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

AI Thread Summary
The problem involves a 65 kg girl walking towards the shore on a 20 kg canoe, which is 2.5m from the shore. The key concept is the conservation of momentum, as there are no external forces acting on the system. As the girl walks towards the shore, the canoe will move in the opposite direction to maintain the center of mass. The discussion emphasizes understanding how the center of mass remains stationary despite the girl's movement. The final distance from the girl to the shore depends on the conservation of the center of mass during her movement.
Huskies213
Messages
34
Reaction score
0
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 ?
 
Physics news on Phys.org
Isn't it just 2.5m?
 
Huskies213 said:
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
 
Re

No outside forces, what is in the problem is all that is given.
 
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.
 
qtp said:
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! :-p

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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top