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Serway Physics Momentum Help

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data
    A cue ball travelling @ 4m/s makes a glancing elastic collision with a target ball of equal bass that is initially at rest. The cue ball is deflected @an angle of 30°. Find the angle between the two velocity vectors; and the speeds of the two balls post collision.


    2. Relevant equations
    [tex] KE= 1/2mv^2 \\
    P=mv
    [/tex]


    3. The attempt at a solution
    [itex] m_1=m_2 \\
    m_1*v_1=m_1*v_{1f} + m_1*v_{2f} \\
    v_{1f}=4-v_{2f}
    [/itex]
    Stuck here. I know I have to use Kinetic Energy, but how?
     
  2. jcsd
  3. Dec 6, 2012 #2
    What's the definition of an elastic collision?
     
  4. Dec 6, 2012 #3
    One where both momentum and kinetic energy are conserved.
     
  5. Dec 6, 2012 #4
    So use conservation of kinetic energy to apply another constraint to the problem.
     
  6. Dec 6, 2012 #5
    [itex]
    .5m_1v_{1o}^2 =.5m_1v_{1f}^2+.5m_1v_{2f}^2 \\
    v_{1o}^2=v_{1f}^2+v_{2f}^2 \\
    v_{1f}=\sqrt{v_1^2-v_2f^2} \\
    [/itex]
    Am I on the right track?
     
  7. Dec 6, 2012 #6
    Yes and also don't forget to separate conservation of momentum into both x and y. Write a conservation equation for both directions and use the results of that as well.
     
  8. Dec 6, 2012 #7
    So the horizontal momentum cancels and the vertical momentum totals the momentum of the cue ball?
    Since the angle of cue ball is 30, I use the two equations to find [itex] v_{1fy} [/itex] and then solve for the horizontal? That's going to be equal and opposite to the struck ball's x velocity, so I have both velocities, use pythagorean theorem and I'm done? Check the logic please!
     
  9. Dec 6, 2012 #8
    well when we write out the momentum in x and y we get

    [itex]x: mv_{1,x} = mv^{'}_{1,x} + mv^{'}_{2,x}[/itex]

    and

    [itex]y: 0 = mv^{'}_{1,y} + mv^{'}_{2,y}[/itex]

    where the prime denotes "after the collision"

    so

    [itex]mv_{1} = mv^{'}_{1} + mv^{'}_{2}[/itex]

    is really only half of it. But can you see how you can use pythagorean's theorem to relate the x and y components to the magnitude of the vector?
     
  10. Dec 6, 2012 #9
    I think your X and Y's are flipped, but yeah I get the point. But I only know two of the 4 quantities, the ones before the collision, and one KE equation. How does that work?
     
  11. Dec 6, 2012 #10

    gneill

    User Avatar

    Staff: Mentor

    Since all the masses are equal we can do away with them in the various expressions. Similarly, we can do away with the (1/2)m's in the kinetic energy terms. Call the initial velocities v and final velocities u, with appropriate subscripts for the x and y components, of course.

    The given departure angle for the cue ball gives you a relationship between its y and x velocities as it departs. Namely, the ratio is tan(30°), or ##1/\sqrt{3}##, so that:
    $$\frac{u1_y}{u1_x} = \frac{1}{\sqrt{3}}$$
    Now, what does conservation of momentum have to say about the relationship between the y-velocities u1y and u2y ?

    And what does it say about about the sum of the x-velocities?

    That's three equations already, and conservation of KE hasn't been considered yet...
     
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