Question

Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. The boys pull themselves together along the rod. When they meet the 40-kg boy will have moved what distance?.

Answer 1

When they meet the 40-kg boy will have moved a distance of 6 m.

Displacement of the 40 kg boy

The displacement of the 40 kg boy is calculated from the principle of center mass.

X(40 kg) = (60 x 10 m    +  40 x 0)/(40 kg + 60 kg)

X(40 kg) = (600)/(100) = 6 m

X(60 kg) = (60 x 0    +  40 x 10 m)/(40 kg + 60 kg)

X(60 kg) = (400)/(100) = 4 m

Thus, when they meet the 40-kg boy will have moved a distance of 6 m.

Learn more about center mass here: https://brainly.com/question/13499822

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