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Physics Elastic Collision Question

  1. Mar 15, 2010 #1
    1. The problem statement, all variables and given/known data
    a .060kg tennis ball, moving with a speed of 2.5 m/s, collides head-on with a .09kg ball initially moving away from it at a speed of 1.15m/s. Assuming a perfect collision, what are the speed and direction of each ball after the collision


    2. Relevant equations

    MaVa + MbVb = MaV’a + MbV’b (conservation of momentum)

    ½MaV²a +½ MbV²b = ½MaV’²a + ½MbV’²b (?)

    3. The attempt at a solution
    I plugged in the numbers into both equations and got....

    1. (.06kg)(2.5m/s) + (.09kg)(1.15m/s) = (.06kg)(V1f) + (.09kg)(V2f)

    and...

    2. 1/2(.06kg)(2.5m/s)^2 + 1/2(.09kg)(1.15m/s)^2 = 1/2(.06kg)(V1f)^2 + 1/2(.09kg)(V2f)^2

    i then started to solve the second equation...

    .25 = 1/2(.06kg)(V1f)^2 + 1/2(.09kg)(V2f)^2

    this is where I became confused....I got rid of the first part (1/2(.06kg)(V1f)^2) and just solved the second part [ 1/2(.09kg)(V2f)^2]

    .25 = 1/2(.09kg)(V2f)^2

    .25 = .05(V2f)^2

    .25/.05 = (V2f)^2

    5 = (V2f)^2

    sqrt(5) = V

    2.24 = V

    I then went back and plugged 2.24 into the first equation ......and then rearranged the equation:

    (.06kg)(2.5m/s) + (.09kg)(1.15m/s) = (.06kg)(V1f) + (.09kg)(V2f) =

    - (.06kg)(V1f) = (.09kg) (2.23) - (.2523)

    -.06kg (V1f) = -.0528

    (-.0528) / -(.06) = V1f

    v= .88

    I know the answers are supposed to be .88 and 2.23 (back of the book), but it seems like I just randomly solved this (I don't get exactly how i got this), is this right?

    I really need to pass physics (architecture major) but i don't understand it, any help would be greatly appreciated.
     
    Last edited: Mar 15, 2010
  2. jcsd
  3. Mar 16, 2010 #2

    ehild

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    "this is where I became confused....I got rid of the first part (1/2(.06kg)(V1f)^2) and just solved the second part [ 1/2(.09kg)(V2f)^2]"

    You can not do that. You can not just omit a term from an equation.

    You have two equations with two unknowns.
    Solve the first equation for v1f and substitute the expression you get in the second equation.


    ehild
     
  4. Mar 16, 2010 #3
    when you say to solve the first equation for V1f do you mean:

    (.06kg)(2.5m/s) + (.09kg)(1.15m/s) = (.06kg)(V1f) + (.09kg)(V2f)

    .2535 = (.06kg)(V1f) + (.09kg)(V2f)

    .06kg(V1f) = .2535 + .09kg(V2f)


    .06kg/.06(V1f) = .2535/.06 + .09kg/.06(V2f)

    V1f = 4225 + 1.5 V2f
     
  5. Mar 16, 2010 #4

    ehild

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    It is good up to here, but the next steps are wrong:
    Check it.

    ehild
     
  6. Mar 16, 2010 #5
    can you please just get me going in the right direction i cant figure out how to solve it because of the two unknowns
     
  7. Mar 16, 2010 #6

    ehild

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    Lets start from here: 0.2535=0.06*v1f+0.09*v2f

    You want v1f be alone on one side of this equation. Subtract 0.09*v2f from both sides. Do it now.

    ehild
     
  8. Mar 16, 2010 #7
    is this right....

    -.06(V1f) = .09 (V2f) - .2535
     
    Last edited: Mar 16, 2010
  9. Mar 16, 2010 #8

    ehild

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    Yes. Now get rid of the coefficient of v1f: divide both sides by -0.06.
     
  10. Mar 16, 2010 #9
    i got this:

    V1f = - 1.5 (V2f) + 4.225

    ....but, how do i plug this into the second equation?
     
