A stone is dropped at t = 0 s. A second stone, with a mass 2.00 times that of the first, is dropped from the same point at t = 0.220 s. Assuming that neither stone has yet reached the ground, how far from the release point is the centre-of-mass of the two stones at t = 0.450 s?
What is the speed of the centre-of-mass of the two-stone system at that time?
d = 1/2 * g * t^2
The Attempt at a Solution
When t = 0.45s, the first stone has been in the air 0.450/0.22 as long as the second one, so it has travelled [1/2*g(tstone2^2) ] / [1/2 * g(tstone1^2)] = [1/2*g(0.45^2) ] / [1/2 * g(0.22^2)] = 4.183884, times more!
D is the distance of the second stone from the standing point at t= 0.45s,
x_cm = (m1 * x1 + m2*x2)/(m1 + m2)
x_cm = (m1 * 4.183884D + 2m1*D)/(3m1) , (recall that m2 = 2*m1)
x_cm = ( 4.183884*D + 2*D)/(3)
Ok now we just need to find D, the distance of the second stone =1/2 * 9,8 * (0.45/2)^2 = 0.25
Alright, plug that back into the equation above to get x_cm = 0.511.. yet that is incorrect!
ugh, been at this one for awhile.. any help would be great.