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Mass Momentum problem

  1. Jul 15, 2005 #1
    Tow identical masses are released from rest in a smooth hemisherical bowl of radius R. One mass is on the top of the bowl a height R from the bottom and the other mass is on the bottom when they are released. There is no friction. If the masses stick together when they collide, how high above the bottom of the bowl wil the masses go after colliding?

    My work so far is:

    stage one:

    K1 + U1 = K2 + U2

    0.5*mAVa2(squared) = mAgR
    Va2 = sqrt(2*g*R) Va2 the speed of mass A on the bottom

    stage two:
    K1 + U1 = K2 + U2

    Mgy2 = 0.5*M*2*g*R where M = 2*m

    so y2 = R


    The right answer is y2=R/4
    I can't see what I am doing wrong or not doing.
    Could someone please give me a hint to this problem?
     
  2. jcsd
  3. Jul 15, 2005 #2
    Break the motion into 3 stages.
    Before the collision (energy is conserved).
    During the collision (momentum is conserverd.)
    After the collision (energy is conserved.)

    you forgot the collision itself. Consider this to be
    immediately before the blocks collide untill immediately
    after they collide. What's the momentum of the system
    in each case?
     
  4. Jul 17, 2005 #3
    O.K. I split the motion into 3 stages

    1. K1 + U1 = K2 + U2

    Va2 = sqrt(2gR)

    2. mVa2 + 0 = 2mVt (where Vt is the speed of both block after they stick together). So Vt = (sqrt(2gR))/2

    3. K1 + U1 = K2 + U2
    2mgy = mVt(squared) - 2mgR

    From this I get y = (6/8)R
    when I should get R/4

    What am I doing wrong. I consider the top of the bowl to be the origin of the y-axis so the bottom is -R

    Could someone please give me a hint.
     
  5. Jul 17, 2005 #4

    Päällikkö

    User Avatar
    Homework Helper

    1. Energy is conserved:
    [tex]mgh = \frac{1}{2}mv^2[/tex]
    2. Collision:
    [tex]m_{b1}v_{b1} + m_{b2}v_{b2} = m_{a1}v_{a1} + m_{a2}v_{a2}[/tex], where b is for before collision and a after it, and the numbers are for particle 1 and particle 2.
    The above equation solves quite nicely.
    3. Energy is conserved:
    [tex]mgh = \frac{1}{2}mv^2[/tex]


    From your equations:

    3. K1 + U1 = K2 + U2
    2mgy = mVt(squared) - 2mgR, this is incorrect, if I understood your markings right.
     
  6. Jul 17, 2005 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Good.
    Good.

    Careful. Remember you are measuring from the top of the bowl. When you solve for y, you should actually get y = - (6/8)R = -3/4R. Don't forget the minus sign! Now you have to restate your answer in the form that the question asked: If the mass rises to y = -3/4R, how high above the bottom does it get?

    Of course, you would have had an easier time if you measured PE from the bottom of the bowl, but it should work either way. :smile:
     
  7. Jul 17, 2005 #6
    Got it. Thanks guys
     
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