The discussion revolves around the conservation of momentum in a scenario involving two boats and a man moving between them. Participants explore the implications of this movement on the velocities of the boats, considering various methods of transfer and the resulting momentum changes.
Participants express multiple competing views, and the discussion remains unresolved regarding the exact implications of the man's movement on the boats' velocities and the application of conservation of momentum.
Limitations include assumptions about the method of the man's transfer between the boats and the definitions of the system boundaries, which affect the analysis of momentum changes.
If you were able to simply step from boat to boat with no inbetween stage. I have had a rethink would it be that boat a stays at the same velocity because no net force is applied to it and although it will loose momentum the mass will decrease also due to the man stepping across leaving the velocity unchanged. Then when the man steps on to boat b he will have momentum acting in an opposite direction to the mans momentum. Therefore a force is required to change the mans velocity to equal boat bs velocity the oposite and equal force of this action will consequently decrease the momentum and therefore the velocity of the boat.Orodruin said:This would depend on the particular setup and how the man moved between the boats.
Alevelman said:Say you had two boats (a and b) moving past each other without colliding and a man moved from boat a to boat b. Is it correct that boat b would decelerate and boat a would accelerate due to the conservation of momentum
This might hurt due to sudden acceleration when he is caught up by the second boat. A far less hurtful method would be to try to match the speed of boat B as good as he is able. This is why we have been arguing that the formulation really does not imply either method and it is impossible to solve without further specifying how the man goes to boat B (i.e., what momentum he brings with him).anorlunda said:I presume that the man steps over without pushing himself forward or back.
Shyan said:But I think boat A should change velocity too. Because we assumed that the man doesn't exert a force on boat A, we have F_{C\to A}=0=\frac{\Delta p_A}{\Delta t} \Rightarrow \Delta p_A=0 \Rightarrow m_A (v_A+dv_A)-(m_A+m_C)v_A=0 \Rightarrow dv_A=\frac{m_C}{m_A}v_A.
Orodruin said:This might hurt due to sudden acceleration when he is caught up by the second boat.
Shyan said:But I think boat A should change velocity too. Because we assumed that the man doesn't exert a force on boat A
Shyan said:Otherwise there is a change in momentum which is caused by no force!
No no, this is wrong. Because \vec F=\frac{d\vec p}{dt}. Although I have no explanation or example, this equation simply rejects what you say.anorlunda said:You're missing the central fun point of this problem. Momentum m*v may be changed either by changing m or by changing v. Only changing v requires force.
Shyan said:\vec F=\frac{d\vec p}{dt}. .
jbriggs444 said:Forces are not the only things that can change the momentum of a system. Re-drawing the boundaries of the system can also do so.
Shyan said:Although I have no explanation or example
Shyan said:But even in case of Newton's 2nd law.
Sorry...I'm really confused right now. My brain refuses to work for me! You're saying there is a non-zero force on boat A? Can you calculate it?Orodruin said:What about the current problem? It serves as a perfect example.
Orodruin said:F = dp/dt is Newton's second law. You normally define it for constant mass objects, in which case it turns into F = ma. For variable mass objects you can also use F = dp/dt, see also http://en.wikipedia.org/wiki/Variable-mass_system
The answer is that it depends on how you define your system (and therefore on what forces act on the different systems). If you consider A+C as one system D and then at some time redefine D to be just A, then the momentum of D will change. If this happens instantly, the corresponding force will be a delta in time, based on the momentum C carries away with it. If C has the same velocity as A after the split, the velocity of D will remain the same.Shyan said:Sorry...I'm really confused right now. My brain refuses to work for me! You're saying there is a non-zero force on boat A? Can you calculate it?
Yeah, I know about that. But as I said, I'm confused now and this confused brain is telling me something is different here. But I can't figure it out.
Its ridiculous I should still struggle with these things in confusion after so many years of being a physics student and studying all those advanced math and physics.
Now could you explain things clearly? I know I'm supposed to be better at this but I'm now like I know nothing!
Yeah, I knew this part.Orodruin said:The answer is that it depends on how you define your system (and therefore on what forces act on the different systems). If you consider A+C as one system D and then at some time redefine D to be just A, then the momentum of D will change. If this happens instantly, the corresponding force will be a delta in time, based on the momentum C carries away with it. If C has the same velocity as A after the split, the velocity of D will remain the same.
This I understand well too. But I don't understand how the "extended 2nd law" relates to this problem. It seems to me that variable mass systems aren't that simple to treat and there is something behind all those calculations that I don't see. I'm now reading this paper which seems to address the issue.Orodruin said:If you just see A as the system all the time, it is a matter of what force C is exerting on A. If C has the same velocity as A and does not change velocity, the momentum transfer between them is zero and the velocity of A and C, which are systems of constant mass, will be unchanged.
Yeah, I was going off road!Orodruin said:The relation is simply saying what you think it is, it defines force as the rate of momentum change. If you have a system without external forces, momentum will be conserved and there really is not much more to it than that. In fact, in this case it is much easier to simply work with momentum conservation - we do not know if the acceleration is constant (assuming C is accelerated) or over what time it occurs.
Orodruin said:The relation is simply saying what you think it is, it defines force as the rate of momentum change. If you have a system without external forces, momentum will be conserved and there really is not much more to it than that. In fact, in this case it is much easier to simply work with momentum conservation - we do not know if the acceleration is constant (assuming C is accelerated) or over what time it occurs.
Quantum Defect said:The total angular momentum of the system should also be conserved. This is like a scattering experiment in AMO or Chemical Physics. Working in a coordinate system with the origin at the center of mass helps.
jbriggs444 said:Ideally, angular momentum is conserved. However, if the boats have keels so that sideways movement is ignored then the water can act as a source or sink for angular momentum and it may fail to appear to be conserved.
jbriggs444 said:Ideally, angular momentum is conserved. However, if the boats have keels so that sideways movement is ignored then the water can act as a source or sink for angular momentum and it may fail to appear to be conserved.
CWatters said:It's possible for the man to speed up or slow down both boats depending on the direction in which he jumps. If he jumps early (eg forwards) he will slow down both boats. If he jumps late (eg backwards) he will speed up both boats. Therefore at some optimal point he will have no effect on the speed of either boat (except to push them sideways).