Two friends, one at each end of a 4.00 m long raft over water, are playing with a 9.07-kg medicine ball. One of them throws the ball to the other. If the total mass of the people and raft is 131 kg and the viscosity between the water and raft is ignored, how far does the raft move before it comes to a stop again?

25.9 cm

Explanation: From the question, one can put the left end of the raft at the origin and the medicine ball at x = 4.00 m. Also, one can as well assume that the CM of the people and raft is at the point "d" from the origin (we can as well assume that the CM is at the midpoint).

ΣM = (131kg×d + 9.07kg×4) / 140.07kg

The system slides to the right as the ball moves to the left:

finally

ΣM = (131kg*(d+x) + 9.07*x) / 140.07kg

These are equal, so

131kg×d + 9.07kg×4 = 131kg*(d+x) + 9.07×x

131d+36.28 = 131d+131x+9.07x

131d- 131d+131x+9.07x = 36.28

36.28 = (131+9.07)x

36.28 = 140.07x

x = 36.28/140.07

x = 0.259 m

Converting to centimeters

0.259×100

= 25.9 cm

d is the correct answer