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Consider an elevator moving down with uniform velocity. A person standing inside watches an object fall from the ceiling of the elevator to the floor. Say the height of the elevator is $h$. Then the work done by gravity in that frame of reference should be $mgh$. But consider this same event being watched by someone else in the stationary frame of reference. In his reference frame, the object travels a larger distance as it falls from the ceiling to the floor of the elevator because the floor itself is moving downwards (one can calculate this extra distance covered to be $u \sqrt{\frac{2h}{g}} $) and hence the change in kinetic energy should be more in that frame than in the moving frame!

I just can't seem to figure out where I'm going wrong here. I'm probably missing something very obvious.

So I would be very grateful if anyone could explain this to me.

Edit: Okay, let's say the object is a clay ball and it collides with the floor inelastically such that it's kinetic energy is converted into heat. In the moving frame of reference the heat would be simply equal to $\frac{1}{2}mv^2$ which is equal to $mgh$. In the stationary frame of reference it would be equal to $\frac{1}{2}mv^2-\frac{1}{2}mu^2$ since the ball after colliding is moving with speed $u$. This can be calculated to be equal to $mgu\sqrt{\frac{2h}{g}} + mgh$ which is clearly greater than the heat produced in the frame attached to the elevator and this is a contradiction because the heat measured in any frame should be the same.

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The energy dissipated is not equal to $\frac{1}{2}mv^2 - \frac{1}{2}mu^2$. The correct expression is $\frac{1}{2}m(v-u)^2$. –  Johannes Dec 24 '12 at 18:04
    
@Johannes What makes you say that? I'm pretty sure it's the other way round. –  Alraxite Dec 24 '12 at 18:18
    
The energy dissipated in a fully inelastic collision is determined by the impact velocity in the center-of-mass frame. In any other frame one has to consider also the kinetic energy transferred. –  Johannes Dec 24 '12 at 19:17
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4 Answers 4

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Boy, this was tricky, but the secret is in conservation of momentum.

See, you are assuming that, after the collision, the velocity of the ball-elevator ensemble is $u$, but this is not fully true: it will be $u' = u + \frac{m}{m+M}\sqrt{2gh}$, $M$ being the mass of the elevator. Of course if $M \to \infty$ that reduces to $u' = u$, but when computing the KE, something funny happens:

$$\frac{1}{2}(m+M)u'^2 = \frac{1}{2}(m+M)u^2 + \frac{m^2}{m+M}gh + um\sqrt{2gh}$$

That last term which does not depend on $M$ is the key here. Of course the first term, with the $(m+M)$ dominates the others, but it will be cancelled out by identical terms in the KE before the collision. But if you assume that because $M \to \infty$ you can take $u' = u$, you will be missing this last term, which exactly cancels out that extra energy.

Doing the math for a finite elevator mass, and using conservation of momentum to compute the final velocity, you eventually get to energy lost in an inellastic collision to be $\frac{1}{2}\frac{mM}{m+M}(u-v)^2$, which for $M \to \infty$ reduces to $\frac{1}{2}m(u-v)^2$, as Johannes already pointed out.

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This is the right answer. –  Nathaniel Dec 25 '12 at 1:33
    
Thanks! Just one thing in reference to my question before the edit: so it's true that gravity does more work in the stationary frame than in the elevator frame of reference? I guess the answer is yes, but here's the counterintuitive thing: say I am in space and I apply a force $F$ on an object through a distance $s$. In my frame of reference, I lost potential energy equal to $F.s$. In a frame moving with speed $u$, the object moved an extra distance of $u\sqrt{2.\frac{m}{F}.s}$ and hence more food in my stomach got converted into mechanical energy in that frame. Isn't this a paradox? –  Alraxite Dec 25 '12 at 14:06
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There is no such thing as conservation of energy between intertial reference frames. (The kinetic energy of a car is larger in any intertial frame that is not it's own restframe)

Considering the observer inside the elevation, the free fall takes $t_f=\sqrt{\frac{2h}{g}}$, after which is has velocity $gt_f$, and thus kinetic energy $mgh$ (which is cheating, as this is what you used to calculate $t_f$ in the first place. However, the total energy of the particle itself is conserved within this frame, between two times.

