Timeline for Is this a way to distinguish between a gravitational field and an accelerated rocket?
Current License: CC BY-SA 4.0
30 events
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| Mar 18, 2020 at 21:32 | comment | added | user255961 | @Dale i sent you a message | |
| Mar 18, 2020 at 20:43 | comment | added | Dale | @Yassin-98 Sorry, it is not personal but I just don’t give out private contact information online, so those two sites are the way to reach me. | |
| Mar 18, 2020 at 20:39 | comment | added | user255961 | @Dale but can i text you there? | |
| Mar 18, 2020 at 20:22 | comment | added | Dale | @Yassin-98 I mostly hang out at www.physicsforums.com which is a discussion-oriented site rather than a question-answer site | |
| Mar 18, 2020 at 20:21 | comment | added | user255961 | @Dale Where can i message you? You answered me very well but i probably found another way to distinguish between an accelerated frame and a gravitational field. | |
| Mar 14, 2020 at 0:34 | comment | added | Dale | @Yasdin-98 “if U2(velocity of ball before collision)=0, and V1=15 m/s, then V2=30 m/s” Yes, but I have no idea what you are talking about for the rest of your comment. | |
| Mar 13, 2020 at 4:12 | comment | added | user255961 | @Dale so if U2(velocity of ball before collision)=0, and V1=15 m/s, then V2=30 m/s, passing one second, V1 will be 24'8 m/s, Does that mean the distance(height) will be 5'2 meters?. | |
| Mar 13, 2020 at 3:31 | comment | added | Dale | @Yassin-98 in your notation it is V2=-U2+2*U1 which is different than what you wrote. You asked “Doesn't that mean the ball will reach different heights in eyes of the inside observer, through time?” No, because in the eyes of the inside observer U1=0 always. It is only for outside observers that U1 increases. So we always get V2=-U2, the ball bounces up at the same speed that it landed, just like in gravity | |
| Mar 13, 2020 at 3:23 | comment | added | user255961 | @Dale i have a last question. If V1( velocity of the rocket after collision)=U1(velocity of the rocket before collision), then V2=2×U1, that means the velocity of the ball will always double the velocity of the rocket. Now, Doesn't that mean the ball will reach different heights in eyes of the inside observer, through time? Because the time of the floor catching a ball going at 20 m/s won't be the same than catching a ball going at 40 m/s or 60 m/s. | |
| Mar 11, 2020 at 1:10 | comment | added | Dale | @Yassin-98 because if you use that information now you find only the “miss” solution. You need to leave it as an unknown for now so that you can find both solutions and discard the miss one. | |
| Mar 11, 2020 at 0:44 | comment | added | user255961 | @Dale But why V2' should be unknow if we know it will be 15 m/s since the ball won't reduce the speed of the rocket? | |
| Mar 11, 2020 at 0:11 | comment | added | Dale | @Yassin-98 you solved for a miss. V1’ and V2’ should both be unknowns. Use both conservation of momentum and conservation of KE. That gives you two equations so you can solve for V1’ and V2’. You will get two solutions. One is a miss, take the other solution. See en.m.wikipedia.org/wiki/Elastic_collision | |
| Mar 10, 2020 at 23:50 | comment | added | user255961 | @Dale (1×0)+(1000×15)=(1×V1')+(1000×15) -> V1'=0, V2 is the velocity before collision of the rocket and V2' is the final, so if V2=15m/s, and if V2=V2', then V1'(final velocity of the ball)=0. (I made a mistake in writing the equation in the last coment, it isn't M2M2, it's M2V2) | |
| Mar 10, 2020 at 23:30 | comment | added | Dale | @Yassin-98 you must have made a mistake. That equation (conservation of momentum) is included inside the equation I posted (elastic collision). | |
| Mar 10, 2020 at 23:02 | comment | added | user255961 | @Dale, Can you tell me why if i use this equation, M1V1 + M2M2 = M1V1' + M2V2', the result of final velocity is 0? | |
| Mar 10, 2020 at 14:55 | comment | added | eagle275 | @Yassin-98 I doubt that .. the rockets mass is replaced by the apparent mass which is "responsible" for 1 G = earths mass and therefore constant ... | |
| Mar 10, 2020 at 14:22 | history | edited | Dale | CC BY-SA 4.0 |
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| Mar 10, 2020 at 14:07 | history | edited | Dale | CC BY-SA 4.0 |
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| Mar 10, 2020 at 12:32 | comment | added | David Z | I've deleted a number of inappropriate comments and responses to them. | |
| Mar 10, 2020 at 3:44 | history | edited | Dale | CC BY-SA 4.0 |
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| Mar 10, 2020 at 3:37 | history | edited | Dale | CC BY-SA 4.0 |
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| Mar 10, 2020 at 3:26 | history | edited | Dale | CC BY-SA 4.0 |
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| Mar 10, 2020 at 2:13 | comment | added | Dale | Comments are not for discussion. If you need discussion then you should go to a discussion forum, not a question and answer forum. In any case, the conservation of momentum formula is also included in the formula I posted. Both KE and momentum are conserved (that is how the elastic collision formula is derived). Your idea that they somehow don’t apply here is flat out wrong. You should discard it immediately. For now, you really need to focus on learning the very basics. | |
| Mar 10, 2020 at 1:00 | comment | added | user255961 | The equation of conservation obviously it's true, but It gives us a result which doesn't seem right. That's because we used the wrong tool to do the job, as the standard collision formula, it doesn't work here. But logicaly, i think the rocket with more mass will make the ball go further, and as i said, again, more kinetic energy and force isn't the same as less kinetic energy and force. So assuming the floor will hit the ball, it's just that my mind can't undestand how something with more energy equals something with less energy. | |
| Mar 10, 2020 at 0:57 | comment | added | user255961 | I undestand what you put in your answer, as Einstein said, acceleration will not change no matter what. I'm acctually think that's the standard collision formula doesn't work here, because here, it's different, since v1=u1 as you said that means 0. But what if we use the conservative equation? M1V1 + M2V2 = M1V1' + M2V2'. Using this equiation the result of the final velocity of the ball is 0, which means the ball will just stuck to the floor and won't bounce, because, V1 of M1 = V1' of M1. But here, we assume that the ball will be hitted by the floor and will bounce | |
| Mar 10, 2020 at 0:08 | history | edited | Dale | CC BY-SA 4.0 |
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| Mar 9, 2020 at 23:55 | comment | added | Dale | The KE and force is already included in the standard collision formula that I posted. All of those effects are considered, and what I described is the result. You seem to forget that although the more massive rocket brings in more KE it also carries more KE away (in other frames). In the end you get what I posted by considering the mass. | |
| Mar 9, 2020 at 23:14 | comment | added | user255961 | Even if the ball don't reduce the acceleration of the rocket, Don't you think a rocket with, for example, 5000kg going at 1G, will make the ball go further(go at more speed) than another with 500kg going at 1G ( assuming in both cases the balls don't reduce the acceleration)? Because kinetic energy and force are more in the one with 5000kg, than the one with 500kg. | |
| Mar 9, 2020 at 23:11 | history | edited | Dale | CC BY-SA 4.0 |
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| Mar 9, 2020 at 23:04 | history | answered | Dale | CC BY-SA 4.0 |