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(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.