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Collision of two balls where some kinetic energy is lost

  1. Aug 31, 2012 #1
    1. The problem statement, all variables and given/known data[/b]

    Two smooth uniform spheres, A and B, are of the same size, but A has twice the mass of B. They are placed at rest on a smooth horizontal table and A is projected directly towards B with speed U. The spheres collide, and then B strikes a barrier perpendicular to its path, rebounds, and hits A again. Find the velocities of A and B immediately after this second collision between A and B, given that the coefficient of restitution is 1/2 at all three impacts, and neglecting air resistance.

    2. Relevant equations

    Conservation of linear momentum
    law of restitution

    3. The attempt at a solution

    Defining some variables :

    [itex]U_{0}[/itex]= velocity of A before collision
    [itex]U_{1}[/itex]= velocity of A after collision with B(which was stationary)
    [itex]U_{3}[/itex]= velocity of A after collison with B for the second time (where B was moving)

    [itex]V_{0}[/itex]=0= velocity of B initially
    [itex]V_{1}[/itex]=velocity of B after collision with A for the first time
    [itex]V_{2}[/itex]= Velocity of B after collision with the barrier
    [itex]V_{3}[/itex]= Velocity of B after the collision with A for the 2nd time.

    Now here is the solution :

    Momentum is conserved in the First collision thus :

    1) 2[itex]U_{0}[/itex] = 2[itex]U_{1}[/itex]+[itex]V_{1}[/itex]

    by law of restitution for first collision :

    2) 0.5 = -[itex](U_{1}-V_{1})/U_{0}[/itex]

    3) by law of restitution in the collision between ball B and the barrier :
    4) conservation of momentum between ball A and ball B
    5)newton's law of restitution in the 2nd collision between ball A and B.


    basically we have to find [itex]V_{3}[/itex] and [itex]U_{3}[/itex]
    After solving all the five equations i get :

    [itex]V_{3}[/itex]= 0.5[itex]U_{0}[/itex]
    [itex]U_{3}[/itex]= 3/8[itex]U_{0}[/itex]

    But the correct answers are :

    [itex]V_{3}[/itex] same as mine
    [itex]U_{3}[/itex]= 0

    I have checked my working for any arithmetic mistakes many times, but i dont think its that kind of a mistake.
    It seems to me that the error is related with +/- signs which i apply when forming these five equations specially when i use the velocity of B after colliding with the barrier, because it changes direction.

    Equations 1 and 2 will always carry correct signs. (becuase i know they are correct)
    its the equations 3,4 and 5 which involve [itex]V_{2}[/itex] and that is where i get confused .

    can someone form equations 3 ,4 and 5 for me. For example equation 4 as i formed it was :

    2[itex]U_{1}[/itex]-[itex]V_{2}[/itex] = 2[itex]U_{3}[/itex] + [itex]V_{3}[/itex]

    Now in the above equation i used the -ve sign for [itex]V_{2}[/itex], is that right ?

    And if use -[itex]V_{2}[/itex] instead of +[itex]V_{2}[/itex], i arrive at my result which i just quoted above (which is wrong )
     
    Last edited: Aug 31, 2012
  2. jcsd
  3. Aug 31, 2012 #2

    TSny

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    Review the definition of coefficient of restitution to see if you have a sign error here.
     
  4. Aug 31, 2012 #3
    Yes, that was just type error, the problem still remain.
    i have edited the first post and made it right .

    Can u for equation four correctly for me
     
  5. Aug 31, 2012 #4

    TSny

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    I believe you want to write 2[itex]U_{1}[/itex]+[itex]V_{2}[/itex] on the left side rather than with a minus sign. The direction of [itex]V_{2}[/itex] will be accounted for by the value of [itex]V_{2}[/itex] being positive or negative.
     
  6. Aug 31, 2012 #5
    I am confused about it, whether to write it as positive hence considering [itex]V_{2}[/itex] as a vector or -ve thus considering it as speed (not velocity)

    But if i take it as positive (as a vector) the i must take all the values of [itex]V_{2}[/itex] as positive, right ?

    Here is what happens if i take it as a vector (ie +ve) :

    equation 3)

    2([itex]V_{2}[/itex]-[itex]V_{1}[/itex]) = [itex]U_{1}[/itex]-[itex]V_{1}[/itex]

    From Equations 1 and 2 :

    [itex]U_{1}[/itex]= [itex]\frac{1}{2}[/itex][itex]U_{0}[/itex]
    and
    [itex]V_{1}[/itex]= [itex]U_{0}[/itex]

    Substituting these values in equation 3 gives us :
    [itex]V_{2}[/itex]= 0.25[itex]U_{0}[/itex]

    Now this value doesn't make sense, because [itex]V_{2}[/itex] should be in the -x direction, because it rebounded from the wall, it cant be in the direction of its initial velocity, that is just wrong.

    Can someone help me understand this ...
     
    Last edited: Aug 31, 2012
  7. Aug 31, 2012 #6

    TSny

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    I don't understand this equation. Aren't you trying to express what happens when the second ball collides with the barrier? Ball 1 doesn't take part in this collision. You should be able to get [itex]V_{2}[/itex] from just [itex]V_{1}[/itex] and the coefficient of restitution.
     
  8. Sep 1, 2012 #7
    That is absolutely right and i think this is the reason i was getting a wrong answer.

    Many thanks to you .

    Since the collision is between B and the wall, the "correct" equation 3 should be :

    3) [itex]V_{2}[/itex] = 0.5 [itex]V_{1}[/itex]

    P.S : I am taking [itex]V_{2}[/itex] as a vector, so that i dont have to use -ve signs with the velocities .
    and finally my answer matches !
     
    Last edited: Sep 1, 2012
  9. Sep 1, 2012 #8

    TSny

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    I think this equation still needs a minus sign. (These signs are a nightmare, aren't they.) We know that ball B will change direction when it hits the wall. So, the [itex]V_{2}[/itex] vector must have the opposite direction of the [itex]V_{1}[/itex] vector. So, it must be [itex]V_{2}[/itex] = -0.5 [itex]V_{1}[/itex].

    We can derive this by applying the coefficient of restitution equation to the collision with the wall. We would have

    e = -[itex]\frac{V_{2}-W_{2}}{V_{1}-W_{1}}[/itex]

    where [itex]W_{1}[/itex] and [itex]W_{2}[/itex] are the velocity vectors of the wall before and after the collision and e is the coefficient of restitution. Of course, [itex]W_{1}[/itex] = [itex]W_{2}[/itex] = 0. So, the equation will reduce to [itex]V_{2}[/itex] = -e [itex]V_{1}[/itex].
     
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