1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Momentum help quesiton please!

  1. Apr 5, 2009 #1
    i used the conservation of momentum equation and conservation of energy equation

    got m1.v1i + m2.v2i = m1.v1f + m2.v2f
    and 1/2 m1 . v1i^2 + 1/2m2 . v2i^2 = 1/2 m1.v1f^2+ 1/2 m2.v2f^2

    i know it's a simutaneous equation but i don't know how to solve it!

    1.1 + 0 = 0.11 x + 0.3y
    5.5 + 0 = 0.55x^2 + .15y^2

    (this is for the elastic one)

    http://img27.imageshack.us/img27/2373/masteringphysics.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 5, 2009 #2
    i used the conservation of momentum equation and conservation of energy equation

    got m1.v1i + m2.v2i = m1.v1f + m2.v2f
    and 1/2 m1 . v1i^2 + 1/2m2 . v2i^2 = 1/2 m1.v1f^2+ 1/2 m2.v2f^2

    i know it's a simutaneous equation but i don't know how to solve it!

    1.1 + 0 = 0.11 x + 0.3y
    5.5 + 0 = 0.55x^2 + .15y^2

    (this is for the elastic one)
     
  4. Apr 5, 2009 #3
    anyone please?
     
  5. Apr 5, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    One way is to solve for y (in terms of x) in equation 1 and substitute that into equation 2. Then you'll have an equation with one unknown, which you can solve.
     
  6. Apr 5, 2009 #5
    i got it wrong!
    can someone show me how to do it please!

    i solved Vf1 to be 3.153 ms^-1
    vf2 to be 0.205 ms^-1

    for the elastic collision
     
  7. Apr 5, 2009 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Show exactly what you did. Start by showing how you solved for y in terms of x in the first equation.
     
  8. Apr 5, 2009 #7
    1.1 + 0 = 0.11 x + 0.3y
    5.5 + 0 = 0.55x^2 + .15y^2

    5.5 = (0.55*x^2) + (0.15*((1.1-(0.11*x)-0.3)^2))
    solving for x i got 3.15

    y = 1.1 - (.11*x) -.3

    y = .45

    thanks for helping
     
  9. Apr 5, 2009 #8

    Doc Al

    User Avatar

    Staff: Mentor

    ...
    Correct this.
     
  10. Apr 5, 2009 #9
    dude i still got the wrong awnser

    is it y = 1.1 / ((0.11x)+0.3)
    y = (1.1 + (0.11x)) /0.3?

    i don't know what the hell is wrong with me this morning, it's 3 am now
     
  11. Apr 5, 2009 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Not quite (almost). Try again.
    Solve this for y. Do it slowly, one step at a time.
     
  12. Apr 5, 2009 #11
    y = ( 1.1 - ( 0.11*X) ) / 0.3

    i'm on my last try for this question, it's like high school all over again
    just want to check with you before i go for my final attempt

    5.5 = (0.55 * X^2) + (0.15 * ( ( ( 1.1 - (0.11*X ) ) / 0.3 ) ^2 ) )
     
  13. Apr 5, 2009 #12

    Doc Al

    User Avatar

    Staff: Mentor

    That's more like it.
     
  14. Apr 5, 2009 #13
    I still managed to get it wrong,

    V1f=2.85
    V2f=2.62

    2.85 was from the solved equation from above,
    and v2f = (1.1-(.11*2.85))/0.3
     
  15. Apr 5, 2009 #14

    Doc Al

    User Avatar

    Staff: Mentor

    When you solve a quadratic equation you get two solutions. Often, only one of the solutions will be physically realistic for a given problem. That's the case here. (Note that the solution that you chose has the first mass moving faster than the second, but in the same direction. How could that be possible unless the masses passed through each other?)

    What's the other solution?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Momentum help quesiton please!
Loading...