Momentum Man, Board, Frictionless Surface

[nfg05]

A 100 kg man stands on the left end of a 100 kg board 2 m long which sits on a frictionless surface as shown below. Where will the left end of the board come to rest if the man walks to the right end of the board and stops?

Diagram: Number line with -2, -1, 0, 1, 2. Left end of board at 0, right end of board at 2. Man standing on left end of the board (at 0 on the number line).

A. -2
B. -1
C. 0
D. 2

---

Members don't see this ad.

 

[engineeredout]

Having a little trouble picturing this.


[*_______]
-2 -1 0 1 2

Is that what you mean?

****ing Sdn screwed up my spacing.

 

[AZFutureDoc]

The way I see it, the answer is A. I draw a free body diagram and when the man walks, puts a horizontal force on the board. the board tries to "push" the same way, but because there is no friction, the board moves 2 numbers left, while the man remains at zero.

 

[AZFutureDoc]

Okay, think a skating rink, frictionless ice. Now we set this board on the ice, with no friction. Now someone stands on this board and tries walking to the end of the board. The question is really does the man move the board or does the board move the man. The board can't do it since it cannot get any force in the x direction.

 

[not respect]

I tihnk it's B. Because the man and the board should move the same distance.

 

[hawk I]

I remember this problem. It comes from the Nova book. For the longest time I could not figure out how the answer was B. After thinking for a while, it finally made sense.

Let's say the man takes steps 1m long. If the left end of the board is at 0 and he takes one step, the board will move to the left with a force equal to what it would take the man to take a 1m step. So the board moves 1m to the left. The left end will be at the -1m mark. You would think that the man would now be at the midpoint of the board and have to take another step (which would lead to the left end of the board finishing at the -2m mark making the answer A). But when the man takes the first step (1m) the board is moving backwards at the same amount (1m) so the step actually takes the man to the right end of the board (2m).

In other words, no matter how long his steps are, the man is covering twice the distance per step.

 

[AZFutureDoc]

I just cannot see B. I have drawn a FBD and cannot see it!

Imagine that this board is on on the ground, on a side walk. The man is at one end. He walks to the other by using frictional forces between his shoes and the board. The reason that the board stays in place is because the frictional forces between the board and side walk are able to equal the walking force. Now that there is no force keeping the board motionless relative to the ground/table it must move the same distance he walks. Can someone explain why I'm wrong, or why the other answer is correct in a little more detail?

 

[AZFutureDoc]

I remember this problem. It comes from the Nova book. For the longest time I could not figure out how the answer was B. After thinking for a while, it finally made sense.

Let's say the man takes steps 1m long. If the left end of the board is at 0 and he takes one step, the board will move to the left with a force equal to what it would take the man to take a 1m step. So the board moves 1m to the left. The left end will be at the -1m mark. You would think that the man would now be at the midpoint of the board and have to take another step (which would lead to the left end of the board finishing at the -2m mark making the answer A). But when the man takes the first step (1m) the board is moving backwards at the same amount (1m) so the step actually takes the man to the right end of the board (2m).

In other words, no matter how long his steps are, the man is covering twice the distance per step.

Seems to me like your explanation makes sense if the man and the board also experience no frictional forces. If the man takes one step, he has moved one meter. The board does not slide under his shoe...

 

[nfg05]

The answer given is B, and I actually came across it in EK although I don't doubt it was in the Nova book as well. I'm not sure I understand Hawk's explanation although it does arrive at the right answer.

The way I thought about it, if the man takes a 0.5 m step the board moves 0.5 m left. The force required for the 100 kg man to move himself 0.5 m right would move the 100 kg board 0.5 m left as well. So basically the board moves left as much as the man (tries to) move right, making the answer A. The simultaneous motion of the board means that although the man is trying to move right and actually IS making his way towards the right end of the board, his position on the number line does not change since the simultaneous leftward motion of the board cancels out his displacement WITH RESPECT TO THE NUMBER LINE.

The man taking a 1 m step and actually traveling 2 m to the end of the board doesn't make much sense to me. If he took a 1 m step right, he is now at the midpoint of the board, period. Yes the board also moved left 1 m, but that just means that the man ends up staying at 0 on the number line (he has however moved 1 m right WITH RESPECT TO THE BOARD), the left end of the board is now at -1 and the right end of the board is at 1. Thus, after the next 1 m step the end result is the man still at 0 on the right end of the board and the left end of the board is at -2.

