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Two identical pucks are on an air table. Puck A has an initial velocity of 2.0m/s in the +x-direction. Puck B is at rest. Puck A collides elastically with Puck B and A moves off at 1.0m/s at an angle of 60 degrees above the x-axis. What is the speed and direction of puck B after the collision?

  • Physics -

    You have two unknowns: the final x and y components of the velocity of Puck B: Vbx and Vby. The two momentum equations (x and y) will let you solve for each. You should not have to use the energy conservation equation at all; they have provided more information than you need. Mathematically speakking, the problem is overdetermined

    M*Vo = M*Va*cos60 + M*Vbx
    0 = M*Va*sin60 -M*Vby

    2 = 1/2 + Vbx

    Vbx = 3/2

    Vby = sqrt3/2

    Vbx^2 + Vby^2 = 9/4 + 3/4 = 3

    After collision
    Vb^2 + Va^2 = 4 = Vo^2
    so kinetic energy really is conserved.

  • Physics -

    What kind of drug is this guy on. How would anyone understand that answer

  • Physics -

    You literally explained nothing. Your numbers just spill onto the page with no explanation of where they came from

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