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Elastic collision dilemma

  1. Nov 2, 2006 #1
    I also cross-posted this on another website (dance.net), so I'll just copy & paste.

    I got three completely different answers while using equally legitimate formulas (formulae)?

    This is on elastic collisions. Here's the original problem. It is a purely elastic collision, so there's no energy lost to heat or any of that...

    A 7 kg bowling ball traveling at an unknown speed knocks over a 2 kg bowling pin at rest. After the collision, the bowling ball is traveling at 1.8 m/s, while the bowling pin flies off at 3.0 m/s.



    I used the conservation of momentum equation:

    m1v1 + m2v2 = m1v1' + m2v2', and found v1 (the desired) to be 2.66 m/s, which is the correct answer.

    But...

    I also tried the conservation of kinetic energy equation:

    (1/2)(m1v1^2) + (1/2)(m2v2^2) = (1/2)(m1v1'^2) + (1/2)(m2v2'^2), and found v1 to be 2.4 m/s

    I tried the equation derived from both the conservation of momentum and the conservation of kinetic energy:

    v1 + v1' = v2 + v2' (Do I have to show how I derived it? It's in both my physics book from Giancoli and Barron's review book for AP Physics B)

    and found v1 to be 1.2 m/s.

    I don't think it's a directional problem, since the direction of motion is the same for both. My AP Physics teacher thinks she needs to "mull" over the question a bit. My mom thinks there has to be some limitation on the versatility of the v1 + v1' = v2 + v2' equation...Well, I just can't seem to give a reasonable explanation for getting three different answers out of perfectly legitimate equations. Help, anyone?

    ~H.
     
  2. jcsd
  3. Nov 2, 2006 #2

    OlderDan

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    Only the second of these three has to be correct because it is the only scalar equation of the three. The first and third equations are vector equations that you have applied as if velocity was a scalar.

    What makes you think the first answer is the correct answer?
     
  4. Nov 2, 2006 #3
    That was the answer given in the answer booklet. This particular problem was not from Giancoli; my teacher copied it out of another book (which I can't seem to recall the name of). I am officially lost in the fog.

    Also, I don't see a vector problem, since both the ball and the pin travel in the positive direction.
     
    Last edited: Nov 2, 2006
  5. Nov 2, 2006 #4

    OlderDan

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    If this collision is truly elastic, then the answer book is officially wrong. Can you see how you would show that momentum is conserved if you did not assume the ball and pin were moving in the same direction after the collision?
     
  6. Nov 2, 2006 #5
    ^I'm thinking that the pin can fly off at an angle, and momentum would still be conserved. Then I need to use the x (or y) component of the final velocity of the pin... then everything would be completely complicated. I think I'll stick with the conservation of KE; things get messy with vectors.

    I get it now. I was just not very confident of my answer from the conservation of KE equation, because it was different from the answer book. Thank you so much.
     
  7. Nov 2, 2006 #6

    OlderDan

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    You got it. Since you don't really need it, I'm going to give it to you

    MV = M*v_b*cos θ + m*v_p*cos φ
    M*v_b*sin θ = m*v_p*sin φ

    M is the mass of the ball. m is the mass of the pin. θ is the angle of deflection of the ball from its original direction and φ is the angle the pin's velocity makes with the initial ball direction. V is the initial ball velocity and v_b and v_p are the final velocities. You know V from the energy calculation. The two angles can be found from these two equations.
     
  8. Nov 2, 2006 #7
    Or, here is a handy little equation for future similar problems. This is ONLY for head on collisions with M2(mass 2) initially at rest.
    V1f=((m1-m2)/(m1+m2))V1i , where V1f=final velocity of mass 1,V1i=initial

    V2f=((2m1)/(m1+m2))V1i ,V2f=final velocity of mass 2
     
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