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Homework Help: Angle between two pool balls in an elastic collision

  1. Mar 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider an elastic collision (ignoring friction and rotational motion).
    A queue ball initially moving at 2.6 m/s strikes a stationary eight ball of the same size
    and mass. After the collision, the queue ball’s final speed is 1.2 m/s.
    Find the queue ball’s angle (theta) with respect to its original line of motion.
    Answer in units of ◦.

    I've attached a picture to make it a little clearer

    2. Relevant equations

    kinetic energy = 1/2mv^2
    momentum = mv

    3. The attempt at a solution

    I've tried a couple of things, but its really the math that's tripping me up I think.
    First I found the velocity of the ball that was hit. I got 3.504...etc. using the kinetic energy equation.

    The next thing I did was try to solve it in terms of the two angles, theta and phi. Since the two y-components must be equal to zero (because there was no y component to the original velocity),

    Eq. 1: 0 = 1.2 sin(theta) + 3.504 sin(phi)

    And because the x components must add up to the original x velocity,

    Eq. 2: 2.6 = 1.2 cos(theta) + 3.504 cos(phi)

    When I try to make these into a system, I get a very unruly equation that leaves me dumbfounded.

    My alternative was to separate each of the two velocity vectors into their components, and this ended up getting me a four equation system which again, left me dumbfounded.

    By similar logic to above, the four equations I came up with were:

    vxcue + vxother = 2.6
    vy cue + vy other = 0
    (vx cue)2 + (vy cue)2 = 1.2
    (vx other)2 + (vy other)2 = 3.504

    Note: Everytime I typed 3.504 I actually used the full calculator value.

    So my question, I guess, is whether or not there is a simpler way to solve this, and if not, if anyone could give me a hand with the math, its confusing me big time.
     

    Attached Files:

  2. jcsd
  3. Mar 16, 2010 #2
    The speed of the 8-ball cannot be greater than the speed of the queue ball.
    I have no idea about the math, but i would guess that the angels of the two balls to the original direction after collision, should be proportional to their speeds.
    A special case of hitting the 8-ball is that both balls roll away at angles of 45 to the original direction and at a speed of 1.3m/s each.
     
    Last edited: Mar 16, 2010
  4. Mar 16, 2010 #3
    Thanks, I'll rework and see what I can do. I recalculated the second balls velocity to be 2.306512519, don't know how I screwed that up the first time through.
     
  5. Mar 16, 2010 #4
    Well, that changes the value, so thanks for pointing that out, but it doesn't change the math :/ Thanks though
     
  6. Mar 16, 2010 #5
    no idea where you get that velocity from
    The speed of the 2nd ball is 2.6-1.2 = 1.4 !
    as the masses are equal, the sum of all speeds before the collision must be = the sum of all speeds after the collision.

    http://en.wikipedia.org/wiki/Collision#Billiards
     
  7. Mar 16, 2010 #6
    But can you do that since they go off at different angles?

    I solved for it using conservation of energy..

    so .5(8)(2.6)^2 = .5(8)(1.2)^2 + .5(8)(x)^2

    x = 2.306... no?
     
  8. Mar 16, 2010 #7
    remember [tex]cos{^2}x+sin{^2}x=1[/tex]

    With a little bit of algebra you should be able to get a quadratic equation in terms of cos (phi).
    1) can be rewritten as 1.2 sin (theta) = -3.504 sin (phi)
    2) can be rewritten as 1.2 cos (theta) = 2.6-3.504 cos (phi)
    (although you still need to correct those "3.504" to the correct value)
     
    Last edited: Mar 16, 2010
  9. Mar 16, 2010 #8
    I can get that far...what I'm confused about is, from there you have to solve for phi in one of the equations, and then plug that into the other right? So, if I solve for phi in the first equation, I get arcsin([1.2sin(theta)]/-2.04). When I plug that into the other equation, I can't see what to do from there. I don't know how to simplify the arcsin and sin combo, and I'm having a hard time finding out how any of this could be changed into a quadratic form...

    Edit: 2.04 is my corrected velocity value, approximately
     
  10. Mar 17, 2010 #9
    You should get an equation containing only phi without theta.
    1.2 sin (theta) = -2.04 sin (phi)
    1.2 cos (theta) = 2.6-2.04 cos (phi)

    What happens when you square both equations and add them together?
     
  11. Mar 17, 2010 #10
    As i pointed out before the correct speeds are:

    queue ball = 1.2
    8-ball = 2.6 - 1.2 = 1.4
     
    Last edited: Mar 17, 2010
  12. Mar 17, 2010 #11
    Bartleby, you're missing...quite a bit here. The speeds you're giving don't describe an elastic collision, or even one where momentum is conserved.
     
  13. Mar 17, 2010 #12
    Now that I've actually got a calculator, I'm not getting 2.04 m/s for the 8 ball.

    sqrt(2.6^-1.2^2)=?
     
    Last edited: Mar 17, 2010
  14. Mar 17, 2010 #13
    You really think the speeds after the collision are greater ? You think those balls accelerate for the rest of their life on that table ?

    http://www.real-world-physics-problems.com/physics-of-billiards.html
     
    Last edited: Mar 17, 2010
  15. Mar 17, 2010 #14
    You clearly don't know the distinction between speed and velocity.

    I suggest you get the nearest introductory physics book, and start reading chapter 1 where it discusses the difference between vectors and scalars.
     
  16. Mar 17, 2010 #15
    Really sorry, I mistyped, don't know where I got that 2.04. Meant approximately 2.306. Again though, its the math beyond that that's causing me a problem. I read somewhere that the angle between the two balls will be 90 degrees, but I couldn't find any more info on it, could that be used to solve the problem?
     
  17. Mar 17, 2010 #16
    [tex]1.2 sin \theta = -2.306 sin \phi [/tex]

    [tex]1.2 cos \theta = 2.6 - 2.306 cos \phi[/tex]

    Square both sides of both equations:

    [tex]1.44 sin{^2}\theta = 5.318 sin{^2} \phi[/tex]


    [tex]1.44 cos{^2}\theta = 6.76 -11.991 cos \phi + 5.318 cos{^2} \phi[/tex]

    add them together and you get:

    [tex]1.44=6.76 - 11.991 cos \phi +5.318[/tex]

    The rest should be simple.
     
    Last edited: Mar 17, 2010
  18. Mar 17, 2010 #17
    Ah, can't believe I didn't see that. Thanks a ton, lemme give it a whirl.
     
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