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

The center of mass of the two stones at t = 360 ms is 0.437 m below the release point.

Step-by-step explanation:

To find the center of mass of the two stones, we first need to determine their individual positions at t = 360 ms. Since the first stone was dropped at t = 0, it will have fallen for 360 ms. Using the formula s = ut + (1/2)gt^2, where s is the distance, u is the initial velocity (0 in this case), g is the acceleration due to gravity, and t is the time, we find that the distance fallen by the first stone is:

s = 0 + (1/2)(9.81)(0.36)^2 = 0.637 m

The second stone was dropped 181 ms later, so it will have fallen for 360 ms – 181 ms = 179 ms. Using the same formula, we find that the distance fallen by the second stone is:

s = 0 + (1/2)(9.81)(0.179)^2 = 0.158 m

The center of mass of the two stones at t = 360 ms can be found using the formula:

x_cm = (m1x1 + m2x2) / (m1 + m2)

Where x1 and x2 are the positions of the individual stones and m1 and m2 are their masses. Plugging in the values, we get:

x_cm = (0.150 x 0.637 + 0.100 x 0.158) / (0.150 + 0.100) = 0.437 m

Therefore, the center of mass of the two stones at t = 360 ms is 0.437 m below the release point.

Answer 2

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

Step-by-step 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+(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)=(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)=(1)/(2) (9.81m/s^2)(179)^2=157,161.1m`

And finally using the equation for the center of mass:

`CM=(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=(mh_(1)+2mh_(2))/(m+2m)`

`CM=(m(h_(1)+2h_(2)))/(3m)=(h_(1)+2h_(2))/(3)`

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

`CM=(635,688m+2(157,161.1m))/(3)=316,670m`

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