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.(adsbygoogle = window.adsbygoogle || []).push({});

(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 m_{1}for the first stone dropped and 2m_{1}for the second stone dropped, y_{1}for how far the first stone dropped and y_{2}for how far the second stone dropped.

y = .5gt^{2}

y_{1}= .5(9.8)(0.850)^{2}

y_{1}= 3.54 m

y_{2}= .5(9.8)(0.350)^{2}

y_{2}= 0.60 m

For part a I got the equation

y_{com}= (m_{1}(3.54) + 2m_{1}(0.60))/3m_{1}

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}= (m_{1}v_{1}+ 2m_{1}v_{2}) / 3m_{1}

Help please.

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# Homework Help: Find center of mass

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