Question: A stone is dropped at t = 0 s. A second stone, with a mass 4.0 times t

Question

Question: A stone is dropped at t = 0 s. A second stone, with a mass 4.0 times that of the first, is dropped from the same point at t = 0.24 s. How far from the release point is the center of mass of the two stones at t = 0.40 s? Assume that neither stone has yet reached the ground.

Attempted Solution: I solved for where the first stone would be at the time stone two was dropped (2.76m), then found where the first stone would be at 0.40s (1.229m) and added 2.76 m to it to get the position of the second (3.989m). I multiplied these by their respective weights, added them, and divided by the mass of the two stones together to get 1.781, but this was incorrect.

Explanation / Answer

distance traveled from the release point by the 1st stone in (0.4s) = .5*9.8*.4^2 = 0.784m distance traveled from the released point by the 2nd stone in (0.4-0.24)s = .5*9.8*(0.4-0.24)^2 = 0.125 m so, the center of mass is at = (m*.784 + 4m*.125)/5m = 0.257 m

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