# Kinetic Energy in different reference frames

Good morning,

I've got a strange little paradox I thought of that I just can't figure out. Imagine that you are building a machine that lets a ball fall in vertical direction from a height h, and converts the potential energy the ball loses in kinetic energy along a horizontal direction (e.g. a ramp). The ball has the Energy $E_0=mgh$ , therefore it will reach the velocity $v_0 =\sqrt{\frac{2E_0}m}$. No problem so far. But now imagine the machine is put in a car with velocity w. The energy the ball gains is still (from the non-moving frame of reference) $E_0$ and the final velocity that the ball has relative to the car is $$v_1= \sqrt{\frac{2(E_0+\frac12mw^2)}m}-w=\sqrt{\frac{2E_0}m+w^2}-w\not=v_0$$ That is in contradiction with Galilian reletivity, because in the inertial frame of the car the process is still the same as at the beginning, so the velocities should be equal. Where is my mistake? (I might calculate the same again in special relativity, I wonder what happens there)

• I have updated my answer – Farcher Sep 30 '17 at 13:59

You assume the ball changes its kinetic energy by $E_0$ in both cases. This is not what happens.

To change the horizontal momentum of the ball you need horizontal force. This force comes from the machine. Newton's third law states there's another force applied to the machine.

In the first case you silently assume the machine is so massive it barely moves. The $\text{force} \cdot \text{displacement}$ is its energy change and it may be neglected because the displacement is tiny. In this case you may say the kinetic energy of the ball changed by $E_0$.

In case of the moving car the machine barely changes its speed but its displacement in the new reference frame is huge. This means the machine loses or gains energy; it exchanges energy with the ball and now you cannot neglect it. So the change of the kinetic energy of the ball is not $E_0$ in this case.

### Strict calculations (classical)

Let's define the following:

• $m$ – mass of the ball.

• $M$ – mass of the machine (with car and everything else rigidly connected).

• $w$ – initial horizontal speed of the machine and the ball. The problem is now one-dimensional so there's no need to introduce vectors. Speed towards right will have opposite sign than speed towards left. This applies to all "vectors" from now on.

• $E_0$ – energy that is not kinetic energy initially, but finally it is.

• $v$ – final horizontal speed of the ball.

• $V$ – final horizontal speed of the machine.

Conservation of momentum:

$$(M+m)w = MV + mv$$

$$V = \frac{(M+m)w - mv}M \tag{1}\label{1}$$

Conservation of energy:

$$(M+m)\frac{w^2}2 + E_0 = \frac{mv^2}2 + \frac{MV^2}2$$

$$(M+m)w^2 + 2E_0 \stackrel{\eqref{1}}= mv^2 + M\left(\frac{(M+m)w - mv}M\right)^2$$

$$…$$

$$v^2 - 2wv + w^2 - \frac{2M}{m(M+m)}E_0 = 0$$

This is a quadratic equation w.r.t. $v$. The discriminant:

$$D = 4w^2 - 4\left(w^2 - \frac{2M}{m(M+m)}E_0\right) = … = \frac{8M}{m(M+m)}E_0$$

$$\sqrt{D} = 2\sqrt{\frac{2M}{m(M+m)}E_0}$$

And solutions:

$$v = \frac{2w \pm 2\sqrt{\frac{2M}{m(M+m)}E_0}}2 = w \pm \sqrt{\frac{2M}{m(M+m)}E_0}$$

$$V \stackrel{\eqref{1}}= w \mp \sqrt{\frac{2m}{M(M+m)}E_0}$$

You take '$+$' or '$-$' according to the way your ramp works. The important thing is: if you take '$+$' for $v$ then you must take '$-$' for $V$ and vice versa.

The changes of speeds:

$$\Delta v = v - w = \pm \sqrt{\frac{2M}{m(M+m)}E_0} \tag{2}\label{2}$$

$$\Delta V = V - w = \mp \sqrt{\frac{2m}{M(M+m)}E_0}$$

These values don't depend on $w$. According to Galileo this is what we expected, no paradox here.

Finally let's consider $M \gg m$. Then:

$$\Delta v \approx \pm \sqrt{\frac{2E_0}m}$$

$$\Delta V \approx 0$$

The first formula is almost your basic result. Almost, because you used '$=$' and this would be the limit for $M \rightarrow \infty$.

Consider the initial state of the ball. Its horizontal momentum and kinetic energy clearly depend on $w$:

$$p_b = mw$$

$$E_b = \frac{mw^2}2$$

This means we "see" some amounts of them in one frame of reference, some other amounts in another frame of reference. Nothing surprising here.

At this point let's "detach" symbols like $v$ and $m$ from our particular problem for a moment and analyze general formulas

$$p = mv$$

$$E = \frac{mv^2}2$$

like this:

$$dp = m dv$$

$$dE = mv dv \tag{3}\label{3}$$

The interpretation is: to increase speed by $dv$ you need to add some momentum $dp$ that doesn't depend on current speed $v$; but you have to add some energy $dE$ that does depend on current speed $v$. The larger $v$ the more $dE$ you need to add to achieve $v+dv$.

So in various frames of reference you need various amounts of energy.

