Physic C............newton's second law?

Question

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 = 100 ms.

a) Where is the center of mass of the two stones at t = 300 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 1

You need one more piece of information... what is the acceleration due to gravity?  Normal earth surface is 9.8 meters/second^2.  

If that's what is intended, then solve this equation twice, once for 0.3 seconds, and once for 0.2 seconds:

H = 0.5*a*t^2   distance the stone falls in t seconds.

H = 4.9 * .09 = 0.441 meters for the smaller stone.

H = 4.9 * .04 = 0.196 meters for the larger stone.

Do a weighted average of their two positions

( 0.441 * 1 + 0.196 * 2 )/2 = 0.4165 meters

That's enough for me... you should be able to work out the velocities  

Answer 2

a) First, find the position, y(t) of each stone:

Stone 1:

-Known: a(=g), v₀(=0), t

-Unknown: y

y1 = v₀t + ½at² = ½gt²

Stone 2:

-Known: a(=g), v₀(=0), t2(=t-0.1s)

y2 = v₀(t-0.1) + ½g(t-0.1)² = ½g(t-0.1)²

Now use the formula for the center of mas of 2 objects:

y(t),cm = (m1∙y1 + m2∙y2) / (m1 + m2)

= [m1∙½gt² + m2∙½g(t-0.1)²] / (m1 + m2)

= [½mgt² + mg(t-0.1)²] / 3m

= [½gt² + g(t-0.1)²] / 3

y(0.3),cm = [½(9.8)(0.3)² + (9.8)(0.3-0.1)²] / 3

= 0.28 m

b) Differentiate the position with respect to time:

v(t),cm = dy/dt = [gt + 2g(t-0.1)] / 3

v(0.3),cm = [(9.8)(0.3) + 2(9.8)(0.2)] / 3

= 2.3 m/s