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System of particles - find energy

  1. Jan 27, 2005 #1
    Iam having problems with this question i think my technique is right but i,ve made a mistake somewhere

    Particles of mass m and 2m respectively are connected by a light inextensible string. Initially, the particles lie at opposite edges of a smooth horizontal table with the string just taut. One of the particles is then nudged over the edge of the table. Find the ratio between the two possible speeds of the system when the other particle reaches the edge of the table.

    i've let the level of potential energy be the table
    m rep A
    2m rep B
    i've stated of nudging B and keeping A on the table
    energy at start = 0
    energy at finish = KE of A and B and PE of B

    1/2*m*v^2 + 1/2*2m*v^2 + 2m*g*-h
    Energy at start = Energy at finish
    1/2mv^2(1/2 + 1) = 2mgh
    v^2(1/2 + 1) =2gh
    v^2 = 2gh/1.5
    v = (4/3gh)^-1

    now if we nudge A and keep B on the table

    Energy at start = 0
    Energy at finish = KE of A and B + PE of A

    1/2*m*v^2 + 1/2*2m*v^2 + m*g*-h
    Energy at start = Energy at finish
    1/2mv^2 +mv^2 = mgh
    mv^2(1/2 +1) = mgh
    v^2 = gh/1.5
    v = (2/3gh)^-1

    (4/3gh)^-1 : (2/3gh)^-1

    sub in random value for h= 5

    i get a ratio of

    1.4 : 1

    which is an answer i don't have much faith in.
    i could have made mistake in my technique or silly error
    Could anyone help me out please!
    Thanks in advance.
     
  2. jcsd
  3. Jan 27, 2005 #2
    Check your math

    1/2*m*v^2 + 1/2*2m*v^2 + 2m*g*-h
    Energy at start = Energy at finish
    1/2mv^2(1/2 + 1) = 2mgh ...........................here
    v^2(1/2 + 1) =2gh......................................here
    v^2 = 2gh/1.5
    v = (4/3gh)^-1.....................................and here

    and similarly in the second part.
     
  4. Jan 27, 2005 #3
    cheers
    v = (4/3gh)^-1
    should be v = (4/3gh)^1/2, i've got that, but this does not solve the problem

    1/2mv^2(1/2 + 1) = 2mgh
    should be mv^2(1/2 + 1) = 2mgh

    v^2(1/2 + 1) =2gh this is correct

    these are typing errors on my behalf
    the ratio i get is 1.4 : 1
     
  5. Jan 27, 2005 #4
    v = (4/3gh)^(1/2)
    and
    v = (2/3gh)^(1/2)
    so the ratio is sqrt(2):1
     
  6. Jan 27, 2005 #5
    cheers ta thx
     
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