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What is the velocity of mass 2m after the collision?

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Mass of 2m slides on a horizontal plane with the velocity of v0 across the x-axis. And hits the mass m which is in rest. after the collision mass m moves with the velocity of v0/2 in the direction which creates an angle of 30 degrees of the original movmen of 2m.
    a) What is the velocity of mass 2m after the collision?
    b) What is the direction of mass 2m after the collision?
    c) Is the collision described is perfectly ellastic collision?

    15zjt5v.jpg

    2. Relevant equations
    P=m*v
    Pi=Pf
    3. The attempt at a solution
    I have broken down the velocity to x,y


    Pinitialx=Pfinalx
    m1*V1ix+m2*V2ix=m1*V1fx+m2*v2fx
    2m*v0+m*0 = 2m*Vfcosθ+m*v0/2*cos30


    Pintialy=Pfinaly
    m1*V1iy+m2*V2iy=m1*V1fy+m2*v2fy
    0 = -2m*Vfsinθ+m*V0/2sin30

    I dont know what to do from here. i cant simplfy the equations,
    could you help me with that?

    Thank you,
     
  2. jcsd
  3. May 26, 2015 #2
    You have 2 unknowns so you want to get two expressions that arise from conservation of momentum in both x and y directions. You will have two simultaneous equations which you can solve for the unknowns. I am not sure if you have already done that its hard to follow those expressions but once you get your two equations its relatively straight forward to solve by substitution. Remember that your answer will be in terms of [tex] m\ and \ v_0 [/tex]
     
  4. May 26, 2015 #3
    I think you meant "V0/2*sin30" as in (V0/2)*sin30.

    Hint: sinx/cosx = tanx
     
  5. May 26, 2015 #4
    My problem is that i just cant simplfy it. could you show me what to do with the parameters? V0 and the θ angles.
    Thank you for helping me.
     
  6. May 26, 2015 #5

    SammyS

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    Your figure has Vf as u. That's a bit handier.

    What are the values of sin30° and cos30° ?

    Solve 0 = -2m*u*sinθ+m*(u/2)sin30 for u. (EDIT: This is corrected below.)

    Solve 0 = -2m*u*sinθ+m*(V0/2)sin30 for u.

    Plug that result into the other equation:

    2m*u+m*0 = 2m*u*cosθ+m*(u/2)*cos30


    2m*u+m*0 = 2m*u*cosθ+m*(V0/2)*cos30

    2m*V0+m*0 = 2m*u*cosθ+m*(V0/2)*cos30

    Solve for cotθ (Lots of stuff cancels.)
     
    Last edited: May 28, 2015
  7. May 27, 2015 #6
    U=-m*sin tetha-sin30/2 ???
     
  8. May 27, 2015 #7
    Doesn't u cancel from this equation?
     
    Last edited: May 27, 2015
  9. May 27, 2015 #8

    SammyS

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    DUH! Yes it does!

    Thanks for noticing.

    I had a major typo there!

    It's now been corrected. (I hope correctly!)

    Solve 0 = -2m*u*sinθ+m*(V0/2)sin30 for u.
     
    Last edited: May 27, 2015
  10. May 27, 2015 #9
    why does it cancels?

    umm,
    iam really confused why does it cancels?
    we are left with u/2 dont we?
     
  11. May 28, 2015 #10
    Shouldn't the first u in this equation be V0?
     
  12. May 28, 2015 #11

    SammyS

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    Yup !
     
  13. May 29, 2015 #12
    This is what i got:
    0 = -2m*u*sinθ+m*(V0/2)sin30 for u.
    u=-m*sinθ+0.5v0*sin30
    2m*V0+m*0 = 2m*u*cosθ+m*(V0/2)*cos30 solve for cotθ
    cosθ=u+0.5v0*30

    when plugging u into cosθ to solve for cotθ

    cotθ=-m+0.5v0*sin30+0.5v0*cos30

    What now?
    Thank you for all of your help :)
     
  14. May 29, 2015 #13

    SammyS

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    The mass, m, cancels out of both equations.

    You did not solve for u correctly in the first equation.

    Your result for the second equation is also far from correct.
     
  15. May 29, 2015 #14
    Could you plz show me how to do it correctly,
    I tried to do it, bit by bit... and it still not right.

    Thanks
     
  16. May 29, 2015 #15

    SammyS

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    I'll get you started.

    To solve 0 = -2m⋅u⋅sinθ+m⋅(V0/2)sin30 for u, subtract the first term from both sides. (Add its opposite.)

    2m⋅u⋅sinθ = m⋅(V0/2)sin30

    Then divide both sides by 2m⋅sinθ. Right? What do you get?
     
  17. May 29, 2015 #16
    I got
    u=05V0*sin30/2sinθ
     
  18. May 29, 2015 #17

    SammyS

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    Using the "reply" feature will help you make more readable posts. Then you can cut & paste more accurately. Also, to make a subscript, use the X2 icon on the green strip above the message box you use to form your message to post.

    You should use parentheses around that denominator. The leading 05 is missing a decimal point. You might as well divide 0.5 by 2 .
    u = (0.5)V0⋅sin30/(2⋅sinθ)

    u = (0.25)V0⋅sin30/(sinθ)​

    By the way, do you know the value of sin(30°) ?
     
  19. May 29, 2015 #18
    So
    u = (0.125)V0/(sinθ)
    How is that looking?

    Thank you very much, you have no idea how glad iam, that you are helping me.
     
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