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Homework Help: Particle Elastic Collision problem

  1. Apr 23, 2006 #1
    I have a problem here. I got "answers", but I don't beleive they are correct and I have no idea where I went wrong.

    The question is... "Determine the speed V_2 (V sub 2) of the larger mass after the collision and Theta 1 and Theta 2.

    mass of smaller particle = m
    mass of larger particle = 2m
    Initial speed of m = 3V_0 (3 v knot)
    Initial speed of 2m= 0
    Final speed of m = 5^(.5)V_0
    Final speed of 2m=V_2 ....??
    Theta1= ?
    Theta2= ?

    This is obviously a conservation of momentum/Kinetic energy problem, with X and Y components of direction. Here is what I have.

    3mV_0=5^(.5)V_0Cos(theta1)+(2m)2^(.5)V_0Cos(theta2 )--->Conservation of momentum in the x direction(COMX)

    0=5^(.5)mV_0Sin(theta1)-2*2^(.5)mV_0Sin(theta2)--->Conservation of momentum in the y direction(COMY)

    .5*m(3V_0)^2=(m(5^(.5*V_0)^2)*(.5)+m(V_2final)^2--->Conservation of Kenetic ENergy(COKE)
    (COKE) goes down to (2^(.5))*(V_0)
    (COMY) goes to 0=5^(.5)*Sin(theta1)-2*2^(.5)*Sin(theta2)
    (COMX) goes to 3=5^(.5)Cos(theta1)+2*2^(.5)*Cos(theta2)

    Any mistakes ? Any suggestions? Thanks for looking, and sorry I am not very well schooled in how to write equations on the computer. PS: theta1 will be angle of m (from the horizontal) and theta2 will be angle of 2m(from the horizontal)

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  2. jcsd
  3. Apr 23, 2006 #2

    lightgrav

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    p_x : m 3v_0 = m sqrt(5)v_0 cos(theta1) + 2m ***v_2***cos(theta2)
    . . . . how did you figure out immediately that v_2 = sqrt(2) v_0 ?

    (It might be easier to calculate this in the "center-of-mass" reference frame,
    (the bee in c.o.m. frame sees the total momentum = 0 ; bee has v_x = v_0).
     
  4. Apr 23, 2006 #3
    (KE)im1=(KE)fm1+(KE)fm2--->
    masses cancel out and 2*m*(.5) cancels out to give the following
    (((3Vi)^2)/2)=(.5*(sqrt5Vi)^2)+Velocity^2[<---velocity for mass 2]
    3vi^2=9vi^2,---->sqrt5Vi^2=5Vi^2, so---> Taking .5*sqrt5Vi^2 to the other side
    ((9Vi^2)\2)-((5Vi^2)\2)=(Velocity final for mass 2)^2---> Subtracting Vi from Vi
    then 4/2*Vi^2=(Velocity final for mass 2)^2---> Sqrt'ing whole equation
    Then sqrt2*Vi=(Velocity final for mass 2)
     
    Last edited: Apr 23, 2006
  5. Apr 23, 2006 #4

    lightgrav

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    For a Perfectly Elastic collision, like yours,
    the relative velocity after collision to be opposite the rel.velocity before.
    . . . ( but 3 v_0 - 0 =/= - sqrt(5) v_0 - sqrt(2) v_0 ~ 3.6 v_0 ) . . .
    in the center-of-mass reference frame.
    v_1fy = -2 v_2fy . . . . . and . . . . v_1fx = - 2 ( v_2fx ) .
    so, we can add the velocity triangles and we find that the v_fx = 0 !

    This leads to (in lab frame) : v_1fy = +/- 2 v_0 , v_1fx = - v_0 ;
    so tan(theta1) = +/- 2 (it is symmetric across the x-axis).
     
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