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Not Head-On Collision

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data

    One snooker ball (initially traveling at 0.6 m s-1) hits another of the same mass (0.4 kg) which was initially at rest. They collide with a coefficient of restitution of 0.9. The balls do not collide head-on, and so they come away at different angles. If the final speed of the incident ball is 0.3 m s-1, what is the final speed of the other (originally stationary) ball?

    2. Relevant equations

    conservation of momentum on x,y axis
    law of restitution on the line of impact


    3. The attempt at a solution

    Momentum is conserved on the y axis ( perpendicular to the line of impact )

    so u1cosφ=V1cosθ + V2 1)

    Momentum is conserved on the x axis ( the line of impact )

    u1sinφ= V1sinθ 2)


    Applying the law of restitution on the line of impact :

    e= ( V2-V1cosθ)/(u1cosφ) 3)



    So if we re-arrange a bit we get :

    0.6cosφ = 0.3cosθ + V2

    0.54cosφ = -0.3cosθ + V2

    Αnd if we subtract these two we get :

    cosφ = 10cosθ 4)


    Νow if we plug 4) to 0.6cosφ = 0.3cosθ + V2 we get :

    5.7cosθ=V2

    But we also know that from 1) & 3) V2 is also equal to = ((e+cosφ)u1)/2 so

    i found that cosθ=0.1 and i then i plugged it to 5.7cosθ=V2 and found that V2 is 0.57 m/s .

    Is my solution correct ?


    Thank you :)
     
  2. jcsd
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