Question

A man with mass 87 kg and a woman with mass 54 kg sit at opposite ends of a canoe of length 3.4 m, whose mass 10 kg is spread uniformly over its length. If the man suddenly moves a distance 0.85 m toward the center and the woman moves 1.13333 m toward the center, how far does the canoe move in the water? 1. 0.07467 2. 0.08444 3. 0.04306 4. 0.02439 5. 0.04459 6. 0.02692 7. 0.09665 8. 0.1102 9. 0.08796 10. 0.0625 Answer in units of m.

Answer 1

Consider the system of man + woman + canoe. Since the net external force on the system is zero, Center of mass of this system does not move.

Let the center of canoe be at a distance from a point fixed to earth. Then center of mass of the system is $x_c=\frac{87(L-1.7)+54(L+1.7)+10L}{87+54+10}$

now after the persons have moved, Let the center of canoe be at a distance from the same point fixed to earth.

Now the center of mass is at $x_c=\frac{87(L'-0.85)+54(L'+1.7-\frac{3.4}{3})+10L'}{87+54+10}$

since the center of mass has not moved, $87(L'-0.85)+54(L'+1.7-\frac{3.4}{3})+10L'=87(L-1.7)+54(L+1.7)+10L$

$(87+54+10)(L-L')=87\times0.85-54\times\frac{3.4}{3}$

Hence distance moved by canoe is $L-L'=\frac{87\times0.85-54\times\frac{3.4}{3}}{(87+54+10)}=0.08444\,m$

answer is 2.