Now consider the external observer. It sees an increase in kinetic energy of $$ \Delta K = \frac{1}{2}m(u+\sqrt{2gh})^2-\frac{1}{2}mu^2 $$ Which simplifies to: $$ \Delta K = \frac{1}{2}m\left(u^2+2u\sqrt{2gh}+2gh\right)-\frac{1}{2}mu^2$$ $$ = mu\sqrt{2gh}+mgh$$

Where the first term is related precisely to the additional difference in height that you calculated!

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You may want to read my edit. –  Alraxite Dec 24 '12 at 13:34
    
@Alraxite The point here, is that in-elasticity is not really well-defined. You have to take into account that in the second case, the clay-ball hits a moving surface. In other words, you should look at it in terms of momentum and momentum-loss rather than energy. –  Bernhard Dec 24 '12 at 13:43
    
Relative to the elevator, the ball approaches the floor with the same speed in both the reference frames. One could define an inelastic collision with a stationary object as a collision in which the moving object loses all its energy to heat. I don't see a problem... –  Alraxite Dec 24 '12 at 14:23
    
@Alraxite Hmm. Another example. Two cars colliding heads on traveling 30 km/h each in opposite direction, or one standing still and the other traveling 60 km/h. These situations are different. Have to think about the relation to your problem. –  Bernhard Dec 24 '12 at 14:38
    
As it so happens I have seen this question on this site before, here's the link: physics.stackexchange.com/questions/45578/… The highest-rated answer nicely explains that why the change in kinetic energy is same in both the reference frames. –  Alraxite Dec 24 '12 at 14:43
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An observer in the lift and an observer at rest in the building observe the same energy being transferred in heat:

A clay ball with mass $m$ drops from the ceiling in an elevator and hits the floor. The elevator has height $h$ and moves downward with uniform speed $u$. An observer at the ground floor observes the clay ball to fall with speed $u+g t$. When it hits the floor, the clay ball has spend a time $\sqrt{\frac{2h}{g}}$ falling and moves with speed $v = u+\sqrt{2hg}$. The lift floor moves with speed u, and the energy dissipated in the inelastic collision corresponds to the velocity difference between the two: $\frac{1}{2} m (v-u)^2 = \frac{1}{2} m (2hg) = mgh$. The velocity $u$ has dropped from this equation, expressing the fact hat he energy transferred is independent of the frame of reference chosen.

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Why do you square the difference of velocities and not take the difference of the squares? –  Jaime Dec 24 '12 at 17:53
    
The speed at which the clay ball and the lift floor collide is $v-u$. The energy dissipated corresponds to the square of this collision speed. –  Johannes Dec 24 '12 at 17:58
    
You are right in the collision energy depending on $(u-v)^2$, but that is a result far from obvious that comes from including conservation of momentum in the derivation, see my answer. –  Jaime Dec 24 '12 at 19:28
    
Hmm.. I think this is pretty obvious. In treating an inelastic collision in any frame other than the center-of-mass frame one has to consider the kinetic energy transferred. –  Johannes Dec 24 '12 at 19:34
    
It isn't obvious at all - that's the entire point of the question. –  Nathaniel Dec 25 '12 at 1:29
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I think the thing to realise here is that changes in kinetic energy aren't independent of the reference frame.

To see this, consider a mass of $2\,\text{kg}$ that accelerates from rest to a speed of $1\,\text{ms}^{-1}$ over some period of time. Its kinetic energy has changed by $\frac{1}{2}m(1^2-0^2)=1\,\text{J}$. But now consider the same mass viewed by someone travelling at $10\,\text{ms}^{-1}$ in the opposite direction. Now the mass's kinetic energy has changed by $\frac{1}{2}m(11^2-10^2)=21\,\text{J}$.

This is what's happening when you consider the two different reference frames regarding the object falling in the lift. It does indeed gain more kinetic energy in the moving frame, and I'm sure that if you calculated $\frac{1}{2}m(v_1^2-v_0^2)$ in the moving versus the stationary frames of reference, the difference would also come out to $u\sqrt{\frac{2h}{g}}$. This is counterintuitive, but it has to be that way in order for energy to be conserved.

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What's the downvote for? I admit this doesn't add much beyond the other answers, but it was written before they were posted, and also before the edit in the question. –  Nathaniel Dec 25 '12 at 1:23
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