Answer to Question #23057 in Mechanics | Relativity for khakan

Question #23057
Two skaters, one with mass 65 kg and the other with mass 40 kg, stand on an ice rink holding a pole of length 10 m and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the 40 kg skater move?
Expert's answer
Assuming the pole remainsstationary, the forces of the pull of the both skaters have to be equal:
F1 = F2 = F (unknown)
Taking the skaters' masses into account, their accelerations will be
a1 = F/m1, a2 = F/m2
Starting from rest at the ends of the pole, they will travel correspondingly
l1 = (1/2)*a1*t^2 and l2 = (1/2)*a2*t^2
until they meet (at moment of time t).

As we know the total length of the pole, we can state that
l1 + l2 = L
(1/2)*(F/m1)*t^2 + (1/2)*(F/m2)*t^2 = L

As F and t are shared, we find a proportion
l1 / l2 = m2 / m1

Thus the lighter skater (40 kg) travels
l1 = L*m2/(m1+m2) = 10 m * 65 kg / (40 kg + 65 kg) = 6.19 m.

Answer: 6.2 m

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