Find Center of Mass of Dropped Stones

[musicfairy]

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
y_com = (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 v_com = (m₁v₁ + 2m₁v₂) / 3m₁



Help please.

 

[LowlyPion]

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
y_com = (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 v_com = (m₁v₁ + 2m₁v₂) / 3m₁

Help please.

Your velocities are functions of time,

V = a*t

so at .85 sec what are each doing?

 

[musicfairy]

I figured out. The masses cancel. That's what threw me off.