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?
1
Expert's answer
2013-01-29T07:42:04-0500
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
distances
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
or
(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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