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Homework Help: Is this the right way to work out this pool ball momentum problem?

  1. Nov 14, 2012 #1
    Please help me! Is this problem worked correctly...? If not, why?

    1. In a pool ball game, consider the collision of two .5kg balls. One ball is at rest initially, the other approaches with a speed vi= 10m/s. The collision is not head-on, and after collision, ball #1 moves with a speed of 6.0m/s in a direction 60deg away from the initial line of motion. What is the speed and direction of ball#2?

    Okay, so I know this is Elastic Collision, so momentum and energy are conserved.

    Pi = Pf
    Pi(x) =Pf(x)
    Pi(y) = Pf(y)

    m(1,i)(v) + m(2,i)(v) = m(1,f)(v) + m(2,f)(v)

    P(x): (.5)(10) + (.5)(0) = (.5)(cos(60))(6) + (.5)(cos(θ))(vf)

    P(y): (.5)(10) + (.5)(0) = (.5)(cos(60))(6) + (.5)(sin(θ))(vf)

    Now I'm doing y/x to figure out the θ.

    y/x = 5/5 = (1.5 + .5sinθ(vf))/(1.5 + .5cos(vf))

    1 = tanθ
    θ = 45deg

    Put the 45deg back into x-comp. momentum:

    (.5)(10) + (.5)(0) = (.5)cos(60)(6) + (.5)(cos45)(vf)
    vf = 9.90m/s

    ....I'm thinking this is incorrect, though. It seems the vf is too high...
    Please help me figure this problem out! Thank you! :)
  2. jcsd
  3. Nov 14, 2012 #2


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    The way you are writing things, I think your second equation is wrong:

    You say

    P(y): (.5)(10) + (.5)(0) = (.5)(cos(60))(6) + (.5)(sin(θ))(vf)

    But it appears to me that both the functions should be sin, and in the y-direction, perpendicular to the initial motion I think you mean, both balls have a zero velocity.

    I am not certain all the supplied information is consistent either - but that will show up later.
  4. Nov 14, 2012 #3
    Okay, I found where this problem was worked out previously in my notes...
    Aparantly, the answer is v=8.0m/s and the θ= 40.5deg

    All the teacher showed us to do was to use the conservation of kinetic energy formula since this is elastic collision.

    I knew this was elastic collision, but I thought we also had to take into account the θ when finding the velocity.
    This is what I THOUGHT it had to be:

    1/2(.5)(10)^2 + 1/2(.5)(0) = (1/2)(.5)cos(60)(6)^2 + (1/2)(.5)cosθ(vf)

    This is what it REALLY IS:

    1/2(.5)(10)^2 + 1/2(.5)(0) = (1/2)(.5)(6)^2 + 1/2(.5)(vf)

    And that's easy to figure out.
    vf = 8m/s

    And the θ we can figure out to be:
    Pi(y) = Pf(y)
    (.5)(0) + (.5)(0) = (.5)sin(60)(6) + (.5)sin(θ)(8) *SIGN SUPPOSED TO BE (-) HERE. WHY?
    θ = 40.5deg

    *@PeterO: You were right! In y direction, it's supposed to be 0m/s for velocity! Of course! And it's supposed to be sinθ always in y-direction! Thanks! :)

    So yes... that's how it is really worked out, but how am I supposed to know when I should include the sin(θ) or cos(θ) before the velocity???
    Are you only supposed to add the sin or cos θ when dealing with momentum and NOT kinetic energy?

    Do you EVER include sinθ or cosθ before the velocity when calculating kinetic energy?
    Thanks SO much!!!
  5. Nov 14, 2012 #4


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    You draw a diagram, and that shows whether you use sin or cos. It varies depending which angle you define.
    Consider the ball #1. the 60 degree angle was between the initial direction [if it had continued on] and the direction it took.
    Once you sketch those two directions, and mark in the 6 m/s, you see that when you add the components, the right angle triangle formed requires cos to calculate the component in the original direction, and sin to calculate the component perpendicular to the original direction.
    If the angle had been defined differently, the sine and cosine function use may have been the other way round.

    The data supplied is inconsistent - it can't happen that way.

    Before collision, we had only one ball travelling at 10 m/s, after collision, we had one ball travelling at 6 m/s, and the other at some other speed.

    In all collisions, momentum is conserved, so we can draw the momentum vectors as a triangle, with the two separate "after" vectors adding to the single "before" vector.

    Vectorially mV1 + mV2 = mVi

    This collision is elastic, so Kinetic energy is conserved also, so..

    1/2mV12 + 1/2mV22 = 1/2mVi2

    divide each of those equations by m and 1/2m respectively and we get.

    As Vectors V1 + V2 = Vi

    numerically V12 + V22 = Vi2

    That is true ONLY if the triangle we are dealing with is a right angled triangle.

    Indeed, in an elastic 2-D collision between billiard balls, the two balls always depart at right angles to each other.

    The velocity values of 6, 8 and 10 you teacher calculated [which is consistent with conservation of energy] form the classic Pythagorean triangle, which has angles of approx 53 degrees and 37 degrees.

    So the statement that the first ball, moved with speed 6 m/s at an angle of 60 degrees with the original direction is impossible!!
    If the angle is 60 degrees, the speed is not 6 m/s
    If the speed is 6 m/s, the angle is approx. 53 degrees.

    We have a faulty question!! Which could have been corrected by omitting the reference to 60 degrees.

    Note : You never use the sin and cos components in the energy equation - only in the (vector) momentum equation.
  6. Nov 14, 2012 #5


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    Perhaps the assumption that the collision is elastic isn't true. So calculate the speed of the and direction based on momentum only. Staying with the convention that the ball is initially moving in the x direction at 10 m/s:

    Pi(x) = .5 (10 m/s)
    Pi(y) = .5 (0 m/s)

    Then what is Pf(x) and Pf(y)?

    You'll end up with 2 equations with 2 unknowns (angle and speed of ball #2). You'll also be able to calculate the energy lost during the collision.
    Last edited: Nov 14, 2012
  7. Nov 14, 2012 #6


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    Agreed. You should always start with momentum - then check later whether the energy had been conserved.
  8. Nov 15, 2012 #7


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    This should be:

    Code (Text):

    P(x): (.5)(10) + (.5)( 0) = (.5)(cos(60))(6) + (.5)(cos(θ))(vf)
    P(y): (.5)( 0) + (.5)( 0) = (.5)(sin(60))(6) + (.5)(sin(θ))(vf)
    Note that θ will be negative. If you want θ to be positive then use:

    Code (Text):

    P(x): (.5)(10) + (.5)( 0) = (.5)(cos(60))(6) + (.5)(cos(θ))(vf)
    P(y): (.5)( 0) + (.5)( 0) = (.5)(sin(60))(6) - (.5)(sin(θ))(vf)
    Last edited: Nov 15, 2012
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