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This would depend on the particular setup and how the man moved between the boats.

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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.This would depend on the particular setup and how the man moved between the boats.

Is this in any way correct or have I gone off on one?

Cheers

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FactChecker

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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.

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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).I presume that the man steps over without pushing himself forward or back.

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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].

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].

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jbriggs444

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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].

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.

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anorlunda

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This might hurt due to sudden acceleration when he is caught up by the second boat.

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

But I think boat A should change velocity too. Because we assumed that the man doesn't exert a force on boat A[/QUOTE]

Shyan, your analysis makes no sense. No force is exerted on boat A but A changes velocity; how does that satisfy F=ma?

OK, it violates KISS, but we can make the problem more complex and still solve it. It can be solved using only conservation of momentum. Bringing forces into it is an unnecessary complication.

Consider two independent events. (1) Man C leaps from boat A. (2) Man C lands on boat B.Momentum is conserved separately for each event.You can assume anything you like about C pushing off, but your assumption is contained in the assumed velocity vC of the man mid-air. vC can equal vA or vC can be faster or slower.

Leap event:

(mA+mC)*vA = mA*(vA+dA) + mC*vC

Solve for dA.

Land event:

mB*vB + mC*vC = (mB+mC)*(vB+db)

Solve for dB.

I still think it is a fun problem because we must apply conservation of momentum to a whole system, not it's constituent parts. In this problem we have five "systems" and we shift among them with time. It is a trap to write an equation that inappropriately mixes systems.

A+C pre-leap

A alone post-leap

C alone in mid air

B alone pre-landing

B+C post-landing

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ShayanJ

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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!

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anorlunda

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Otherwise there is a change in momentum which is caused by no force!!!

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.

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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.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.

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.

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anorlunda

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[itex] \vec F=\frac{d\vec p}{dt} [/itex]. .

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

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.

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Although I have no explanation or example

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

But even in case of Newton's 2nd law.

F = dp/dt

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

Orodruin said:F = dp/dtisNewton'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

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!

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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.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!

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.

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ShayanJ

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Yeah, I knew this part.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.

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ShayanJ

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Yeah, I was going off road!!!

Thanks

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Quantum Defect

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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.

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jbriggs444

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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.

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.

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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.

Experiment with two boats with initial velocities equal to zero. :)

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anorlunda

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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.

It is still conserved, but you must also include the water into "the system" and the water will be moved.

Please please, adding complications does not make the problem easier to understand. The OP said it well; let's stick with that.

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I think this is the scenario the op set up. I made it as simple as possible by setting all masses and velocities to either +1 or -1. The system consists of boat A, boat B, and the man.

boat A mass = 1m

boat B mass = 1m

man mass = 1m

boat A velocity = +1v

boat B velocity = -1v

The man is in boat A and both boats are moving parallel to the x axis on frictionless water obeying Newton's first law, assuming all motion of the boats is restricted to the x axis.

The initial momentum of the system is:

boat A = (1m + 1m) * +1v = +2

boat B = 1m * -1v = -1

So the total initial momentum of the system is: 2 + -1 = +1

When the boats are side by side, the man jumps in a direction 90 degrees to the x axis. So his mass is effectively transfered from boat A to boat B. Since the man has a momentum of +1 (1m * +1v) and boat B has a momentum of -1 (1m * -1v), then the final momentum of boat B after the jump will be 0. However, boat A continues on unimpeded.

The momentum of the system after the jump is:

boat A = 1m * +1v = +1

boat B = (1m + 1m) * 0v = 0

So the total momentum of the system after the jump is: 1 + 0 = +1

boat A mass = 1m

boat B mass = 1m

man mass = 1m

boat A velocity = +1v

boat B velocity = -1v

The man is in boat A and both boats are moving parallel to the x axis on frictionless water obeying Newton's first law, assuming all motion of the boats is restricted to the x axis.

The initial momentum of the system is:

boat A = (1m + 1m) * +1v = +2

boat B = 1m * -1v = -1

So the total initial momentum of the system is: 2 + -1 = +1

When the boats are side by side, the man jumps in a direction 90 degrees to the x axis. So his mass is effectively transfered from boat A to boat B. Since the man has a momentum of +1 (1m * +1v) and boat B has a momentum of -1 (1m * -1v), then the final momentum of boat B after the jump will be 0. However, boat A continues on unimpeded.

The momentum of the system after the jump is:

boat A = 1m * +1v = +1

boat B = (1m + 1m) * 0v = 0

So the total momentum of the system after the jump is: 1 + 0 = +1

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As we have

FC→A=0=ΔpAΔt⇒ΔpA=0⇒mA(vA+dvA)−(mA+mC)vA=0⇒dvA=mCmAvA.

So conservation of momentum equation is:

(mA+mC)vA+mBvB=mA(vA+dvA)+(mB+mC)(vB+dvB).

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jbriggs444

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That is an enticing theory, however... If both boats retain the same forward velocity then, by conservation of momentum, so does the man. If he starts stationary on one boat and ends stationary on the other then the only scenario that fits is one in which neither boat is moving at all.

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