1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Elastic Collision

  1. Jun 7, 2016 #1
    1. The problem statement, all variables and given/known data
    The problem is pretty simple, however I don't understand which value to use after using quad formula to solve. See below.

    V1(initial) = 2.5 m/s
    V2(Initial) = -5.9 m/s

    A pool ball moving with a speed of 2.5 m/s makes an elastic head-on collision with an identical ball traveling in the opposite direction with a speed of 5.9 m/s. Find the velocities of the balls after the collision.

    2. Relevant equations
    ΣP(initial) = ΣP(final)
    M1V1 + M2V2 = M1V1' + M2V2'
    M1 = M2
    2.5 + (-5.9) = V1' + V2'

    V2' = -V1' - 3.4

    Elastic collision: KE(initial) = KE(final)
    (Mass cancels out)
    V12+V22 = V1'2+V2'2

    3. The attempt at a solution
    From above, I got:
    0 = 2V1'2 - 6.8V1' - 29.5
    by substituting "V2' = -V1' - 3.4" for V2'

    After applying quadratic equation I got:
    V1' = -2.5 m/s, 5.9 m/s

    My problem is I do not know which one is the correct answer. My answer key says V1' is equal to -2.5 m/s, but I have no idea why.
    Any help is appreciated!
  2. jcsd
  3. Jun 7, 2016 #2


    User Avatar
    Homework Helper

    Check the signs of the solutions. The answer key can be wrong.
  4. Jun 7, 2016 #3
    I don't really understand what you said in the brackets, do you mean if the initial velocity is positive, then the answer must be negative after collision?
    I also have another problem with this question. In my solution I substituted "V2' = -V1' - 3.4", so if V1' is equal to -2.5 m/s then V2' should be -0.9 m/s. However the answer is V1' = -2.5 m/s, V2' = +5.9 m/s, which makes no sense.
  5. Jun 7, 2016 #4
    Ah, I found my mistake right after I posted. Now my final answer is V1' = -5.9 m/s, and V2' = 2.5 m/s. I think that is the right answer.
    Thanks for the help :)
  6. Jun 7, 2016 #5


    User Avatar
    Homework Helper

    The balls just exchanged velocity due to the collision. The other solution v1'=2.5 m/s, v2'=-5.9 m/s means that nothing has changed, both balls keep moving with the original velocity. That means they did not collide. But the equations for conservation of momentum and conservation of energy are valid also in the case if the balls just pass each other and do not collide.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted