Cap'n Refsmmat Posted June 21, 2008 Share Posted June 21, 2008 It isn't. But it's been accelerated in a direction opposite to the direction of the train's travel. From the reference frame of the train, it's moving backwards. From that of the rails, it's moving forwards, just more slowly. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 21, 2008 Share Posted June 21, 2008 It isn't. But it's been accelerated in a direction opposite to the direction of the train's travel. From the reference frame of the train, it's moving backwards. From that of the rails, it's moving forwards, just more slowly. If the ball is moving at 1 m/s in the opposite direction of travel of the train, why is the distance between the ball and the starting line continuously increasing as time increases? If the ball was moving in the opposite direction of travel of the train, the ball would eventually reach the starting line, correct? Link to comment Share on other sites More sharing options...
YT2095 Posted June 21, 2008 Share Posted June 21, 2008 ok, lets try This. Q: you stand on the back deck of the train that`s moving at 20m/sec, you throw a ball off the back deck at 20m/sec in the opposite direction the train is moving in (AKA, along the tracks). what happens to the ball in: 1) the frame relative to You the thrower? 2) the frame relative to the tracks? Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 21, 2008 Share Posted June 21, 2008 ok, lets try This. Q: you stand on the back deck of the train that`s moving at 20m/sec, you throw a ball off the back deck at 20m/sec in the opposite direction the train is moving in (AKA, along the tracks). what happens to the ball in: 1) the frame relative to You the thrower? 2) the frame relative to the tracks? You just gave me the ball's velocity and the direction of travel (back towards the starting line). If the train is traveling 20 m/s away from the starting line, and the ball is 100 meters away from the starting line when you throw it, the ball will reach the starting line in 5 seconds, because you already told me the ball was traveling 20 m/s towards the starting line. 100/20=5. Link to comment Share on other sites More sharing options...
YT2095 Posted June 21, 2008 Share Posted June 21, 2008 FTW???? I didn`t mention a starting line did I? you cannot possibly extrapolate Any such data that you did without pulling it out of thin air! now, Please answer both my questions using ONLY the data given. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 21, 2008 Share Posted June 21, 2008 FTW???? I didn`t mention a starting line did I? you cannot possibly extrapolate Any such data that you did without pulling it out of thin air! now, Please answer both my questions using ONLY the data given. You gave me the ball's direction of travel (opposite the train's direction of travel), and the ball's velocity (20m/s). The velocity of the train in the opposite direction is totally irrelevant, as at the moment of release of the ball, the ball travels 20 m/s away from that POINT on the tracks. What the train does is irrelevant. From the time you throw the ball, both the ball and the train will be 20 meters away from that point (ground zero) in one second but in opposite directions of ground zero. Link to comment Share on other sites More sharing options...
YT2095 Posted June 21, 2008 Share Posted June 21, 2008 yeah yeah, whatever, what will you See as an observer from both frames I asked you about? ("Reference points" as you call them). Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 21, 2008 Share Posted June 21, 2008 yeah yeah, whatever, what will you See as an observer from both frames? ("Reference points" as you call them). If I was on the train throwing the ball, and I observed the distance between myself and the ball after one second, there would be 40 meters between us. My release point would be the halfway point between myself and the ball at all times. Link to comment Share on other sites More sharing options...
YT2095 Posted June 21, 2008 Share Posted June 21, 2008 no no no, that`s NOT what I asked, I didn`t mention a 1 second duration did I? so, if you would simply answer the question, we can probably get this all cleared up Nicely, or at least Identify your problem a bit better. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 21, 2008 Share Posted June 21, 2008 ok, lets try This. Q: you stand on the back deck of the train that`s moving at 20m/sec, you throw a ball off the back deck at 20m/sec in the opposite direction the train is moving in (AKA, along the tracks). what happens to the ball in: 1) the frame relative to You the thrower? 2) the frame relative to the tracks? What do you mean what happens to the ball? The distance between myself and the ball increases as time elapses. I'm not trying to be funny, I just don't understand your question. We are talking about measured motion, correct? Motion occurs over a duration of time, correct? Would you like to know the distance between myself and the ball at a specific duration? Link to comment Share on other sites More sharing options...
YT2095 Posted June 21, 2008 Share Posted June 21, 2008 simple answers; for Q,1 you would see the ball behave as normal, as if you were stationary throwing it. for Q,2 an Outside stationary observer would simply see the ball plop to the floor as if it was just dropped. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 21, 2008 Share Posted June 21, 2008 simple answers; for Q,1 you would see the ball behave as normal, as if you were stationary throwing it.for Q,2 an Outside stationary observer would simply see the ball plop to the floor as if it was just dropped. You gave me the ball's velocity of 20m/s. Are you now telling me the ball's velocity was actually zero? According to your scenario, how far would I be away from the ball after the elapsed time of 1 second from release? Link to comment Share on other sites More sharing options...
