- #1
mbrmbrg
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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 = 250 ms.
(a) How far below the release point is the center of mass of the two stones at t = 600 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?
First and foremost: the times are 0.250s and 0.600s respectively.
I set my co-ordinate system with x=0 at the release point and the x-axis is positive in the downward direction (sorry).
Other than that...
I first found (or perhaps tried to find) the center of mass at 0.250s and said that the COM moves with a constant acceleration of 9.81m/s^2.
PART A:
To find where the first stone was at 0.250 s, I used the equation [tex]x=x_0+v_0t+\frac{1}{2}at^2[/tex], where x_0 and v_0 are both 0. so x = 1/2g(.250s)^2 = 0.307m.
I then said that at 0.250s, [tex]x_{COM}=\frac{(2m)(0)+(m)(0.307)}{m+2m}=0.012meters[/tex]
So at 0.600s, the position of the COM is given by [tex]x=x_0+v_0t+\frac{1}{2}at^2[/tex]. I solved that saying using 0.012m as the initial position and 0m/s as the initial velociy, and got an incorrect answer (1.87m)
PART B:
Not surprisingly, when I used my answer for part (a) in the equation [tex]v^2=v_0^2+2a\Delta x[/tex], I also got the wrong anwer.
And now I am stuck (and feeling stupid, because this problem got one out of three dots for difficulty).
(a) How far below the release point is the center of mass of the two stones at t = 600 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?
First and foremost: the times are 0.250s and 0.600s respectively.
I set my co-ordinate system with x=0 at the release point and the x-axis is positive in the downward direction (sorry).
Other than that...
I first found (or perhaps tried to find) the center of mass at 0.250s and said that the COM moves with a constant acceleration of 9.81m/s^2.
PART A:
To find where the first stone was at 0.250 s, I used the equation [tex]x=x_0+v_0t+\frac{1}{2}at^2[/tex], where x_0 and v_0 are both 0. so x = 1/2g(.250s)^2 = 0.307m.
I then said that at 0.250s, [tex]x_{COM}=\frac{(2m)(0)+(m)(0.307)}{m+2m}=0.012meters[/tex]
So at 0.600s, the position of the COM is given by [tex]x=x_0+v_0t+\frac{1}{2}at^2[/tex]. I solved that saying using 0.012m as the initial position and 0m/s as the initial velociy, and got an incorrect answer (1.87m)
PART B:
Not surprisingly, when I used my answer for part (a) in the equation [tex]v^2=v_0^2+2a\Delta x[/tex], I also got the wrong anwer.
And now I am stuck (and feeling stupid, because this problem got one out of three dots for difficulty).