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) How far below the release point 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

how do we calculate the center of mass?

Answer 2

\(\vec{r}_{cm} = \dfrac{\sum\limits_{i} m_i\vec{r}_i}{\sum\limits_{i}m_i}\)

Answer 3

differentiating gives
\(\vec{v}_{cm} = \dfrac{\sum\limits_{i} m_i\vec{v}_i}{\sum\limits_{i}m_i}\)

Answer 4

and where is the difficulty rising up at?

Answer 5

there is no difficulty as such
my answer is not matching with the textbook, just wanted to take second opinion :)

my answer is 5.7cm
http://www.wolframalpha.com/input/?i=1%2F3%280.3%5E2%2B2*0.2%5E2%29

but textbook says 28cm

Answer 6

is our distance from origin:

d1 = gt^2/2

d2 = g(t-100)^2/2

Answer 7

yes, i used seconds, so i have below in my notes

d1 = gt^2/2

d2 = g(t-0.1)^2/2

Answer 8

then the distance between them is just d1-d2 right?

center of mass to me means the point which would balance the 2 objects on a seesaw

Answer 9

Yes

\[h_{cm} = \dfrac{d_1*(m) + d_2*(2m)}{m+2m} = \dfrac{0.3^2+2*0.2^2}{3}=5.7cm\]

Answer 10

Oh wait, I think I see the blunder... somehow I have cancelled g/2 wrongly

Answer 11

thanks @amistre64 that fixes the problem !
http://www.wolframalpha.com/input/?i=1%2F3%280.3%5E2%2B2*0.2%5E2%29*9.8%2F2

Answer 12

yep, cant forget gravity :)

Answer 13

haha i had it in the equations, but i saw some imaginary "g/2" in the denominator and cancelled in that fraction... now both parts agree with the textbook answer !