scilearner Posted May 26, 2010 Share Posted May 26, 2010 (edited) If I push a spring on a stationary wall and release, the spring would fly in the opposite direction right? I understand that there is elastic potential energy stored and this causes the movement but when I normally compress a spring it springs back to its orginal shape, how does it fly away in other direction. For example when I compress a spring against a wall, why doesn't it regain its original shape and fall down, why does it fly away instead. I'm thinking the elastic potential energy is used to make the spring go back to its orginal shape. Thanks I'm trying to understand newton's third law using this. If I push a book, it will push back on me. How does this occur using spring example. When I push the book does my hand compress a bit and then I move back in the other direction, ok but isn't the book compressed a little bit as well, wouldn't the make the book spring back and move back. How does the book move foward, if I compress it, shouldn't it spring back. Edited May 26, 2010 by scilearner Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted May 26, 2010 Share Posted May 26, 2010 If I push a spring on a stationary wall and release, the spring would fly in the opposite direction right? I understand that there is elastic potential energy stored and this causes the movement but when I normally compress a spring it springs back to its orginal shape, how does it fly away in other direction. For example when I compress a spring against a wall, why doesn't it regain its original shape and fall down, why does it fly away instead. I'm thinking the elastic potential energy is used to make the spring go back to its orginal shape. Thanks In order for it to do this it gains momentum. Similarly, when you leap off the ground you do not stop immediately on cessation of the push against the ground. Link to comment Share on other sites More sharing options...
scilearner Posted May 26, 2010 Author Share Posted May 26, 2010 In order for it to do this it gains momentum. Similarly, when you leap off the ground you do not stop immediately on cessation of the push against the ground. Thanks for the reply Gains momentum from the wall? If the wall is stationarry before and after the collision. It's change in momentum is zero. So how does the ball get a change in momentum? Link to comment Share on other sites More sharing options...
swansont Posted May 26, 2010 Share Posted May 26, 2010 Thanks for the reply Gains momentum from the wall? If the wall is stationarry before and after the collision. It's change in momentum is zero. So how does the ball get a change in momentum? It has an essentially infinite mass, since it's attached to the earth. The speed it has afterwards is immeasurably small. Link to comment Share on other sites More sharing options...
scilearner Posted May 26, 2010 Author Share Posted May 26, 2010 (edited) It has an essentially infinite mass, since it's attached to the earth. The speed it has afterwards is immeasurably small. Thanks swansoft Ok that makes sense. So in this situation does the ball lose or gain momentum. I understood your reason for why the wall doesn't move but this is my problem I don't understand how direction of movement after a collison is determined. EDIT: New question is below. Edited May 26, 2010 by scilearner Link to comment Share on other sites More sharing options...
Leader Bee Posted May 26, 2010 Share Posted May 26, 2010 Let's say object A was travelling at 1m/s and Object B was travelling at 1m/s. They collided after that object A travelled at -1m/s. So the change of momentum of object A is -1-1= -2 . How can an object move at a speed less than nothing? Link to comment Share on other sites More sharing options...
scilearner Posted May 26, 2010 Author Share Posted May 26, 2010 (edited) How can an object move at a speed less than nothing? Velocity I mean. Ignore the previous posts. This is the question I have. I was finally able to word it. Now let's say there was an object with initial momentum of 1 (m=1 v=1)and final momentum of -1(v=-1). That means change of momentum is -1-1= -2. Now if I consider the change in kinetic energy of the object 1-1 is zero. This is my biggest question, how come there is a change of momentum with out a change in energy. Edited May 26, 2010 by scilearner Link to comment Share on other sites More sharing options...
swansont Posted May 26, 2010 Share Posted May 26, 2010 Keep in mind that kinetic energy will only be conserved in elastic collisions. So one must use specific examples — you can't just make up numbers and hope they will work. Two objects, each traveling at 1 m/s, will never collide. Objects of equal mass, moving at 1 m/s and -1 m/s can collide and recoil at the same speed. That would be elastic. or they can collide and come to rest, which would be completely inelastic. In either case, the momentum is zero. Link to comment Share on other sites More sharing options...
