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When do I use Kinetic Energy and Momentum?

  1. Nov 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A billiard ball (Ball #1) moving at 5.00 m/s strikes a stationary ball (Ball #2) of
    the same mass. After the collision, Ball #1 moves at a speed of 4.35 m/s. Find
    the speed of Ball #2 after the collision.

    2. Relevant equations
    Conservation of Kinetic Energy
    Conservation of Momentum

    3. The attempt at a solution
    I used the Pi = Pf equation of conservation of momentum and got 0.65 m/s while the conservation of Kinetic Energy equation yielded the correct answer: 2.47 m/s. What's wrong? Why couldn't I use the momentum equation too?
     
  2. jcsd
  3. Nov 6, 2015 #2
    As the momenta are vectors you need to have also directions to solve it. Imagine your scenario calculated with the conservation of momentum like you did: Your formula only applies if both balls are moving in the same direction after the collision, but the first ball is faster than the second one → the first ball would have to overtake the second one. In this case you can use the conservation of energy (which is scalar and not a vector), because it is assumed that the collision between two billard balls is elastic.
     
  4. Nov 6, 2015 #3
    Did you have an angle of 90° between the two final velocities?
     
  5. Nov 6, 2015 #4
    Would the conservation of momentum equation also be valid because the quantities that I used had - (left) and + (right) signs? I assumed Ball #1 moved to the right. The question did not say whether the problem involved an elastic head-on collision so would I be right in saying that I should be using the conservation of momentum rather than the conservation of kinetic energy equation (KEi = KEf)?
     
  6. Nov 6, 2015 #5
    No, I didn't. I assumed that the masses are colliding on a flat, frictionless surface.
     
  7. Nov 6, 2015 #6

    haruspex

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    The question is certainly ill-defined. You are not told whether it is a head-on collision, nor whether it is elastic. As a result, there is a range of possible answers.
    To get an answer, you will need to assume one of the following:
    1. It is elastic, but the collision is not head on.
    2. It is inelastic, head-on, and the author made a mistake regarding which ball is moving at 4.35m/s afterwards.
    You also have to assume no friction between the balls and the table.
     
  8. Nov 7, 2015 #7
    It is a 2D-problem, you need to take into account the velocity components perpendicular to the initial moving direction of the first ball before the collision; so it is not enough distinguishing the final values only by the sign.

    Billard ball collisions are more or less elastic (not only in theory), therefore they are a commonly used for examples and exercises regarding elastic collisions. You can use the conservation of momentum for solving, but you will not get one definite answer, but an infinite number of possible solutions.
     
  9. Nov 7, 2015 #8

    NascentOxygen

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    If you assume the collision is in 1-dimension and the moving ball rebounds along its path, conservation of momentum will give you a velocity for ball #2. But a check of KE will indicate the two bodies have more combined KE after the collision than before. Such an answer doesn't make sense.
     
  10. Nov 7, 2015 #9
    Me, too. The 90° angle I mentioned is measured in the horizontal plane (surface of billiard ball table).

    I think the intent of this problem is for you discover precisely what you are now discovering. The only way both kinetic energy and momentum can both be conserved is if it's a two-dimensional collision. Your textbook should have it all worked out, probably in a worked example, showing that the only way to conserve both momentum and energy is if the angle between the two balls' paths is 90°.

    Anyway, by using energy conservation you have calculated the correct answer. If you want to do that using momentum conservation you need to consider two dimensions. Resolve the momentum vectors into x- and y-components, or draw the triangle they form. If you go the component route, when you make a choice of the orientations of the x- and y-axes it affects the amount of calculating you have to do to get to the answer.

    In this diagram ##\vec{p_\ ^\ }## is the total momentum before the collision, ##\vec{p_1'}## and ##\vec{p_2'}## are the momenta of Balls #1 and #2 after the collision, respectively. Thus $$\vec{p_\ ^\ }=\vec{p_1'}+\vec{p_2'}.$$

    momentum.png
     
  11. Nov 7, 2015 #10

    Ray Vickson

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    The easiest way to do such problems is to first go into the so-called "center of mass" (CM) frame, which is a new coordinate system that moves at a constant velocity with respect to the original ("lab") frame and in which the initial total momentum = 0. Let ##k = 2.5## (m/sec), so that the initial velocity of ball 1 in the lab frame is ##\vec{v}_{i1} = (2k,0)## and the initial velocity of ball 2 in the lab frame is ##\vec{v}_{i2} = (0,0)##. The velocity of the origin of the CM frame is ##(k,0)##, so the initial velocities of balls 1 and 2 in the CM frame are ##\vec{v}_{i1}' = (k,0)## and ##\vec{v}_{i2}' = (-k,0)##. If ##\theta \in (-\pi,\pi]## is the deflection angle of ball 1 in the CM frame, the final (post-collision) velocities in the CM frame are ##\vec{v}_{f1}' = (k \cos(\theta), k \sin(\theta))## and ##\vec{v}_{f2}' = (- k \cos(\theta), - k \sin(\theta))## if the collision is elastic. Note that total momentum and kinetic energies are preserved automatically, for any value of ##\theta##.

    For a head-on collision we would have ##\theta = \pi##, so ##\vec{v}_{f1}' = (-k,0)##, meaning that ##\vec{v}_{f1} = (-k,0) + (k,0) = (0,0)## in the lab frame; that is, ball 1 would come to rest, and ball 2 would continue forward at speed 2k = 5.0 m/s. This does not match the given problem conditions, so the collision cannot be head-on if it is elastic. However, for general ##\theta## we have ##\vec{v}_{f1} = (k,0) + (k \cos(\theta), k \sin(\theta)) = k (1 + \cos(\theta), \sin(\theta))## in the lab frame. The terminal speed of ball 1 in the lab frame is ##s_{f1} = k \sqrt{(1 + \cos (\theta))^2 + \sin^2 (\theta)} = k \sqrt{2} \sqrt{1 + \cos (\theta)} = 4.35## m/sec. Thus, we have
    [tex] \sqrt{1 + \cos (\theta)} = \frac{4.35}{2.5 \sqrt{2}} \Rightarrow \cos(\theta) = 0.5138000000 = 0.5138 [/tex]
    There are, of course, two values for ##\theta##, corresponding to an upward or downward deflection through the same angle.

    Once you know ##\theta## you can evaluate ##\vec{v}_{f2}'## (CM) and from that get ##\vec{v}_{f2}## (lab) and the speed ##s_{f2} = |\vec{v}_{f2}|## in the lab frame. You can also find the angle of deflection of ball 1 in the lab frame. Can you see how?
     
    Last edited: Nov 7, 2015
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