# Momentums with canoes

1. Nov 15, 2005

### jalapenojam

Can anyone explain to me how to do the following problem?
Ricardo, mass 75 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 20 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 m apart and symmetrically located with respect to the canoe's center. Ricardo notices that the canoe moved 50 cm relative to a submerged log during the exchange and calculates Carmelita's mass, which she has not told him. What is it?

Thanks!

2. Nov 15, 2005

### jalapenojam

i tried using the center of mass for a system of particles formula (m1x1 + m2x2 + ...)/(m1 + m2 + ...) using ricardo's seat as the starting point, but that gave me 3 kg as carmelita's mass, which i'm pretty positive is wrong.

3. Nov 15, 2005

### physicsfox

This question is about the position of the center of mass between the man, the woman and the canoe. Pick an arbitrary x position to measure from, and then determine the center of mass. When they switch positions, the center of mass remains at the same location. The forces that create the movement of the two passengers are considered internal forces and thus do not accelerate (or change the position in this case) of the center of mass.

4. Nov 15, 2005

### physicsfox

It may be easier not to use a spot on the boat as your reference point (since the boat moves). Pick an arbitrary Xo, and then measure everything from that location, both before and after movement.

5. Nov 15, 2005

### nevetsnosaj

i did m1x1 - m3x3 = m2x2 and i got 71.67 as the mass of Carmelita. Is that the right answer?

m1 = Ricardo
m2 = Canoe
m3 = Carmelita

6. Nov 15, 2005

### suspenc3

nevetsnosaj..Can you explain a little more what you did?

7. Nov 15, 2005

### nevetsnosaj

i'm using his formula and just plug the number in. I think when Ricardo moving, he made the canoe moved for a certain distance. (the lake and the canoe is frictionless so that the momentum that ricardo give to the canoe makes the canoe move) When carmelita move, she make the canoe move to the other way since carmelita is going in opposite direction of Ricardo.
From that thoughts and his formula above, i did what i did.

8. Nov 15, 2005

### suspenc3

so you did m1(x1) - m3(x3) = m2(x2)?

What did you use for the x position of the canoe..ricardo..and carmelita?

9. Nov 15, 2005

### nevetsnosaj

x1 = ricardo = 3m
x2 = canoe = 50 cm
x3 = carmelita = 3m
that's what i did.

10. Nov 15, 2005

### Fermat

Draw two diagrams, one for before the movement of ricardo and carmelita and another diagram for afterwards, as physicsfox suggested.

On the 1st diagram, mark the positions of the mass of ricardo, carmelita and the canoe. Now take moments to find the position of the COM of the canoe plus two canoeists.

When ricardo and carmelita exchange position, this will alter the position of the COM wrt the canoe, (but not the lake) and you know that that difference in position is 50 cm.
Now draw the 2nd diagram, marking the positions of ricardo, carmelita and the canoe again. Take moments again to find the COM of the canoe plus canoeists.

You should now be able to eliminate unknowns.

11. Nov 15, 2005

### nevetsnosaj

so with that we can use the conservation of momentum and eliminate the Velocity and then find the mass of Carmelita?

12. Nov 15, 2005

### Fermat

There is no momentum involved. There are no velocities involved. Simply a change in the COM of the canoe.
The movements of the canoeists are internal forces/actions and have no (external) effect upon the canoe. This means that there is no net force on the canoe/canoeists and hence no acceleration and hence its COM does not change position (wrt to its external environment). But since the COM changes inside the canoe then the canoe must move about in order to keep its original COM in the same postion (wrt the lake).

By taking moments, for the before and after cases, and using the fact that the change in COM is 50 cm, the moment equations involved will be all you need.

13. Nov 15, 2005

### nevetsnosaj

i got confused with the COM and wrt. Isn't COM is Conservation of Momentum?

14. Nov 15, 2005

### Fermat

Sorry, COM = Centre of Mass
wrt = with respect to

15. Nov 15, 2005

### nevetsnosaj

ok. I get it. So since the COM change, the canoe moves.
I've figured out the distance between each person to the COM after the switching. What am I supposed to do with that number?

16. Nov 15, 2005

### Fermat

Well, if you have a (numerical) value for the position of the COM, then the moment eqn will give you the mass of carmelita.

What value do you have ?

17. Nov 15, 2005

### nevetsnosaj

Ricardo to COM = 175 cm
Carmelita to COM = 125 cm

and what's the formula of the moment eqn ?

18. Nov 15, 2005

### Fermat

Hmm.
Ricardo is heavier, so the COM should be closer to him than carmelita.
Good point: The numerical values are correct

19. Nov 15, 2005

### nevetsnosaj

opps... that's a typo..... So i get it right, don't i? I don't know what is the moment eqn . Can you tell me the equation?

20. Nov 15, 2005

### Fermat

The moment eqn is the eqn of moments.

Take moments (moment arm - is that term familiar ?) about any point.
e.g. m1*x1 + m2*x2 + ... = 0

Edit: I think you have already done something like this.