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(I previously posted this on https://math.stackexchange.com/questions/2165992/problem-with-kinetic-energy-in-inertial-frames and it has been suggested I ask it here).

Suppose I slide a ball (in response to a now-deleted comment, it's an idealized ball which is frictionless and therefore slides rather than rolling so there is no rotational energy) down the passage and up a ramp at the end. If I release the ball with speed $v$ it acquires kinetic energy $T = (1/2) m v^2$ and, by conservation of energy. when it reaches the ramp it rises to a height $ h = T/mg = (1/2g)(v_{initial}^2 - v_{final}^2 )$ where $v_{final} = 0$ .So far so good.

Putting some numbers into this, I impart a speed of $2 m/s$ and $g = 9.81 m/s^2$. Then the ball will rise up approximately $h = 0.2m$ on the ramp.

What I didn't mention is that I am actually doing this in a train (if trains were good enough for Einstein then they're good enough for me). I'm sliding the ball in the forward direction and the train is moving at $20$ m/s. So now when I calculate the rise I get $ h = (1/2g)(22^2 - 20^2 ) = 4.81m$. How high does the ball actually rise ?

I can see that I probably need to impart more energy to the ball to accelerate it from $20 $ to $ 22 m/s$ than from $0 $ to $ 2$. What bothers me is that this seems to contradict the equivalence of inertial frames, and the speeds involved are hardly relativistic


As noted in the approved answer,the calculation above neglects conservation of momentum. So, take it that the train is moving at $20 m/s$ after I release the ball (otherwise there is an interaction to be considered when I launch the ball) and let the train have mass $M$.

Then the final speed of ball and train is given by $ u=\dfrac{20M + 22 m}{M+m}$ The energy equation is now the potential energy gained by the ball, $V = $ initial kinetic energy - final kinetic energy $ = 1/2 (M20^2 + m22^2) - 1/2(Mu^2 + m u^2) $ If one then slogs through the arithmetic (which I have on paper but don't have the persistence to type in) one ends with

$V = (m/2)(22 - 20)^2 M/(M+m) $

One should then note that the "stationary" situation actually needs a slight correction too for the momentum imparted to the world as the ball moves up the slope. In both cases, the mass of the world and the train are very large compared to the ball, and so $M/(M+m) \to 1$ and the PE gained by the ball is the same in both cases $= m.2^2/2$

A final thought on this is that when I launch the ball in the first place, conservation of momentum determines a small reduction in speed of the train (or the world) and I expect a detailed calculation would show that the amount of energy I need to impart would be the same in both cases.

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    $\begingroup$ Hello Tom, see if you are moving in a train at constant velocity, your frame is inertial, and train is at rest w.r.t you and you are at rest w.r.t train, so if you now throw the ball it would appear to move at $ 2 m/s $ w.r.t you because you think you are at rest( trains velocity contribution is 0 acc to you), so you can do the same calculation as you did before, and I am not a relativity guy but I think that if you measure the KE from outside say a person standing on a road measures it, he would find an increase in KE?? What do you say? $\endgroup$ Mar 1, 2017 at 8:12
  • $\begingroup$ @SarthakSharma Thanks. I'd say in that case how high does the person on the road expect it to rise ? $\endgroup$ Mar 1, 2017 at 8:15
  • $\begingroup$ I think the 2nd calculation that you did is valid from the point of view of the person in the road, two realities for two different viewpoint you see a rise of 2m where the person on the road sees it to be 4m, how does that work in relativity, its crazy $\endgroup$ Mar 1, 2017 at 9:03
  • $\begingroup$ @SarthakSharma I think the truth is emerging in the answer provided by Farcher, and we can expect the laws of physics to hold across inertial frames. $\endgroup$ Mar 1, 2017 at 9:11
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    $\begingroup$ @ASuckerinMaths In a closed system momentum is conserved. When the ball comes to "rest" in the train it is moving with the same speed as the train. So let this final speed be $u$. Then the momentum equation is momentum after = momentum before, i.e. $(M + m)u = M.20 + m.22$ This then gives a value for $u = (M.20 + m.22)/(M + m)$ that can be substituted in the energy equation, energy after = energy before, i.e. $V + (Mu^2 + m u^2)/2 = (M20^2 + m22^2)/2$ where $V$ is the potential enegry gained by the ball rising. $\endgroup$ Mar 2, 2017 at 10:25

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What you have neglected is the fact that momentum has to be conserved in the horizontal plane and as the ball rises it imparts some of its horizontal momentum to the carriage and the missing kinetic energy is the extra kinetic energy that the carriage has collected.

So the final velocity of the ball and the carriage is not $20$ m/s it is more than that.

If the mass of the train is $M$, the mass of the ball is $m$ and the final horizontal velocity of the train and ball is $V$ then applying conservation of momentum in the horizontal plane:

$M20+m22 = (M+m)V \Rightarrow V=\dfrac{20M + 22 m}{M+m}= 20\left( 1+\dfrac{11m}{10M}\right )\left(1-\dfrac mM\right )^{-1}>20$

As an relatively easy to analyse illustration of what is happening consider the ball hitting the end of carriage wall and suffering an elastic collision.
You will find that the rebound speed of the ball relative to the ground not $18$ m/s but greater than that and the final speed of the carriage relative to the ground is greater than $20$ m/s.

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  • $\begingroup$ Ah .... and I presume that if one completes the calculation then the ball rises to the same height as in the stationary case ? $\endgroup$ Mar 1, 2017 at 8:46
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The general assumption that the energy of the ball is conserved is incorrect. The train takes up energy and momentum. The energy it takes is equal to:

$Mv\delta v + 1/2 M \delta v^2 $.

the reason you can ignore this amount of energy sometimes is that due to conservation of momentum $\delta v$ goes like 1/M for large masses and hence the second term is 0 and only the first term contributes for large M than. This is however 0 in the rest frame only, which leads to the apparent contradiction above. So as soon as the mass of the ramp/train/earth is not infinite OR you leave the rest frame of the ramp/train/earth your system (the ball) loses energy. Importantly either one suffices. To see that it all works out correctly you than have to do the explicit calculation you have done above.

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