Solving Momentum Change Problem: Canoe & Girl's Distance to Shore

[Huskies213]

Can anyone help with this problem?

A 65 kg girl stands in the middle of her 20 kg canoe. The canoe is 3m long, and the end that is closet to land is 2.5m from the shore. The girl now walks toward the shore until she comes to the end of the canoe. What is the distance from the girl to the shore ??

I know its the formula x= m1x1+m2x2/ m1+m2 does anyone know how to solve it from there ?

 

[dfx]

Isn't it just 2.5m?

 

[topsquark]

Can anyone help with this problem?

A 65 kg girl stands in the middle of her 20 kg canoe. The canoe is 3m long, and the end that is closet to land is 2.5m from the shore. The girl now walks toward the shore until she comes to the end of the canoe. What is the distance from the girl to the shore ??

I know its the formula x= m1x1+m2x2/ m1+m2 does anyone know how to solve it from there ?

Is there a net external force acting on the canoe-girl system?

-Dan

 

[Huskies213]

Re

No outside forces, what is in the problem is all that is given.

 

[qtp]

this is a center of mass problem. the system will conserve the center of mass and when the girl starts to walk towards the shore the canoe will move away from the shore to conserve the center of mass.

 

[topsquark]

this is a center of mass problem. the system will conserve the center of mass and when the girl starts to walk towards the shore the canoe will move away from the shore to conserve the center of mass.

"conserve center of mass" I don't think I've ever heard that phrase before! :tongue2:

Specifically if there is no net external force on the system then the total momentum of the system is conserved. That means that the center of mass is not accelerating. So if the CM was stationary before she started walking where is it when she's done walking?

-Dan