scilearner Posted May 26, 2010 Author Share Posted May 26, 2010 Keep in mind that kinetic energy will only be conserved in elastic collisions. So one must use specific examples — you can't just make up numbers and hope they will work. Two objects, each traveling at 1 m/s, will never collide. Objects of equal mass, moving at 1 m/s and -1 m/s can collide and recoil at the same speed. That would be elastic. or they can collide and come to rest, which would be completely inelastic. In either case, the momentum is zero. Hey thanks swansont. Why is in either case momentum is zero. -1 -1=-2. Let's say if a ball was coming towards you and you hit it a with a bat, then the ball flies in opposite directions. If the balll came at you at 1 m/s (direction towards you) and went away from you at -1m/s why is change of kinetic energy zero. Could you show me from examples or otherway. Thanks Link to comment Share on other sites More sharing options...
swansont Posted May 26, 2010 Share Posted May 26, 2010 Hey thanks swansont. Why is in either case momentum is zero. -1 -1=-2. In the example I gave, the balls had equal magnitudes of momentum, but were traveling in opposite directions. The vector sum is zero. In your example, in order to change the direction, there must have been a collision. You need to include the other object in the discussion. In that case, it only works if the mass of the other object approaches infinity, and thus has a velocity close to zero. In any event, the final speed of the first object can only be arbitrarily close to -1. Let's say if a ball was coming towards you and you hit it a with a bat, then the ball flies in opposite directions. If the balll came at you at 1 m/s (direction towards you) and went away from you at -1m/s why is change of kinetic energy zero. Could you show me from examples or otherway. Thanks You have to define your system properly. F = dP/dt, so if there is no net external force on the system, the momentum will not change. You hit it with a bat, which exerts a force. The momentum must change in that situation. Link to comment Share on other sites More sharing options...
scilearner Posted May 26, 2010 Author Share Posted May 26, 2010 Ok forgetting all this. Swansont I basically want to know why Newton's third law occurs. If I push a box that was at rest. The box started to move. Then if I work out the change of momentum I realize 2 opposing forces have acted. Is this the explanation for Newton's third law? Link to comment Share on other sites More sharing options...
swansont Posted May 26, 2010 Share Posted May 26, 2010 Ok forgetting all this. Swansont I basically want to know why Newton's third law occurs. If I push a box that was at rest. The box started to move. Then if I work out the change of momentum I realize 2 opposing forces have acted. Is this the explanation for Newton's third law? Yes, basically. I can't push on an object any harder than it pushes on me, nor can I exert the force for any longer than it is exerted on me. Forces internal to the system cancel out and do not change the momentum of that system. Link to comment Share on other sites More sharing options...
scilearner Posted May 26, 2010 Author Share Posted May 26, 2010 Yes, basically. I can't push on an object any harder than it pushes on me, nor can I exert the force for any longer than it is exerted on me. Forces internal to the system cancel out and do not change the momentum of that system. Ok thanks then I have a final question. If a gun and bullet were at rest. Then when the bullet was fired gun recoiled. The energy for this came from gun powder. The gun powder provides energy only for the bullet so why does the gun recoil back. Gun hasn't lost any energy. Thanks Link to comment Share on other sites More sharing options...
swansont Posted May 26, 2010 Share Posted May 26, 2010 The gun powder provides energy only for the bullet This is your misconception. The powder burns/explodes and provides energy to both (we ignore everything else, like the gases present, in the ideal case). Conservation of momentum dictates how it will be divided up between the two; because the gun and person holding it are quite massive the bullet will have a large speed and thus a large kinetic energy. But the gun does recoil, and will have some kinetic energy as well. Link to comment Share on other sites More sharing options...
