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Physics Center of Mass, driving me nuts

  1. Oct 10, 2013 #1
    1. The problem statement, all variables and given/known data


    A stone is dropped at t = 0 s. A second stone, with a mass 2.00 times that of the first, is dropped from the same point at t = 0.220 s. Assuming that neither stone has yet reached the ground, how far from the release point is the centre-of-mass of the two stones at t = 0.450 s?

    What is the speed of the centre-of-mass of the two-stone system at that time?

    2. Relevant equations

    d = 1/2 * g * t^2

    3. The attempt at a solution


    When t = 0.45s, the first stone has been in the air 0.450/0.22 as long as the second one, so it has travelled [1/2*g(tstone2^2) ] / [1/2 * g(tstone1^2)] = [1/2*g(0.45^2) ] / [1/2 * g(0.22^2)] = 4.183884, times more!

    D is the distance of the second stone from the standing point at t= 0.45s,

    x_cm = (m1 * x1 + m2*x2)/(m1 + m2)

    x_cm = (m1 * 4.183884D + 2m1*D)/(3m1) , (recall that m2 = 2*m1)

    m1 cancels

    x_cm = ( 4.183884*D + 2*D)/(3)

    Ok now we just need to find D, the distance of the second stone =1/2 * 9,8 * (0.45/2)^2 = 0.25

    Alright, plug that back into the equation above to get x_cm = 0.511.. yet that is incorrect!


    ugh, been at this one for awhile.. any help would be great.
     
  2. jcsd
  3. Oct 10, 2013 #2
    How much time has elapsed between the time that stone 2 is released (t = 0.22 sec) and the time t = 0.45 seconds? How far did stone 2 fall during this time interval? How far had stone 1 fallen during the 0.45 sec since it was released?
     
  4. Oct 10, 2013 #3
    Thank you, I think I got it =)
     
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