Topic
A physics problem
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A pitcher can throw a baseball at 40 m/s. If the ball weighs 0.145 kg the ball has an energy of
0.145 * 40 * 40 / 2 = 116 Joule. Now put the same pitcher in a supersonic airplane going at 1000 m/s. The ball has an energy of 0.145 * 1000 * 1000 / 2 = 72.5 kJ. What happens if the pither throws the ball forward in the plane 40 m/s relative to the plane? The energy of the ball is now 0.145 * 1040 * 1040 / 2 = 78.416 kJ. The ball gained 5916 J when it was thrown. That's a lot. Where did it all come from? Was the ball heavier to throw just because he was in a plane? Remember, energy must be preserved. Yes, I know the answer. Do you? |
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Edited by Moonshadow on 24/02/12 15:05 (GMT)
What happens if the pither throws the ball forward in the plane 40 m/s relative to the plane? you answered the question yourself :3 The extra gain comes from the way exponential functions behave at high values. The difference between (x-1)^2 and x^2 becomes bigger and bigger as X increases. There is no extra weight or anything of that matter. Just mathematical things. |
No, it's not that simple. It's not a math question. It's pure physics, and one of the laws of physics is the conservation of energy. The ball really did gain that much energy when the pitcher threw it, and it's a lot of energy, much more than any human can produce in that short time. The energy is measurable, for example by letting the ball hit a target or something. This sounds like a paradox, and that why I think it's a fun problem. :-) How did the ball get all that energy? |
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I already told you.
you said the ball has 40 m/s relative to the plane since the plane moves at 1000 m/s, it means the ball has 1040 m/s The pitcher already had kinetic energy, so did the ball, because they were in the plane, so when the pitcher throws it, the ball gains extra speed. |
I already told you. So how much energy did the pitcher spend in the throw? |
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116
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Then there are 5800 joules missing. Where did they come from?
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It was there all along.
It didn't come from anywhere. |
It was there all along. The ball gained 5916 joules in the moment of the throw. They were not there before. |
yes they were. the ball was already moving when the throw occurred. they were in a plane remember? |
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And I thought this was a nerd forum. Nerds know physics.
Here is the answer: The energy comes from the airplane losing speed. It will be moving less than 1000 m/s after the throw. This lost kinetic energy is where the 5800 missing joules come from. The lesson to be learned is that while the total energy must be preserved, energy transfer is not. In the reference frame of the plane the ball gains 116 joule, and in the reference frame of the ground it gains 5916 joule. In another reference frame it even lost 1000000000 joule, etc. You can get any number you want. But the total energy of the system never changes. |
I'm sorry. what? how is the plane losing any speed? perhaps you mean relative speed to the ball? The pitcher is inside the plane (in the plane?), throwing the ball inside the plane automatically gives it the same kinetic energy the plane has plus the boost from the throw.
and your point is...? if you actually read what i said, you will notice i said the exact same thing. and just for teh nerds: there is no such system that doesn't lose energy. your system is losing energy all the time, which is why the conservation of energy is not applied in practical physics. |
I'm sorry. No relative to the ground. The plane is loosing speed because in a closed system the total momentum cannot change. If the ball gains speed something else has to lose speed. The pitcher pushes the plane back when he throws the ball.
No. You said that the energy didn't come from anywhere, which is wrong. |
aahhh, I see its a play on words. Yes. Or I would say it's a play with reference frames. :-) |
It was there all along. what i said here was that the ball already had some speed thus some kinetic energy when it was thrown. the fact that the ball was standing still (in relation to the frame, just so you understand) made it look as if it had none, but it actually did have some in relation to the ground. When the throw occurred, the ball already had some speed, and the added one was the one gained from the throw, thus it had:
in relation to the plane and you said the ball has 40 m/s relative to the plane I don't quite get what you didn't understand. I explained it quite clearly here... just so that you know, in physics, when the reference point is not mentioned, it is always the ground, by convention. |
I don't quite get what you didn't understand. No you didn't. That the ball has the speed of the plane plus some gained by the throw is quite obvious. It's also very simple to calculate how much energy the ball gained. But the question was where did the energy come from. It didn't come from the person throwing it (who added only 116 joule in ALL reference frames), so it must come from somewhere else. That was the whole point of the exercise, and you didn't answer it. And yes, I'm using the ground as reference. |
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Edited by stonberg on 15/03/12 17:16 (GMT)
If so, then the speed of the ball relative to the ground will not be 1040 m/s, which is what you stated above. If the mass of the plane is Mp kg, then its speed, Vp, after the ball is thrown will be given by: Vp = ((1000 * Mp) - (40 * 0.145)) / Mp So your kinetic energy calculation of the ball after the throw in the plane should be: Ke Ball = 0.5 * 0.145 * (40 + (((1000 * Mp) - (40 * 0.145)) / Mp)) ^ 2 Sorry, but your question was inaccurate and misleading. |
I disagree about the misleading part. The figures are accurate taking the number of significant digits into account. If you assume some reasonable weight for the supersonic plane (say 5000 kg) the speed of the plane after the throw will be 999.99884 m/s. I have no problem writing this as 1000 m/s. It's a physics problem after all, and rounding the figures doesn't change the answer to "where does the energy come from". |
