Question

A dog, with a mass of 9.0 kg, is standing on a flatboat so that he is 22.0 m from the shore. He walks 6.9 m on the boat toward the shore and then stops. The boat has a mass of 37.0 kg. Assuming there is no friction between the boat and the water, how far is the dog from the shore now?

Answer 1

$$x_c=\frac{9(22)+37(6.9)}{9+37}=9.85\ m\\ \frac{9(22-6.9+9.85)+37(-x+9.85)}{9+37}=9.85\\x=3.67\ m$$