Answer
At $t = 300~ms,$ the speed of the center of mass of the two stones is $~~2.29~m/s$
Work Step by Step
We can find the speed of the first stone after $0.30~s$:
$v = v_i+at$
$v = 0+(9.8~m/s^2)(0.30~s)$
$v = 2.94~m/s$
We can find the speed of the second stone after $0.20~s$:
$v = v_i+at$
$v = 0+(9.8~m/s^2)(0.20~s)$
$v = 1.96~m/s$
Let $M$ be the mass of the first stone.
Let $2M$ be the mass of the second stone.
We can find the speed of the center of mass at $t=300~ms$:
$v = \frac{(M)(2.94~m/s)+(2M)(1.96~m/s)}{M+2M}$
$v = \frac{(M)~(6.86~m/s)}{3~M}$
$v = 2.29~m/s$
At $t = 300~ms,$ the speed of the center of mass of the two stones is $~~2.29~m/s$
