Question

Two people, one of mass 76.4 kg and the other of mass 58.0 kg, sit in a rowboat of mass 84.4 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat 2.14 m apart, now exchange seats. Friction between the boat and water is negligible. How far will the boat move?

Answer 1

Given that,

Mass of the heavy person = M1 = 76.4 kg; that of the light person = M2 = 58 kg

Mass of the boat = Mb = 84.4 kg;

Distance between the two persons = d = 2.14 m.

In this problem, the center of mass (COM) is not moving and there are no external forces acting on the system. If we assume, d be the distance of COM and P Q R be the distance of M1, M2 and Mb from the com, then we can write:

d =( 76.4P + 58 Q + 84.4R) / (76.4+58+84.4)

d = ( 76.4P + 58 Q + 84.4R)/ 218.8

After, they have exchanged their seats and if x be the distance moved then, d will be

d = [58(P+x) + 76.4(Q+x) + 84.4 (R+x)] / 218.8

We can equate the above values of d

( 76.4P + 58 Q + 84.4R)/ 218.8 = [58(P+x) + 76.4(Q+x) + 84.4 (R+x)] / 218.8

76.4P + 58 Q + 84.4R = 58P + 58x + 76.4Q + 76.4x + 84.4R + 84.4x

18.4P - 18.4Q = 218.8 x

x = 18.4(P - Q ) / 218.8

We find that, P - Q is the distance between two persons, that is (P - Q) = 2.14

x = 18.4 x 2.14 / 218.8 = 0.18 m

Hence, The distance moved by the boat = x = 0.18 meters.