Two Falling Stones and Their Center of Mass

[mbrmbrg]

A stone is dropped at t = 0. A second stone, with twice the mass of the first, is dropped from the same point at t = 250 ms.
(a) How far below the release point is the center of mass of the two stones at t = 600 ms? (Neither stone has yet reached the ground.)
(b) How fast is the center of mass of the two-stone system moving at that time?

First and foremost: the times are 0.250s and 0.600s respectively.
I set my co-ordinate system with x=0 at the release point and the x-axis is positive in the downward direction (sorry).

Other than that...

I first found (or perhaps tried to find) the center of mass at 0.250s and said that the COM moves with a constant acceleration of 9.81m/s^2.

PART A:
To find where the first stone was at 0.250 s, I used the equation x=x_0+v_0t+\frac{1}{2}at^2, where x_0 and v_0 are both 0. so x = 1/2g(.250s)^2 = 0.307m.

I then said that at 0.250s, x_{COM}=\frac{(2m)(0)+(m)(0.307)}{m+2m}=0.012meters

So at 0.600s, the position of the COM is given by x=x_0+v_0t+\frac{1}{2}at^2. I solved that saying using 0.012m as the initial position and 0m/s as the initial velociy, and got an incorrect answer (1.87m)

PART B:
Not surprisingly, when I used my answer for part (a) in the equation v^2=v_0^2+2a\Delta x, I also got the wrong anwer.

And now I am stuck (and feeling stupid, because this problem got one out of three dots for difficulty).

 

[lotrgreengrapes7926]

For part A, you should simply find where the 2 stones are at .6 sec, and find their center of mass.

 

[mbrmbrg]

Thanks, that got the right answer for part a.
For part b, I wrote the equation for the position COM purely symbolically then took the first derivative and used that to find velocity. So that's that!