Find Center of Mass of Dropped Stones

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To find the center of mass of two stones dropped at different times, the first stone falls for 850 ms, reaching a distance of 3.54 m, while the second stone falls for 350 ms, reaching 0.60 m. The center of mass can be calculated using the formula ycom = (m1(3.54) + 2m1(0.60))/3m1, where the mass cancels out, simplifying the calculation. For the velocity of the center of mass, the velocities of the stones at 850 ms must be determined, with the formula vcom = (m1v1 + 2m1v2) / 3m1 also simplifying due to mass cancellation. Understanding that the velocities are functions of time is crucial for solving both parts of the problem.
musicfairy
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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 = 500 ms.

(a) How far below the release point is the center of mass of the two stones at t = 850 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?



I started by solving for how far they fell. I'll use m1 for the first stone dropped and 2m1 for the second stone dropped, y1 for how far the first stone dropped and y2 for how far the second stone dropped.

y = .5gt2

y1 = .5(9.8)(0.850)2
y1 = 3.54 m

y2 = .5(9.8)(0.350)2
y2 = 0.60 m



For part a I got the equation
ycom = (m1(3.54) + 2m1(0.60))/3m1

But how do I solve this when I'm not given the value of m?


I'm not sure how to solve part b. Is it the same idea except I replace y with v?

I'm saying vcom = (m1v1 + 2m1v2) / 3m1



Help please.
 
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musicfairy said:
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 = 500 ms.

(a) How far below the release point is the center of mass of the two stones at t = 850 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?
I started by solving for how far they fell. I'll use m1 for the first stone dropped and 2m1 for the second stone dropped, y1 for how far the first stone dropped and y2 for how far the second stone dropped.

y = .5gt2

y1 = .5(9.8)(0.850)2
y1 = 3.54 m

y2 = .5(9.8)(0.350)2
y2 = 0.60 m
For part a I got the equation
ycom = (m1(3.54) + 2m1(0.60))/3m1

But how do I solve this when I'm not given the value of m?I'm not sure how to solve part b. Is it the same idea except I replace y with v?

I'm saying vcom = (m1v1 + 2m1v2) / 3m1

Help please.

Your velocities are functions of time,

V = a*t

so at .85 sec what are each doing?
 
I figured out. The masses cancel. That's what threw me off.
 
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