scilearner Posted May 26, 2010 Author Share Posted May 26, 2010 (edited) This is your misconception. The powder burns/explodes and provides energy to both (we ignore everything else, like the gases present, in the ideal case). Conservation of momentum dictates how it will be divided up between the two; because the gun and person holding it are quite massive the bullet will have a large speed and thus a large kinetic energy. But the gun does recoil, and will have some kinetic energy as well. Thanks swansont Now we atleast know my misunderstanding, but I'm still bit uncertain. How does the gun powder divide the energy equally between the gun and the bullet, how come bullet doesn't get excess energy or something. Also I'm thinking gun powder applies a force at the bullet, so bullet must apply a force at the gun powder not the gun. My other misconception is I don't understand much the difference between momentum and kinetic energy. I know one is a scalar quantity but let's say there was a momentum collision and the 2 objects united and went off together after that. Then the kinetic energy is not conserved but momentum is I don't understand why? Thanks for any help in advance Edited May 26, 2010 by scilearner Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted May 26, 2010 Share Posted May 26, 2010 Thanks swansont Now we atleast know my misunderstanding, but I'm still bit uncertain. How does the gun powder divide the energy equally between the gun and the bullet, how come bullet doesn't get excess energy or something. Also I'm thinking gun powder applies a force at the bullet, so bullet must apply a force at the gun powder not the gun. My other misconception is I don't understand much the difference between momentum and kinetic energy. I know one is a scalar quantity but let's say there was a momentum collision and the 2 objects united and went off together after that. Then the kinetic energy is not conserved but momentum is I don't understand why? Thanks for any help in advance It doesn't. The bullet, because it is less massive, accelerates during the interaction over a longer distance, and therefore receives more of the energy. Same, but opposite, momentum transfer, but different energy. Link to comment Share on other sites More sharing options...
scilearner Posted May 27, 2010 Author Share Posted May 27, 2010 (edited) It doesn't. The bullet, because it is less massive, accelerates during the interaction over a longer distance, and therefore receives more of the energy. Same, but opposite, momentum transfer, but different energy. Oh are you saying energy is W=fx. So bullet travels further and has more energy but the energy from the explosion gives the same force to both. EDIT: But I still don't understand. Why does gun powder give a forward force to the bullet and backward force to the gun. Also why is this question not about gun powder and bullet. If gun powder provides the force, the reaction force should be towards gun powder, not the gun. EDIT AGAIN: Let me think this again. If force is rate of change of momentum. Since gun powder is part of the gun, the gun changed the momentum of the bullet. Oh so you can't change the momentum of something else without that momentum coming from you, meaning losing it. Oh so the gun retaliates. Oh so basically if you want to displace something, the person who is trying to displace the object is going to get displaced as well atleast a tiny bit. May be that is what momentum is saying. Is this right? Edited May 27, 2010 by scilearner Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted May 27, 2010 Share Posted May 27, 2010 Oh are you saying energy is W=fx. So bullet travels further and has more energy but the energy from the explosion gives the same force to both. EDIT: But I still don't understand. Why does gun powder give a forward force to the bullet and backward force to the gun. Also why is this question not about gun powder and bullet. If gun powder provides the force, the reaction force should be towards gun powder, not the gun. EDIT AGAIN: Let me think this again. If force is rate of change of momentum. Since gun powder is part of the gun, the gun changed the momentum of the bullet. Oh so you can't change the momentum of something else without that momentum coming from you, meaning losing it. Oh so the gun retaliates. Oh so basically if you want to displace something, the person who is trying to displace the object is going to get displaced as well atleast a tiny bit. May be that is what momentum is saying. Is this right? You're on the right track. There will be no net momentum change to a system without an outside force. In terms of displacement, if the centre of gravity of a closed system is not moving, there is nothing that can happen in that closed system to displace the centre of gravity...so at the point the bullet is, say, 20 yards to the right, there must be an equivalent displacement of mass X distance to the left, whether it is the gun, you, or the ground etc. Link to comment Share on other sites More sharing options...
scilearner Posted May 27, 2010 Author Share Posted May 27, 2010 (edited) You're on the right track. There will be no net momentum change to a system without an outside force. In terms of displacement, if the centre of gravity of a closed system is not moving, there is nothing that can happen in that closed system to displace the centre of gravity...so at the point the bullet is, say, 20 yards to the right, there must be an equivalent displacement of mass X distance to the left, whether it is the gun, you, or the ground etc. Hey thanks a lot for the reply J.C.MacSwell That makes me think Newton's third law can be also derived from balancing centre of mass in situations like this. Like a seesaw. I have never thought about it that way. Thanks Edited May 27, 2010 by scilearner Consecutive posts merged. Link to comment Share on other sites More sharing options...
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