  11. Mar 16, 2010 #10
    1/2(.06kg)(2.5m/s)^2 + 1/2(.09kg)(1.15m/s)^2 = 1/2(.06kg)[(-1.5)(V2f) +4.225)^2 + 1/2(.09kg)(V2f)^2

    .25 = 1/2(.06kg)[(1.5)(V2f) + 4.225]^2 + 1/2(.09kg)(v2f)^2



    ....would this be right?
     
  12. Mar 17, 2010 #11

    ehild

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    There should be a - sign in front of (1.5) (v2f).

    Keep at least 4 significant digits on the left side of the equation.

    It is not needed to use units in the equation. If you do, use them everywhere.

    ehild
     
  13. Mar 17, 2010 #12
    .2470 = 1/2(.06)[-(1.5)(V2f) + 4.225]^2 + 1/2(.09)(v2f)^2


    .2470 = .03(-(1.5)(V2f) + 4.225)^2 + .045(V2f)^2


    .2470 = .03((-1.5V2f) + 4.225)^2 + (.045V2f)^2


    Is this right so far?
     
  14. Mar 17, 2010 #13

    ehild

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    Why do you square 0.045? It has to be outside the parentheses.

    ehild
     
  15. Mar 17, 2010 #14
    .2470 = .03((-1.5V2f) + 4.225)^2 + .045(V2f)^2

    Do I distribute the .03 and .045, or am I missing something?
     
  16. Mar 17, 2010 #15

    ehild

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    You can distribute a sum: a(b+c) = ab+ac. (v2f)^2 is not a sum.



    ehild
     
  17. Mar 17, 2010 #16
    Oh, I see. So would I just leave the equation as is?


    .2470 = .03((-1.5V2f) + 4.225)^2 + .045(V2f)^2

    How do I go about answering when V2f is in there twice? Would I have to subtract one of them like so...

    .2470 - .045(V2f)^2 = .03((-1.5V2f) + 4.225)^2

    Sorry for all the questions I'm just really confused about this problem.
     
  18. Mar 17, 2010 #17

    ehild

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    You can do that but it has not much sense.

    First evaluate the square. Apply the identity: (a+b)^2=a^2+2ab+b^2

    Next distribute 0.03.

    Arrange the equation so one side is 0. Group like terms.

    It is a quadratic equation. Use the quadratic formula to solve it.

    You can find step-by step explanations for solving equations at

    http://www.sosmath.com/algebra/solve/solve0/solve0.html.

    ehild
     
  19. Mar 17, 2010 #18
    this is what I have so far:


    0 = 6.75(V1f)^2 + 53.55 + .045(V2f)^2 - .2470

    Not to be so negative......but I think I did something wrong
     
    Last edited: Mar 17, 2010
  20. Mar 17, 2010 #19

    ehild

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    It is not correct.

    Evaluate the square by using the identity (a+b)^2=a^2+2ab+b^2: a=*-1.5 (v2f), b=4.225.

    (-1.5(V2f) + 4.225)^2= 2.25 (v2f)^2-3(4.225) (v2f)+4.225^2=2.25 (v2f)^2-12.675 (v2f)+17.8506.

    This expression is multiplied by 0.03. Distribute.

    ehild
     
    Last edited: Mar 18, 2010
  21. Mar 18, 2010 #20
    After distributing I got:

    .0675 V2f^2 - .3802 V2f + .5355


    The original equation was:

    .2470 = .03((-1.5V2f) + 4.225)^2 + .045(V2f)^2

    I added .045 V2f^2 to .0675 V2f^2 and then subtracted .2470 from .5355

    and got:

    .1075 V2f^2 - .3802 V2f + .2885

    Using the quadratic formula:

    .3802 +/- √-.3802^2- 4 (.1075) (.2885) / 2(.1075)

    adding:

    .3802 + √.1445 - .1240 / .215

    .3802 + √.0205 / .215

    .3802 + .1431 / .215

    .5233 / .215 = 2.43

    subtracting:

    .3802 - .1431 /.215

    .2371 / .215 = 1.10


    Did I do something wrong because my book says the answers should be .88 and 2.23
     
    Last edited: Mar 18, 2010
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