physics- woman and canoe?

Question

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?

Answer 1

§ Paul, Mr. SC’s approach is correct! Since you have not yet flagged him as the best answer, I clarify him with more details.
♠ since total momentum of the system is 0 it will stay 0 no matter what woman is doing in the boat; and position of the mass center of the system will also be unchangeable; let x-axis be along the boat, x=0 being in the middle of the boat and attached to the ground, woman’s position being x1=-5/2 +1=-1.5; thus the equation for mass center through torques is:
♣ before T1=0*60 +x1*45 = -67.5; after T2= x*60 +x2*45; where x2 =x+1.5;
since (see ♠) T1=T2, then –67.5 =x*60 +(x+1.5)*45, hence x=-1.2857m, minus showing to the back.
☻if you are still doubtful. Here is another approach.
♠ momentum conservation law: 0=m*w + n*u, here m=45kg, w(t) is woman’s speed, n=60kg, u(t) is canoe’s speed; note this equation is valid for any time tick t no matter how functions w(t) and u(t) are actually looking;
♣ now all talk above requires integration to find coordinates:
m∫w(t)dt +n∫u(t)dt ={for t=0 until T} = m*(x2+1.5) +n*(x-0) =0,
while x2 =x+1.5; then 45*(x+3) +60x=0; x=-135/105 =-1.2857m;

Answer 2

The person and the canoe make a system and there is no external force to this system. Therefore, the center of mass stays the same. You locate the CM in the beginning and you'll know the rest