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Final speed and direction after a collision (elastic+inelastic)

  1. Aug 4, 2017 #1
    1. The problem statement, all variables and given/known data

    A billiard ball moves at a speed of 4.00 m/s and collides ELASTICALLY with an identical stationary ball. As a result, the stationary ball flies away at a speed of 1.69 m/s. Determine

    a. the final speed and direction of the incoming ball after the collision

    b. the direction of the stationary ball after the collision

    2. Relevant equations

    p = p'
    px = px'
    py = py'

    3. The attempt at a solution

    I assume the net external force is zero, therefore, momentum is conserved in both the x and y direction.

    I can say:

    m1*v1x + m2*v2x = m1*v1x' + m2*v2x'

    Since the two balls are identical, masses cancel. Furthermore, the v2x is 0 because it is stationary.


    v1x = v1x' + v2x' --> I want v1x' therefore, v1x' = v1x - v2x'

    I know v1x is 4.00 m/s (the initial velocity is only in x direction) so I can write: v1x' = 4 - v2x' (this is equation one)

    I did the same for the y-direction --> py = py'

    The initial velocity for both particles is 0 (particle one is only horizontal and then particle two is stationary), I can now say:

    0 = m1*v1y' + m2*v2y' --> This becomes m1*v1y' = -m2*v2y'

    The masses cancel because they are identical, and I am left with v1y' = -v2y' (this is equation two)

    Then, since the collision is elastic, I know that momentum is conserved as well as kinetic energy. So I can write,

    1/2m1*v1^2 + 1/2m2*v2^2 = 1/2m1*v1'^2 + 1/2m2*v2'^2

    Again, it condenses to:

    1/2m1*v1^2 = 1/2m1*v1'^2 + 1/2m2*v2'^2 (masses cancel and 1/2 cancels to give):

    v1^2 = v1'^2 + v2'^2 (I know v1 and also v2', so can plug it into the equation to give):

    16 = v1'^2 + 2.8561

    Therefore, v1' = 3.63 m/s

    After getting the speed of particle one after the collision, i am lost and can not seem to get the x and y components to do the tangent relationship to find the direction?

    b) I am lost here too...

    Any help will be GREATLY appreciated. Thank you in advance.
  2. jcsd
  3. Aug 4, 2017 #2


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    How is the speed related to the x and y components of velocity?
  4. Aug 4, 2017 #3
    v1x' = 3.63 cos (theta1)

    v1y' = 3.63 sin (theta 1)


    v2x' = 1.69 cos (theta 2)

    v2y' = -1.69 sin (theta 2)

    Are these correct? Where do I go from here though? How do I find the angle?
  5. Aug 4, 2017 #4


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    I was thinking about how to express the speed v1', say, in terms of v1'x and v1'y. If you can determine the x and y components of the velocity, you can then determine the angle.
  6. Aug 4, 2017 #5
    Im so confused.

    Do you mean use pythagreom theorem to say that the velocity that I found is equal to the square root of the (v1x')^2 + (v1y'^2)?
  7. Aug 4, 2017 #6


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  8. Aug 4, 2017 #7
    If I use that process, then I have:

    3.63 = the square root of ( v1x'^2 + v1y'^2)

    I know that v1x' = 4 - v2x' and that v1y' = -v2y'....

    How would I go from there? I still have two unknowns? :/

    Thank you so much!
  9. Aug 4, 2017 #8


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    OK. Similarly for the second ball.
  10. Aug 4, 2017 #9
    It comes to 13.1769 = 16 - 8v2x' + v2x'^2 + v2y'^2

    I still have two unknown variables?
  11. Aug 4, 2017 #10


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    You have four equations for the four unknowns v1'x, v1'y, v2'x, v2'y. Two of these equations are the momentum equations, and the other two equations are the pythagorean equations.
  12. Aug 4, 2017 #11
    Okay. So...

    I have the two momentum equations that I found to be:

    v1x' = 4 - v2x'
    v1y' = -v2y'

    then I have for the pythagorean equations:

    3.63 = the square root of (v1x'^2 + v1y'^2)
    1.69 = the square root of (v2x'^2 + v2y'^2)

    If I plug one of the momentum into the pythagorean equations, I still have two variables though? Or do I solve one of the pythagorean equations for, lets say, v1y' and then set that equal to v1y' in the momentum equation? But then I still have two variable??
  13. Aug 4, 2017 #12


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    I would square the pythagorean equations to get rid of the square roots. Then use your momentum equations to write the the v1'-squared-pythagorean equation in terms of v2'x and v2'y. You can then work with this equation and the other squared-pythagorean equation.
  14. Aug 4, 2017 #13
    Okay so after a couple pages of computations, i got theta for the first particle to be 65 degrees, does that seem right? That you for your help on that part. Would I just plug back in the equations to find the second part?
  15. Aug 4, 2017 #14
    My only issue after solving is that the angle for the stationary object is smaller than the other. In the diagram that I am given, it seems that theta 2 is larger?
  16. Aug 4, 2017 #15


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    It looks to me like you have the answers switched. Are you sure you're not getting 65 degrees for ball 2 (that was initially stationary)?
  17. Aug 7, 2017 #16
    After checking everything...

    I got 65.2 degrees below the horizontal for particle 2 and 25 degrees above the horizontal for particle 1. I originally did the tan^-1 of (x/y) instead of (y/x), which gave me the wrong answer originally for particle 1, making it seem larger than the particle two angle after collision.
  18. Aug 7, 2017 #17


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    OK. It turns out that you can show that whenever you have an elastic collision with equal mass balls and one of the balls is initially at rest, it always turns out that the angle between the final velocities of the two balls is 90o. Thus, as a check, you are getting 65o + 25o = 90o.
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