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Conservation of momentum of two boats

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  1. Jan 11, 2015 #1
    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
     
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  3. Jan 11, 2015 #2

    Orodruin

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    This would depend on the particular setup and how the man moved between the boats.
     
  4. Jan 11, 2015 #3
    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.

    Is this in any way correct or have I gone off on one?
    Cheers
     
  5. Jan 11, 2015 #4

    FactChecker

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    If all of the man's change of velocity happens in boat b, you are correct. If he jumps backward from boat a to get to b, he would actually speed up boat a and he would slow boat b down less.
     
  6. Jan 11, 2015 #5

    Orodruin

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    As FactChecker has already stated, the outcome depends completely on what momentum the man takes with him, which in turn depends on how he steps/jumps out of the boat.
     
  7. Jan 12, 2015 #6

    anorlunda

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    I think you found a fun conservation of momentum problem. I presume that the man steps over without pushing himself forward or back.

    Let's say man C leaves boat A and enters boat B. The velocity of A will not change because it does not accelerate or decelerate the man, but the momentum of A will decrease because the mass of the (boat A plus man B) system decreases. Boat B will change velocity because it must change the velocity man C to match the velocity of boat B. The mass of boat B will be increased by the mass of man C.

    Note: It is important to use the same sign convention for vA and vB. If the boats travel in opposite directions, these velocities will have opposite signs.

    The initial momentums are (mA+mC)*vA and mB*vB

    The final momentums are mA*vA and (mB+mC)*(vB+dv)

    Since momentum is conserved, (mA+mC)*vA + mB*vB = mA*vA + (mB+mC)*(vB+dv)

    Now you can solve for dv (the change in velocity of boat B).

    dv = (((mA+mC)*vA + mB*vB - mA*vA)/(mB+mC)) -vB

    The right hand side of that equation includes only initial quantities. If all those are known, you can compute the final value of dv.

    The reason this makes a fun problem is because it involves mass changes as well as velocity changes.
     
  8. Jan 12, 2015 #7

    Orodruin

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    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).
     
  9. Jan 12, 2015 #8

    ShayanJ

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    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 [itex] 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 [/itex].
    So conservation of momentum equation is:
    [itex](m_A+m_C)v_A+m_B v_B=m_A(v_A+dv_A)+(m_B+m_C)(v_B+dv_B)[/itex].
     
    Last edited: Jan 12, 2015
  10. Jan 12, 2015 #9

    jbriggs444

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    If I understand the derivation of the above equation, you are looking at the momentum of the boat A plus man C. Since there is no external force, momentum is conserved when the man steps off the boat. You interpret this to mean that the momentum of A alone after man C steps off is the same as the momentum of boat A plus man C prior to his having stepped off. But that's just plain false.

    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.
     
  11. Jan 12, 2015 #10

    anorlunda

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    You want to inject "hurt" into a physics problem. What a spoilsport.

     
  12. Jan 12, 2015 #11

    ShayanJ

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    Yeah...sorry guys...you're right, that's pretty confused!!!
    I now get it where I was wrong. I was thinking this way A changes momentum with no force but actually at first its A+C and then its A alone and so its really wrong to think like that!
     
  13. Jan 12, 2015 #12

    anorlunda

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    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.
     
  14. Jan 12, 2015 #13

    ShayanJ

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    No no, this is wrong. Because [itex] \vec F=\frac{d\vec p}{dt} [/itex]. Although I have no explanation or example, this equation simply rejects what you say.
    I edited my last post and explained why I was wrong. Its the point that jbriggs444 mentioned. I was thinking before and after the jump, the system is the same and we should have conservation of momentum but this is wrong. Before jump its A+C and after jump its A alone.
    But even in case of Newton's 2nd law. It seems to me that you're somehow right. Its only the change in velocity that needs force but...ohh god...I'm confused.
     
  15. Jan 12, 2015 #14

    anorlunda

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    That applies only when mass is constant. jbriggs444 said it another way.


     
  16. Jan 12, 2015 #15

    Orodruin

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    What about the current problem? It serves as a perfect example.

    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
     
  17. Jan 12, 2015 #16

    ShayanJ

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    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!
     
  18. Jan 12, 2015 #17

    Orodruin

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    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.

    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.
     
  19. Jan 12, 2015 #18

    ShayanJ

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    Yeah, I knew this part.
    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.
     
  20. Jan 12, 2015 #19

    Orodruin

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    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.
     
  21. Jan 12, 2015 #20

    ShayanJ

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    Yeah, I was going off road!!!
    Thanks
     
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