Let's get back to your light ball and the massive machine. In the first case ($w = 0$) both objects gain velocity, so they gain kinetic energy. Almost all of the additional energy $E_0$ is transferred to the ball, the remaining tiny part is transferred to the machine.

In the second case, however, you can have $w$ arbitrarily large. Let the machine (ramp) work in a way that accelerates the ball further (i.e. $\Delta v > 0$). We already know from $\eqref{2}$ that $\Delta v$ doesn't depend on $w$. Still from $\eqref{3}$ we have:

$$\Delta E_b \approx mw\Delta v$$

With large enough $w$ it's easy to have $\Delta E_b \gg E_0$. Where does this energy come from?

As I said in my short answer above, this energy comes from the machine. In this case the machine acts as a source of energy because its velocity in this reference frame is reduced, so its kinetic energy is decreased.

In other words:

• In a frame of reference where $w = 0$ the machine gains tiny amount of kinetic energy and the ball gains moderate amount (about $E_0$) of kinetic energy. The net change in kinetic energy is exactly $E_0$.
• In a frame of reference where $w$ is large (in comparison to, say, the absolute value of $\Delta V$) the machine loses (or gains, depending on which way the ramp works) huge amount of kinetic energy and the ball gains even more (or loses almost as much, depending on which way…). Still the net change in kinetic energy is exactly $E_0$.
• (trivia) In a frame of reference where $\Delta V = -2w$ the speed of the machine changes from $w$ to $-w$, its kinetic energy remains the same and all $E_0$ is converted to kinetic energy of the ball.

It's now clear to me your assumption that the kinetic energy of the ball increases by $E_0$ was approximately good in your first case but totally wrong in general second case.

• One may say it's crucial that your ramp converts energy perfectly, without friction, without inducing ball rotation etc. I say it's not that important. If only a part of available energy is converted to kinetic energy then we shall simply call this part $E_0$.

• $E_0$ may be any kind of energy that can be converted to horizontal kinetic energy as long as this conversion doesn't alter horizontal momentum. E.g. it can be the energy of compressed spring or gunpowder.

• I think the whole phenomenon is similar to perfectly inelastic collision played in reverse. This answer to another question discusses energy in such collision and the conclusion is the same: in reference frame where $M$ is already moving you cannot neglect its kinetic energy.

• This is an interesting and instructive question and answer. If you can spare the time to do it, I think we could benefit from a detailed analysis. – garyp Sep 29 '17 at 13:21
• I tried to include this effect in my calculation, but I still can't make it work out. First, on may recognize that the car moves only horizontally, so only the horizontal force will matter. Furthermore, $\Delta p = \int F dt$ and $$E=\int F ds = \int F \frac{ds}{dt} dt=w\Delta p=wm v_0$$ – Intergalakti Sep 29 '17 at 15:09
• But inserted in the formula above, $$v_1=\sqrt{\frac1m(2E_0+wmv_0)+w^2}-w$$ Is there something else I miss or is my calculation just wrong? – Intergalakti Sep 29 '17 at 15:16
• @Intergalakti There's some math now in my expanded answer. – Kamil Maciorowski Sep 29 '17 at 23:45

An updated answer with the analysis of the problem considerably simplified by using the frame of the centre of mass of the ball and the machine first and allowing the ball to roll off the slope and move in a horizontal direction.

Let $m$ be the mass of the ball, $M$ be the mass of the machine and $w \hat i$ be the velocity of the centre of mass of the ball and the machine relative to some observer.

In the centre of mass frame, the ball and the machine start off at rest and then have velocities $v \hat i$ and $V \hat i$ such that using the conservation of momentum in the centre of mass frame

$m v \hat i + M V \hat i =0 \Rightarrow mv + MV =0$

The increase in kinetic energy of the ball and the machine in the centre of mass frame is $\frac 12 m v^2 + \frac 1 2 MV^2$ and this comes from the gravitational potential energy lost by the ball $mgh$.

Note that this shows that the $v_0$ in the question is not the same as $v$ in this answer.
If $M\gg m$ the difference is small but (possibly) significant.

Jumping out of the centre of mass frame the observer noted that the initial kinetic energy of ball and machine was $\frac 12 (m+M) w^2$ and the final kinetic energy of the system was

$\frac 12 m (v\hat i + w \hat i)^2 + \frac12 M ( V \hat i + w \hat i)^2 = \frac 12 (m+M) w^2 + \frac 12 m v^2 + \frac 12 M V^2 + mvw + MVw$

$mvw +MVw = (mv+MV) w = 0$ as momentum in the centre of mass frame was conserved.

${\rm KE}_{\rm final} = \frac 12 (m+M) w^2 + \frac 12 m v^2 + \frac 12 M V^2 = \frac 12 (m+M) w^2 + mgh$

${\rm KE}_{\rm initial} = \frac 12 (m+M) w^2$

So the energy books balance.

If $M \gg m$ then the machine acquires away a very small amount of kinetic energy compared with the machine.

Your $E_0$, the kinetic energy acquired by the ball, is not $mgh$, it is $\frac 12 mv^2 = mgh - \frac mM \frac1 2 m v^2$