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 = 181 ms. How far below the release point is the center of mass of the two stones at t = 360 ms? (Neither stone has yet reached the ground.)

Answer 1

Answer:

the center of mass is 316,670m bellow the release point.

Explanation:

First, we must find the distance at which the objects are in time t = 360s

We will use the formula for vertical distance in free fall

$h=v_{0}t+\frac{1}{2} gt^2$

$v_{0}$ is the initial velocity, is 0 for both stones since they were just dropped.

g is acceleration of gravity and t is time ($g=9.81m/s^2$)

At $t=360s$ the first stone has been falling for the entire 360 seconds, its position h1 is:

$h_{1}=\frac{1}{2} (9.82m/s^2)(360s^2)=635,688m$

And at 360 seconds the second stone has been fallin for$t= 360s -181 s = 179s$, so its position h2 is:

$h_{2}=\frac{1}{2} (9.81m/s^2)(179)^2=157,161.1m$

And finally using the equation for the center of mass:

$CM=\frac{m_{1}h_{1}+m_{2}h_{2}}{m_{1}+m_{2}}$

We know that the mass of the second stone is twice the mass of the first stone so:

$m_{1}=m\\m_{2}=2m$

replacing these values in the equation for the center of mass

$CM=\frac{mh_{1}+2mh_{2}}{m+2m}$

$CM=\frac{m(h_{1}+2h_{2})}{3m}=\frac{h_{1}+2h_{2}}{3}$

Finally, replacing the values we found fot h1 and h2:

$CM=\frac{635,688m+2(157,161.1m)}{3}=316,670m$

the center of mass is 316,670m bellow the release point.