Walking in a Boat, PHYSICS HELPPPPPp?

Question

A 55.0 kg} woman stands up in a 68.0 kg} canoe of length 5.00 m}. She walks from a point 1.00 m} from one end to a point 1.00 {\rm m} from the other end. If the resistance of the water is negligible, how far does the canoe move during this process?

thank you

Answer 1

Let the center of the canoe be at a distance 'd' from a refernce point on the line joining the ends of the canoe.

=> woman is at a distance of d - 1.5 from the reference point

=> center of mass of the system of canoe and woman is at a distance

= [68d + 55(d - 1.5)] / (68 + 55) initially   ... (1)

If, after the woman has moved, center of canoe is at a distance 'd1', then woman is at a distance d1 + 1.5 and center of mass of the system is at a distance

= [68d1 + 55(d1 + 1.5)] / (68 + 55)   ...   (2)

As no external force acted on the system, eqns. (1) and (2) should represent the same distance

=> 68d + 55(d - 1.5) = 68d1 + 55(d1 + 1.5)

=> 68(d - d1) + 55(d - d1) = 55 * 3

=> d - d1 = 165 / (68 + 55)

= 1.34 m.

Answer 2

Not sure, but it sounds like an explosion problem.  Anything going away from anything alse is supposed to be an explosion.  So it should be (entire mass)(velocity)=(mass of woman)(woman's velocity)+(mass of boat)(boat's velocity)

Hmm but that doesn't seem right.  I'm not sure, sorry.

Answer 3

Consider the end from where the original distance of the owner is 1.00 m. Fix origin O in water just below this point.

The mass of the woman = mw = 55.0 kg

The mass of the canoe = mc = 68.0 kg

Considering the canoe as uniform, the CM(center of the mass of the canoe) is at its geometrical center.

Initially,

position of CM of the woman = xw1 = 1.00 m

Position of CM of the canoe = xc1 = length of canoe/2 = 5.00 m/2 = 2.50 m

Let X1 = position of CM of the system (woman + canoe)

X1 = (mw*xw1 + mc*xc1)/(mw+mc)

Or X1 = (55*1 + 68*2.5)/(55+68)

Or X1 = 225/123 m

Or X1 = 1.83 m --------------------------(1)

There is no external force on the system. Therefore, its CM remains the same. When the woman moves in one direction, then the canoe moves in opposite direction so that the CM remains the same.

Suppose the canoe moves by distance d.

When the woman is 1.00 m from other end, then she is 5.00m(i.e. length of canoe) - 1.00 m from first end = 4.00m from first end.

Therefore, she is 4.00m - d from O.

The canoe's center is 2.50 m - d from O.

Finally,

position of CM of the woman = xw2 = 4.00 m - d

Position of CM of the canoe = xc2 = 2.50 m - d

Let X2 = position of CM of the system (woman + canoe)

X2 = (mw*xw1 + mc*xc1)/(mw+mc)

Or X2 = {55*(4-d) + 68*(2.5-d)}/(55+68)

Or X2 = (55*4 - 55d + 68*2.5 - 68d)/123

Or X2 = (220 - 55d + 170 - 68d)/123

Or X2 = (390 - 123d)/123

Or X2 = 3.17 - d---------------------(2)

CM remains the same. Therefore, X1 = X2

Therefore, from (1) and (2)

1.83 = 3.17 - d

Or d = 3.17 - 1.83 = 1.34 m

Ans: 1.34 m