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In: Physics

Two people, one of mass 76.4 kg and the other of mass 58.0 kg, sit in...

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?

Solutions

Expert Solution

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.

genius_generous answered 5 months ago
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