 

[TieuBachHo]

This is very confusing but let me try to explain it intuitively this way. As a man move 1/2 of his step forward, the board slides 1/2 step backward. However, at this critical movement, the man body is actually 1/2 step forward as the lagging leg steps half way to the right. So basically, one step a man moves forward, the board slides a half of a step backward and a man body also moves a half of a step forward (positionally). As the man reaches his position in 1, the left side of the board reaches position -1. Make sense LOL

 

[AZFutureDoc]

The answer given is B, and I actually came across it in EK although I don't doubt it was in the Nova book as well. I'm not sure I understand Hawk's explanation although it does arrive at the right answer.

The way I thought about it, if the man takes a 0.5 m step the board moves 0.5 m left. The force required for the 100 kg man to move himself 0.5 m right would move the 100 kg board 0.5 m left as well. So basically the board moves left as much as the man (tries to) move right, making the answer A. The simultaneous motion of the board means that although the man is trying to move right and actually IS making his way towards the right end of the board, his position on the number line does not change since the simultaneous leftward motion of the board cancels out his displacement WITH RESPECT TO THE NUMBER LINE.

The man taking a 1 m step and actually traveling 2 m to the end of the board doesn't make much sense to me. If he took a 1 m step right, he is now at the midpoint of the board, period. Yes the board also moved left 1 m, but that just means that the man ends up staying at 0 on the number line (he has however moved 1 m right WITH RESPECT TO THE BOARD), the left end of the board is now at -1 and the right end of the board is at 1. Thus, after the next 1 m step the end result is the man still at 0 on the right end of the board and the left end of the board is at -2.

Wait it seems like you are saying -1 is right, but then explaining that -2 is. I can ask a Princeton Review tutor when I'm there on Friday. But man this is making mad! I want to understand this!

 

[nfg05]

Wait it seems like you are saying -1 is right, but then explaining that -2 is. I can ask a Princeton Review tutor when I'm there on Friday. But man this is making mad! I want to understand this!

What I'm saying is that -1 is right according to the book, but explaining why I think -2 is right. The book explanation sucks and given that it's EK the answer could even be wrong.

 

[AZFutureDoc]

Ya. true... I did so good on the EK 30 minute test on this material too! Walking like this, on ice, we know that you really dont go anywhere. I am just trying to see how the board changes this. Seems like the result should be the same, as if you were wearing shoes with soles the size of the board. The only thing I may not be seeing are any effects that the board having the same mass as the man has on the system.

 

[not respect]

well since the title says momentum I just thought about it like that, so if the guy moves to the right with velocity m he has momentum mv to the right which the board must have to the left to conserve momentum... thus he moves 1 m to the right, the board must move 1 m to the left. In your situation you have the guy not moving at all, but somehow the board has moved 2m to the left, that doesn't conserve momentum.

 

[AZFutureDoc]

Got it. Exact (almost) example given on Audio Osmosis, Phy lecture 4 on reverse collisions.

Use conservation of momentum. Before, p=0, so after, p=0. That means that, since, both masses are 100 kg,

MmanVman=MboardVboard

Vman=Vboard. If the board moves left, the man MUST move right.



I'll add reverse collisions to the list of things Kaplan failed to ever talk about...

 

[davjan]

Let me chime in here :]

conservation of momentum says that if the person moves with +1m/s, then the board must move with -1m/s : same magnitutde in opposite direction. So, the momentum is conserved in the system of board + person.

now, if the person is moving to the right at 1m/s and the board is moving to the left at 1m/s, then the rate of displacement between the board and the person is 2m/s.
Example would be: if two people are running at 100m/s toward each other, they will be approaching each other at 200m/s and vice versa.

so, the person would have traveled 2 meters on the board after walking for 1 second; while in 1 second, the board and the person would have each moved only 1 meter with respect to the ground, since their velocity with respect to the ground is 1m/s while their velocity with respect to each other are 2m/s.

 

[Cawolf]

Thanks for chiming in on a 6+ year old thread! :]

 

[davjan]

Thanks for chiming in on a 6+ year old thread! :]

you're welcome