Answer to Question #23057 in Mechanics | Relativity for khakan
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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