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2 balls involved in 1-D Elastic Collision

  1. Jul 16, 2005 #1
    A ball of mass 0.200 kg has a velocity of 1.50i m/s; a ball of mass 0.300 kg has a velocity of -0.400i m/s (where "i" is supposed to be that unit vector along x-axis). They meet in a head-on elastic collision. (a) Find their velocities after the collision.

    I know I can use the conservation of both momentum and kinetic energy to solve this problem. I've done that. I looked in the solution manual and the person who wrote the solution used a different method I don't comprehend. First, they state the conservation of linear momentum, then they related the final velocities and way I dont understand.

    The following 2 equations are used to solve the problem in the solution manual:

    conservation of momentum for the two-ball system gives us:
    0.200 kg (1.50 m/s) + 0.300 kg(-0.400 m/s) = 0.200 kg V1f + 0.300 kg V2f

    Relative Velocity Equation:
    V2f - V1f = 1.90 m/s


    The reason this second equation confuses me is because, if both the balls had negative velocities after the collision, then V2f - V1f would be a negative velocity, wouldn't it? ( dont know, that's why I'm here, asking :-p ) So, how do they come to that conclusion?
     
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  3. Jul 16, 2005 #2

    HallsofIvy

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    "The reason this second equation confuses me is because, if both the balls had negative velocities after the collision, then V2f - V1f would be a negative velocity, wouldn't it?"

    Not necessarily: (-6)- (-7)= 1. The difference between two negative numbers can be positive.
     
  4. Jul 16, 2005 #3

    Gokul43201

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    Just because both balls have negative velocities does not mean their difference must also be negative. -3 and -4 are negative numbers but -3 - (-4) is not.

    Nevertheless, there is a relevant question here : "Why is this second equation valid?" OR "Why must the relative velocity be reversed ?" Notice that this second equation is basically saying : v2 - v1 = u1 - u2 (assuming the first ball is the lighter one)

    This second equation is often called an "empirical law" and goes by the name of Newton's Law of Restitution. It is an experimentally determined law and as yet, the best derivation I can think of (have not seen an elementary derivation of it) goes along the following lines :

    Imagine you are an observer sitting on ball #2. We shall describe the dynamics of the collision relative to your frame of observation. Writing the energy conservation equation (in your frame, you are at rest and only ball #1 moves) ,we have :

    [tex]KE~(before) = \frac{1}{2}m_1 ~u_{1-rel}^2~~~~~~~--~~~(1) [/tex]

    and

    [tex]KE~(after) = \frac{1}{2}m_1 v_{1-rel}^2 ~~~~~~~--~~~(2)[/tex]

    where

    [tex]u_{1-rel} = u_1 - u_2 ~~and~~v_{1-rel} = v_1 - v_2 [/tex]

    Sunbstituting and equating (1) and (2) gives :

    [tex] \frac {1}{2} m_1(u_1 - u_2)^2 = \frac {1}{2} m_1(v_1 - v_2)^2 [/tex]

    [tex](u_1 - u_2)^2 = (v_1 - v_2)^2 [/tex]

    This allows two possibilities


    [tex]u_1 - u_2 = v_1 - v_2 [/tex]
    or
    [tex]u_1 - u_2 = v_2 - v_1 [/tex]

    Of these two, the second one must be correct as the first would lead to a direct contradiction of momentum conservation.

    It is important to note that this form of Newton's Law of Restitution is valid only for an elastic collision. The more general form is :

    [tex]\frac {v_1 - v_2}{u_1 - u_2} = e ~(coef.~of~restitution)~,~~-1 \leq e \leq 0[/tex]

    For a perfectly elastic collision, e = -1 and for a perfectly inelastic collision, e=0.
     
    Last edited: Jul 16, 2005
  5. Jul 16, 2005 #4
    Thanks a lot for the assistance you guys. this has abeen huge help :)
     
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