YT2095 Posted June 21, 2008 Share Posted June 21, 2008 well, if you were the ball thrower, after 1 second it would be 20m away from you, and assuming the ground and train are all normal height etc... it will also be on the floor at that time too, and Perfectly stationary if it wasn`t a bounce ball that is. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 21, 2008 Share Posted June 21, 2008 well, if you were the ball thrower, after 1 second it would be 20m away from you, and assuming the ground and train are all normal height etc... it will also be on the floor at that time too, and Perfectly stationary if it wasn`t a bounce ball that is. So where does the 20m/s velocity of the ball come from? Link to comment Share on other sites More sharing options...
YT2095 Posted June 21, 2008 Share Posted June 21, 2008 the train moving away from the ball. all your doing by Throwing it, is Canceling the energy that the train gave it in the 1`st place. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 21, 2008 Share Posted June 21, 2008 the train moving away from the ball. That's the train's velocity, not the ball's. While you are holding the ball on the train moving at 20m/s, the ball is traveling 20m/s in the same direction. How did you come up with 20m/s in the opposite direction? If the ball hit the ground at release, the ball didn't travel any distance, except to the ground. the train moving away from the ball.all your doing by Throwing it, is Canceling the energy that the train gave it in the 1`st place. That's acceleration, not velocity. If the train was traveling to the right (side view) at 20m/s, and the ball was traveling in the opposite direction (left) at 20 m/s, they would be 40 meters apart after the elapsed time of 1 second. Link to comment Share on other sites More sharing options...
Sayonara Posted June 21, 2008 Share Posted June 21, 2008 I have already clearly shown that the ball can simultaneously have two different velocities, which are both correct answers. The posts from about #104 onwards are making this problem less than clear. Link to comment Share on other sites More sharing options...
YT2095 Posted June 21, 2008 Share Posted June 21, 2008 The posts from about #104 onwards are making this problem less than clear. don`t you mean Perfectly clear!? or do you have some problem with it? Link to comment Share on other sites More sharing options...
Sayonara Posted June 21, 2008 Share Posted June 21, 2008 You both seem to be going in different directions. I understand the problem we are discussing and I still found it confusing. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 21, 2008 Share Posted June 21, 2008 (edited) If the train is traveling East at 20m/s with a guy on the back of the train with a ball, the ball is also traveling 20m/s in the same direction. According to your observation of the scenario, all you did was "decelerate" the ball to 0m/s when you "threw it." You told me the ball had a velocity of 20 m/s in the opposite direction. That means the ball traveled WEST at 20m/s. Big difference there. Your 0 m/s ball falls to the ground, and the 20 m/s ball travels in the opposite direction of travel 20 meters per second. Excluding gravity and friction, the ball will continue to travel 20 m/s in the opposite direction. The ball will be 20 meters from the point of release after 1 second has elapsed. Edited June 21, 2008 by Motor Daddy Link to comment Share on other sites More sharing options...
Sayonara Posted June 21, 2008 Share Posted June 21, 2008 Has anyone else forgotten what we are trying to demonstrate? Link to comment Share on other sites More sharing options...
pioneer Posted June 21, 2008 Share Posted June 21, 2008 All these examples of relative motion are all showing how the same event looks different, depending on the reference. But there is a preferred reference which allows one to better know the relative motion within all the references in terms of energy conservation. For example, someone watching the train will see the train in motion. From the reference inside the train it looks like the country side is in motion. The second reference creates the illusion their reference is stationary even though all experiments will indicate it has kinetic energy. The amount of energy used by the trains engine will never be enough to make the entire country side move. It violate energy conservation if we consider inside the train a valid reference for anything beyond what is going on inside the train. But even that is suspect in terms of reality energy. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted June 21, 2008 Share Posted June 21, 2008 No. All inertial frames are equivalent and equally valid. Link to comment Share on other sites More sharing options...
Klaynos Posted June 21, 2008 Author Share Posted June 21, 2008 There is one reference point and that is the starting line. Nope, there are an infinite number of reference frames all of which are as valid as each other. There's 2 sensible ones for this situation that of the train and that of the platform. BOTH are equally valid, one says the train is moving, the other says that the platform is moving. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 21, 2008 Share Posted June 21, 2008 Nope, there are an infinite number of reference frames all of which are as valid as each other. There's 2 sensible ones for this situation that of the train and that of the platform. BOTH are equally valid, one says the train is moving, the other says that the platform is moving. Am I justified in saying I am stationary and the Sun is moving? Link to comment Share on other sites More sharing options...
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