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Help me prove this conceptual assumption!

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data
    A batter hits two identical baseballs with the same initial speed and from the same initial height but at different initial angles. Prove that both balls have the same speed at any height h if air resistance can be neglected.

    This was an example in the book and was "proven" completely conceptually. I thought it would be very interesting to tackle it mathematically. I probably have a mistake somewhere, and I'm not very familiar with proofs at all so bare with me. I'm 99% sure my attempt at the proof is non conclusive, help will be very appreciated!


    2. Relevant equations
    [tex]w_{grav}=mgy_1 - mgy_2[/tex]
    [tex]w_{grav}=U_{grav1}-U_{grav2}=-Δu[/tex]
    [tex]u_{grav} = mgy[/tex]
    [tex]\frac{1}{2}mv_{1}^2+mgy_1 = \frac{1}{2}mv_{2}^2+mgy_2[/tex]

    3. The attempt at a solution

    [tex]v_{01}=v_{02}=v_1[/tex]
    [tex]h_1=h_2=y[/tex]
    [tex]m_1=m_2=m[/tex]
    For baseball 1
    [tex]\frac{1}{2}mv_{1}^2+mgy_1 = \frac{1}{2}mv_{2}^2+mgy_2[/tex]
    [tex]\frac{\frac{1}{2}mv_{1}^2+mgy_1}{2} = \frac{\frac{1}{2}mv_{2}^2+mgy_2}{2}[/tex]
    [tex]mv_{1}^2=mv_{2}^2 + \frac{Δu}{2}[/tex]
    [tex]\frac{mv_{1}^2}{m} = \frac{mv_{2}^2 + \frac{Δu}{2}}{m}[/tex]
    [tex]v_{1}^2=v_{2}^2+\frac{Δu}{2m}[/tex]
    [tex]v = \sqrt{v_2^2+\frac{Δu}{2m}}[/tex]

    Now let take into account baseball 2

    [tex]\frac{1}{2}mv_{1}^2+mgy_1 = \frac{1}{2}mv_{2}^2+mgy_2[/tex]
    [tex]\frac{1}{2}mv_{1}^2 - \frac{1}{2}mv_{2}= mgy_2-mgy_1[/tex]
    where it implies
    [tex]-Δk = Δu[/tex]
    so from previous equation
    [tex]v = \sqrt{v_2^2+\frac{Δk}{2m}}[/tex]

    I can go on to simplify this more but it wouldn't do us any good. Anyhow the equation implies that final velocity is in a relationship with no variables holding change in y or angle. Therefore both baseballs will have the same speed at any height y at any angle.
     
    Last edited: Nov 10, 2011
  2. jcsd
  3. Nov 11, 2011 #2

    Andrew Mason

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    Isn't that all you need? Since total energy is the same for each ball and since this does not change with time, we can say that at any moment:

    [tex]\frac{1}{2}mv_{1}^2+mgy_1 = \frac{1}{2}mv_{2}^2+mgy_2[/tex]

    where v1 and v2 are the speeds of the respective balls and y1 and y2 are their respective heights.

    If [itex]y_1 = y_2[/itex] then [itex]mgy_1 = mgy_2[/itex] so:

    [itex]\frac{1}{2}mv_{1}^2 = \frac{1}{2}mv_{2}^2 => v_1 = v_2[/itex]

    AM
     
  4. Nov 11, 2011 #3
    [tex] y_1 ≠ y_2 [/tex] because one implies initial height and the other implies final height of one baseball. Not for height of one ball with the height of another ball. Furthermore, different angles will give different heights.
     
  5. Nov 11, 2011 #4

    Andrew Mason

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    y1 and y2 are the respective heights of the balls in my post. The question asks you to show that if the heights of the balls are the same, that is to say: y1=y2, their speeds are the same. The total energy of each ball is .5mv^2 + mgy. Since they have the same mass and have the same initial kinetic energy and the same initial potential energy, and since energy has to be conserved (in the absence of friction) then

    [tex]mgy_1 + \frac{1}{2}mv_1^2 = E_1 = E_2 = mgy_2 +
    \frac{1}{2}mv_2^2[/tex]


    How are you interpreting the question?

