A stone is dropped at t = 0. A second stone, with 6 times the mass of the first, is dropped from the same point at t = 160 ms. (a) How far below the release point is the center of mass of the two stones at t = 290 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?

Answer with Explanation:

Let mass of one stone=

Mass of second stone,m'=6 m

a.

t=160 ms=

Initial velocity==u'=0

Where

Center of mass=

Center of mass=

b.

Using the formula

Velocity of center of mass=

Velocity of center of mass=

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the assestment will probley apele with cause and effect