# Frame of References

1. Feb 11, 2004

### mindcircus

A train moves along the tracks at a constant speed u. A woman on the train throws a ball of mass m straight ahead with a speed v with repsect to herself. What is the kinetic energey gain of the ball as measured by a person on the train? By a person standing by the railroad track? How much work is done by the woman throwing the ball and the train?

I'm not completely sure I have these concepts down.

With respect to the woman, wouldn't kinetic energy just be (1/2)mv^2? I think this because the woman, ball, and reference frame are all moving at the same speed, initially. But for someone on the side of the tracks, I know the same logic can't apply. The woman and ball are moving at the speed of the train, while the reference frame is stationary...I'm not quite sure where to go with this.
Work is just the change in kinetic energy. But I don't know how to deduce the work done by the train. The ball is moving with the horizontal speed of the train. Do I just use that speed in the equation of KE? I'm not really sure how to figure this out.

2. Feb 12, 2004

### HallsofIvy

Yes, relative to the woman, the kinetic energy gained by the ball will be (1/2) mv2. To a person standing on the ground, the original speed of the ball was u and the new speed was u+v. The change in kinetic energy observed by the person on the ground is still (1/2)mv2 since the speed of the ball has still changed by v. (And (1/2)mv2 is the work done by the woman in either reference frame.)

3. Feb 12, 2004

### LURCH

And would this also be the work done by the train? When the woman throws, the force of the throw is translated down her body through her feet to the train. This becomes a rear-ward acceleration equal to the force of the throw. To maintian a constant speed, the train's engine must provide a forward acceleration of equal force, right?

The way the question is phrased, I don't think it could be asking the work done by the train to accelerate the ball from stationary (relative to the otuside observer) to u+v.

4. Feb 12, 2004

### krab

That doesn't follow. To a person on the ground, the change in kinetic energy of the ball is
$$\Delta E_{b,g}={1\over 2}m(v+u)^2-{1\over 2}mu^2={1\over 2}mv^2+mvu$$
The work done by the woman is the same whether observed from the ground or from the train: $(1/2)mv^2$. The extra work done by the train, as pointed out by LURCH, is $mvu$. You can do this by conservation of momentum. The train loses momentum of $mv$ and so in absence of power applied, would slow down by speed increment $mv/M$, where M is the mass of the train. Convert that to energy lost by the train and we find the missing $mvu$.

5. Feb 12, 2004

### HallsofIvy

You know, it occured to me as soon as I clicked the "Submit Reply" button that (u+v)2 is not u2+v2. You are completely correct.