Physics

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?

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  1. 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.

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

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

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  4. Given:
    Ma = M kg, Va = 2 m/s.
    Mb = M kg, Vb = 0.
    V3 = 1m/s[60o] CCW = Velocity of Ma after collision.
    V4 = ? = Velocity of Mb after collision.

    Momentum before = Momentum after.
    Ma*Va + Mb*Vb = Ma*V3 + Mb*V4.
    M*2 + M*0 = M*1[60] + M*V4,
    2M = M[60] + M*V4,
    Divide by M:

    2 = 1[60] + V4,
    2 = 0.5 + 0.866i + V4,
    V4 = 1.5 - 0.866i = 1.732[-30o] = 30o S. of W. = 1.732m/s[210o] CCW. = Velocity of Mb after collision.

    =

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