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2D Collsion

  1. Jun 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Question: A 12kg ball travelling at 5 m/s strikes a stationary ball of 7kg. After the collision the balls are 60 degrees apart, each 30 degrees either side of the original path of the striking ball.

    Calculate the speed of both balls after collision?


    2. Relevant equations Concerned about layout, saw similar question online and tried to follow it through, but not sure of most efficient method or if what I did was right at all....



    3. The attempt at a solution I have noticed online several questions like this do not quote mass. So can I ignore this by simply cancelling out the term 'm' from momentum calculations?

    Here is my attempt, I'm following a similar example without mass and also following an example that states from the outset it is elastic whereas mine does not.

    Would there be a different technique if it is inelastic or elastic? I'm thinking not as momentum is conserved regardless?

    Anyway here is my attempt with different equation layout

    u = velocity of striking ball after collision
    v = velocity of the struck ball after collision

    for the x axis momentum to be conserved, momentum before = momentum after

    m x 5.00 m/s + 0 = (m x u cos 30) + (m x v cos 30)

    My plan is to ignore mass values given so;

    u cos30 + v cos30 = 5.00 m/s

    u (cos 30 / cos 30) = (5.00 / cos 30)

    1.0 u+v= 5.77 m/s (eqn 1)


    Using conservation laws along the y-axis

    m u sin 30 = m v sin 30

    u (sin30/sin30) - v = 0

    1.0 u-v = 0 (eqn 2)

    From eqn 1 and 2 added together we get 2u = 5.77

    so u (the speed of the striking ball is 2.885 m/s

    Is this correct, do I take the angle between the two spheres, or the angle they are from the original path?
     
  2. jcsd
  3. Jun 10, 2012 #2
    Hi Physicphillic! Welcome to Physics Forums :smile:

    You cannot ignore the masses. Momentum is the product of mass and velocity, and you are given two different masses. The examples you saw might have been with equal masses.

    Just conserve the momentum in along both x and y axes (remembering to include mass!) and that should lead you to the answer.
     
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