# 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

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

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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