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Basic 2-D Collision

  1. Jun 9, 2012 #1
    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?

    My attempt to split into x,y coordinates v1(intial) 12kg x 5m/s cos(0) = 60 for x
    v1(initial) 12kg x 5m/s sin(0) = 0 for y

    v2(initial) = 0 for x due to velocity
    0 for y due to velocity

    For after collision: v1(after) = 12kg x v1(after)cos(30) = ? for x
    12kg x v1(after)sin(30) = ? for y

    And v2(after) 7kg x v2(after)cos 30 for x
    v2(after) 7kg x v2(after)sin30 for y

    not sure how to use these equations to get the results, is it to do with vector cross product, or simply simultaneous equations, either way I'm stuck.

    Thanks
     
  2. jcsd
  3. Jun 9, 2012 #2

    HallsofIvy

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    Science Advisor

    For one thing, you need to work on how you write equations! You say "v1(after)= " and then the right side is the momentum, not the x coordinate of velocity.

    You should have two equations, one for the x coordinate and one for the y coordinate, that say that the x component of momentum before the collision equals the the sum of the two x components of momentum after the collision and the same thing for the y component.
     
  4. Jun 9, 2012 #3
    Thanks for reply, this is my very first posting; I have noticed online several questions like this do not quote mass.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


    and v the speed of the struck ball is found from the 1.0 term (in eqn 2) multiplied by the speed of the striking ball now

    So it is also 2.885 m/s

    Is this wrong, if so could you tell me where - if it is right was there a much more efficient way to go about it

    Thanks
     
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