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Really easy problem >.<

  1. Apr 8, 2004 #1
    dont u hate it when the answer is staring u in the face and u just cant see it :S well i can't

    im having problem on a few questions all modeled around the same problem just with different numbers and different bits to work out so hopefully if someone can spot my error i will be able to get them all done

    so the question is

    Two identical particles are moving in the same straight line. Immediately before they collide, they are moving with speeds of 1m/s and 2m/s in the same direction. Given that the coefficient of restitution is 0.8, calculate the speeds of the two particles after the collision.

    (Taken from the ocr m2 book if anyone has it handy)

    now i labeled the particles B and A respectivly

    B = 1m/s and A = 2m/s

    for them to collide the diagram must be sumthing like


    however i chose the top one cos im used to working left to right

    now as far as i remember (hey im on holiday :P)

    Rebound speed = E x Initial speed

    where e is the coeffectient of restitution

    therefore A's rebound speed should be

    R = E x I
    R = 0.8 x 2
    R = 1.6

    and B's

    R = E x I
    R = 0.8 x 1
    R = 0.8

    however when checking my answers i found the book told me 1.9m/s and 1.1m/s

    can n e 1 spot my stupid mistake? >.<

    thanks :)
  2. jcsd
  3. Apr 8, 2004 #2
    If c is the coefficient of restitution then:
    [tex]c = \frac{u_2 - u_1}{v_2 - v_1}[/tex]
    Where u denotes the speed of each particle after the impact and v denotes the speed of each particle before the impact.

    In your case the equation is:
    [tex]0.8 = \frac{u_2 - u_1}{2m/s - 1m/s} = u_2 - u_1[/tex]
    But you have more unknowns than equations, so you can't solve this just yet. You need one more equation that links u1 and u2, and this is where the conservation of momentum comes in:
    [tex]m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2[/tex]
    Since the mass of the particles is the same (the particles are "identical") you can cancel it:
    [tex]v_1 + v_2 = 3m/s = u_1 + u_2[/tex]

    Now can you solve it? :smile:
  4. Apr 8, 2004 #3

    Doc Al

    User Avatar

    Staff: Mentor


    To solve this problem you need to apply two things: (1) the correct definition of the coefficient of restitution, and (2) conservation of momentum.

    For two objects colliding, the coefficient of restitution is the ratio of the speed of separation to the speed of approach. (Hint: the speed of approach is VA - VB = 1 m/s.)
  5. Apr 8, 2004 #4
    lost me :P

    would the coefficient of restitution be

    final seperation (VA - VB)
    Initial seperation (VA - VB)


    if so then the final seperation is 0.8 which is correct to the book but still brings me no closer to the answer :S

    u mention conservation of momentum tho hmmmmm gunna go play with book and calculator for a while :P

    *edit* playing with conservation of momentum

    using the mass M

    2M + 1M = XM + YM

    where X and Y are the final speeds of A and B

    divide through by M

    3 = X + Y (ye thats rite which is gr8 cos once i got one i can just plonk it into there and i got the other)

    unfortunately thats as far as i can get *Scratches head*
    Last edited: Apr 8, 2004
  6. Apr 8, 2004 #5
    Am I being ignored? :frown:
  7. Apr 8, 2004 #6
    The coefficient of restitution compares the approach and separation speeds.

    Both these speeds are relative speeds, as in they depend on the motion of both objects that are involved in the collision.

    e is equal to separation speed/initial speed only if one of the objects has a velocity of zero both before and after the collision. Otherwise, you must compensate for the second object's velocity.

    For instance, in your problem, one ball is moving at 2 m/s and the second at 1 m/s before the collision. Is the approach speed 2 m/s? 1 m/s? Which is it? It will be how quickly the 2m/s ball is catching up to the 1m/s ball, or (2 - 1)m/s = 1m/s.

  8. Apr 8, 2004 #7

    Doc Al

    User Avatar

    Staff: Mentor

    LOL. You guys are killing me. :rolleyes:

    BananaMan, take advantage of what Chen worked out for you!
  9. Apr 8, 2004 #8
    [tex]0.8 + u_2 = u_1[/tex]

    subbing this into second equation gives

    [tex]3m/s = 0.8 + u_2 + u_2[/tex]


    [tex]2.2m/s = 2 u_2[/tex]

    [tex]u_2 = 1.1m/s[/tex]

    subbing back into 2nd equation

    [tex]3m/s = u_1 + 1.1[/tex]

    [tex]1.9m/s = u_2[/tex]

    woot :)

    thanks, soz i didnt use it at first it looked confusing, gunna have to go re-read ur post and understand where numbers come from :P
  10. Apr 8, 2004 #9
    ok just had a play around with it and got that one sorted thanks for ur help, gunna go try apply it to the other problems now :)

    *edit* just to check

    if i find a negative coefficient in a question wat does this mean?

    can it just be reversed to give a positive?
    Last edited: Apr 8, 2004
  11. Apr 8, 2004 #10
    Just use the coefficient as it's given, even if it's negative. It may mean that the bodies change the direction of their movement after the impact. It may also mean other things but I don't have time to think about this too much right now. :smile: So just use it as it is.
    Last edited: Apr 8, 2004
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