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Collisions between 3 particles (perfectly elastic)

  1. Jul 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Three perfectly elastic particles A, B, C with masses 4 kg, 2 kg, 3 kg respectively, lie at rest in a straight line on a smooth horizontal table. Particle A is projected towards B with speed 15 m/s and after A has collided with B, B collides with C. Find the velocities of the particles after the second collision and state whether there will be a third collision.

    2. Relevant equations
    v1-v2= -e (u1-u2)
    Momentum before collision and momentim after collision equations using the principle of conservation of momentum.



    3. The attempt at a solution
    Well I formed the equations using the equations stated above. Should I include the momentum of all 3 before collision together and do the same for after or should I do it separately between A and B first and then B and C? If I do it for all 3 together then how will this equation v1-v2= -e (u1-u2) change? e=1 for a perfectly elastic collision.
     
  2. jcsd
  3. Jul 25, 2016 #2

    haruspex

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    Why do anything more complicated?
     
  4. Jul 25, 2016 #3
    This is what I've done so far.
    Let u1= velocity of A before collision, v1=velocity of A after collision
    u2= velocity of B before collision, v2=velocity of B after collision
    u3= velocity of C before collision, v3=velocity of C after collision

    u1= 15 and u2 = 0
    v1=? and v2=?

    Momentum before collision = 4(15) + 0(2) = 60
    Momentum after collision = 4v1 + 2v2 = 60 since momentum before collision = momentum after collision

    AND v1-v2 = -e(u1-u2) where e=1 for a perfectly elastic collision.
    v1-v2 = -1(15-0) so v1-v2 = -15 so v1= -15+v2

    Plugging it into 4v1+2v2=60 gives us 4(-15+v2) + 2v2 = 60 so v2 = 60 m/s and therefore v1 = 45 m/s

    Now I know the velocity with which B hits C so u2 = 45 and u3=0
    And v2-v3= -1(u2-u3) and the same procedure...
     
  5. Jul 25, 2016 #4

    haruspex

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    Anything strike you as surprising about those answers?
     
  6. Jul 25, 2016 #5
    They're huge compared to the initial velocities

    It's because of the coefficient of restitution
     
  7. Jul 25, 2016 #6
    Btw, I'm still not sure of part g of the gravitational potential question :frown:
     
  8. Jul 25, 2016 #7

    haruspex

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    No matter how good the restitution, the KE cannot increase,
     
  9. Jul 25, 2016 #8
    So I'm getting all the velocities as positive which means they're all travelling in the same direction after collision so a third collision is unlikely?
     
  10. Jul 25, 2016 #9

    haruspex

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    Too early to say. The velocities you quote are clearly wrong. Can you spot your mistake?
     
  11. Jul 25, 2016 #10
    Is it the equations?
     
  12. Jul 25, 2016 #11
    Ok I see it 6 v2=120 so v2=20
     
  13. Jul 25, 2016 #12
    So now I get v1 = 5m/s, v2= 20 m/s and v3=16m/s
     
  14. Jul 25, 2016 #13
    V2 = 20m/s is velocity of B after first collision ? What about velocity of B after second collision ? Will there be a third collision ?
     
  15. Jul 25, 2016 #14
    Velocity of B after second collision is -4 m/s
     
  16. Jul 25, 2016 #15
    So there is a possibility of a collision between A and B
     
  17. Jul 25, 2016 #16
    Right
    Yes . But why use the word "possibility" ? They surely will collide .
     
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