# Center of Mass of canoe

## Homework Statement

A 45.0-kg woman stands up in a 60.0-kg canoe of length 5.00 m. She walks from a point 1.00 m from one end to a point 1.00 m from the other end.

If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?

## Homework Equations

center of mass = (m)(x_1) + (M)(x_2) / (m + M)

m = mass of woman
M = mass of boat

## The Attempt at a Solution

Momentum is conserved right? Because there's no horizontal force. I set my y coordinate where the woman started walking. I plugged in:

center of mass: .9m (I'm not sure about this. I used the woman's weight + boat's weight versus the weight of the boat at another end.)
x_1 = 3m
m = 45kg
M = 60kg
x_2 = What I'm trying to find....

What else am I missing?

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berkeman
Mentor
Peach said:
center of mass = (m)(x_1) + (M)(x_2) / (m + M)
Write one version of that equation for the beginning situation, and one version for the ending situation.

center of mass_1 = (m + M)(x_1) + (M)(x_2) / (m + M + M)

center of mass_2 = (M)(x_1) + (m + M)(x_2) / (m + M + M)

Is this right? Because at the beginning, the woman's weight was added to the boat. At the end, her weight's added to the other end of the boat....

berkeman
Mentor
I'm not sure I'm understanding your equations. Where is your origin? I think it's easiest if you set x=0 at the woman before she starts walking. Then all you have is the boat's mass center at some distance to the right Xbi (boat, intial). At the end, you have the woman at some distance to the right, and the boat's CM has moved to the left. Since the two total mass centers have to be the same, set them equal and solve away....

I think I made the mistake of setting the origin at the center of the boat in the first eqn and then setting it at the woman before she starts walking in the second eqn. I'm confused about what you mean by the total mass centers have to be the same...? Actually...What is the total mass centers? berkeman
Mentor
Actually, the fundamental thing that you are using is that linear momentum is conserved for an isolated system (like this basically frictionless boat movement). Initially the sum of the momentums of the boat and woman is zero, right? So during her movement and afterwards, the total linear momentum must remain zero.

So what does that mean about the distance she moves (considering her mass) and the distance that the boat moves in the opposite direction (considering its mass)?

HallsofIvy
Homework Helper

## Homework Statement

A 45.0-kg woman stands up in a 60.0-kg canoe of length 5.00 m. She walks from a point 1.00 m from one end to a point 1.00 m from the other end.

## Homework Equations

center of mass = (m)(x_1) + (M)(x_2) / (m + M)

m = mass of woman
M = mass of boat

## The Attempt at a Solution

Momentum is conserved right? Because there's no horizontal force. I set my y coordinate where the woman started walking. I plugged in:

center of mass: .9m (I'm not sure about this. I used the woman's weight + boat's weight versus the weight of the boat at another end.)
x_1 = 3m
m = 45kg
M = 60kg
x_2 = What I'm trying to find....

What else am I missing?
What exactly is the question? You say "x_2= what I'm trying to find" but you don't say what x_2 is.

I'm really confused, I thought momentum has to do with velocity and mass. So the only way for momentum to be 0, is if the velocity is 0? Um, does this mean distance doesn't increase at all? It cancels out..?

Edit: Sorry I forgot. The question: If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?

berkeman
Mentor
I'm really confused, I thought momentum has to do with velocity and mass. So the only way for momentum to be 0, is if the velocity is 0? Um, does this mean distance doesn't increase at all? It cancels out..?
No, the vector sum of the momentums need to stay zero throughout. So if the woman walks one way with some mass and velocity, what does the canoe have to do?

The canoe has to have some mass and velocity that will equal the woman's momentum..but the other direction....?

berkeman
Mentor
The canoe has to have some mass and velocity that will equal the woman's momentum..but the other direction....?
Yeah, like that. So at every instant in the transition part of this problem, the sum of the total momentum will be zero, right? So tell us how this goes down....

If the momentum is zero, then it isn't moving at all? This Sparknotes page tells you exactly how to solve it:

sparknotes.com/testprep/books/sat2/physics/chapter9section5.rhtml

type in the http..., I couldn't do it becasue I dont have 15 posts yet, and apparently you have to have 15 posts before you can post urls :P