Answer to Question #23057 in Mechanics | Relativity for khakan
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?
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.