – Jason Mars Mar 9 '17 at 21:48

The two observers will observe different kinetic energies of the ice rink. When the ice rink slows down the mass, it must do so by giving itself momentum, even if only a very tiny amount. The difference in the kinetic energy of the ice rink observed by the two observers makes up for the difference in energy that you calculated, so they will agree on the energy dissipated by the ice rink.

Edit:

As I started to work through the math of this, I realized this is wrong. The energy of the rink partially explains the missing energy, but the dissipated energy measured by the two observers will not be exactly the same. In general, energy in two different reference frames are not equal.

• This really does sound reasonable, but Travis, how does the difference in kinetic energy gained by the ice rink relate to different observers? You see if you have different observers coming at different higher speeds for example $v$, $2v$, $3v$ etc. then we observe the difference in kinetic energy rising. I hope you can relate, clarify, and like to comment? – Jason Mars Mar 9 '17 at 3:08
• Precisely! And I have worked out the successive energies involved. But time and again I have failed to reach a definitive conclusion. – Jason Mars Mar 9 '17 at 4:34

Consider a small block (mass m) sliding along a second, larger one, of mass M. There are no other forces acting on M but the friction at the interface M-m.

1. Consider first a reference frame in which initially M is at rest and m moves with speed v. m moves initially with relative speed v in respect to M and finally both blocks move with $v_f$. We have $v_f=\frac{mv}{m+M}$ and the difference in kinetic energies $\Delta KE= \frac{(M+m)v_f^2}{2}-\frac{mv^2}{2}=- \frac{mv^2}{2}\frac{M}{m+M}$ In the limit M>>m this become $\Delta KE =- \frac{mv^2}{2}$

2. Now in a frame moving with the initial speed (v) of the mass m but in opposite direction. Initially m moves with 2v. M moves with v. Finally they both move with some speed $v'_f$.
$v'_f=\frac{2m + M}{m+M}v$ The change in KE will be $\Delta KE= \frac{(M+m){v'_f}^2}{2}-\frac{M v^2}{2}-\frac{4m v^2}{2}=\frac{(M+m)}{2}(\frac{2m + M}{m+M})^2v^2-\frac{M v^2}{2}-\frac{4m v^2}{2}=\frac{v^2}{2}\frac{4m^2 +M^2+4mM-Mm-M^2-4m^2-4mM}{M+m}=-\frac{mv^2}{2}\frac{M}{m+M}$

Again, for M>>m this reduces to $\Delta KE=- \frac{mv^2}{2}$

You could even consider a reference frame moving with v in the same direction as m. In this case m is accelerated from rest to some final speed and M slows down from v to the same final speed. The overall change in KE will be same.