    AM
     
  6. Nov 11, 2011 #5
    I think you are misunderstanding something.

    They say if they have the same initial height not the same initial and final height. They also say that they will have different angles, different angles will mean different maximum heights due to the different trajectories.

    [tex]mgy_1 + \frac{1}{2}mv_1^2 = E_1 = E_2 = mgy_2 +
    \frac{1}{2}mv_2^2[/tex]

    The equation you posted above pertains to a initial height and a final height of one baseball. Not to both baseballs at different angles? But I don't know if that makes a difference.

    Edit: Your reasoning of kinetic and potential essay is spot on, but it doesn't give an explicit proof. Its kind of like --"hey look at equation and assume it will make no difference". I want something more rigorous. What you showed is not a proof.
     
  7. Nov 11, 2011 #6

    Andrew Mason

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    ??? The equation I posted applies to both baseballs at ANY height.

    [itex]mgy_1 + \frac{1}{2}mv_1^2 = E_1 = E_2 = mgy_2 + \frac{1}{2}mv_2^2[/itex] is always true for any height y1 and y2 that the respective balls attain.

    That equation follows from 1. the fact that the two balls have the same mass, the same speed and the same height initially so that they have the same total energy AND 2. the fact that energy is conserved at all times so that they have the same total energy as each other at ALL times.

    The particular question you are asking, it appears, is to prove that the two baseballs will have the same speed if the balls are at the same height: that is y1 = y2. You cannot prove this mathematically. You need to start with premises which are derived from physics. It follows from the premises in 1. and 2. above, which are derived from physics, that the equation I have written is true. If follows mathematically from that equation that if y1=y2 then v1=v2.

    AM
     
  8. Nov 11, 2011 #7
    So does that mean my proof was completely incosistent and nonsensical?

    I see what your saying, that if they will have the same kinetic energy, and since they have the same same mass then they will always have the same velocity. But its too loose a definition, I'm not completely buying that there is no way to prove it. Are you absoloutely sure? To say there is no possible proof sounds like a pretty shaky statement.
     
  9. Nov 12, 2011 #8

    Andrew Mason

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    They have the same TOTAL energy. They only have the same kinetic energy if they have the same potential energy.

    Initially, they have the same potential energy and the same kinetic energy. That is because they start off at the same height and the same speed. They then take different paths because of the different launch angles. If at some point they again have the same potential energy you know they must have the same kinetic energy (since their total energies are always equal).

    You cannot prove things in physics. Proof only works in mathematics. You must start with premises (Newton's laws, conservation of energy) that are assumed to be true. Math then allows you to determine what consequences necessarily flow from those premises.

    AM
     
  10. Nov 12, 2011 #9
    Yes I realize that. So them having the same total energy implies their velocities will be the same at any height? I thought only the same kinetic energy means the same speed at any height?

    So I guess my "proof" and your "proof" are basically the same. My "proof" seems to put me more at ease though since it puts everything in terms of v. :approve:

    Anyhow, thank you for your help! I completely understand now that this statement can't be proved the same way that other statements could be proved. I just wanted to at least have something that would make sense. Its a tough pill to swallow though.
     
    Last edited: Nov 12, 2011
  11. Nov 12, 2011 #10

    Andrew Mason

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    I am confused by your confusion. Since they have the same mass, if they have the same kinetic energy they will have the same speed. They do not always have the same kinetic energy. They will have the same kinetic energy (and, therefore, the same speed) ONLY IF they also have the same height. You simply have to look at the equation for total energy to see that, as I have explained above. Read my earlier posts. This is not that difficult.

    I am mystified by your disappointment. I am also unable to follow the reasoning in your initial post. Your "proof" and my explanation are definitely not the same. I cannot make any sense of what you are trying to show in your first post.

    You cannot prove mathematically that energy is conserved or that the masses are equal or that they start off at the same height. Those things are premises for the problem.

    The conclusion that the two balls will have the same speed at the same height follows from the premises. That conclusion can be and has been proven mathematically to follow from the premises.

    Your conclusion: "Therefore both baseballs will have the same speed at any height y at any angle." is simply not correct. The balls will have the same speed if and when they are BOTH at the same height and at no other times.

    AM
     
    Last edited: Nov 12, 2011
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