Physics

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Hi guys,
Im a bit stumped on how to solve an online physics homework problem and it's really bothering me. I have no idea how to come up with the angles or even attempt the problem and neither do my friends. We'd all really appreciate it if someone could help us.
Many Thanks,
Pooja
A proton, moving with a velocity of viihatbold, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 2.00 times the speed of the initially at rest proton, find the following.
(a) the speed of each proton after the collision in terms of vi
initially moving proton multiplied by vi
initially at rest proton multiplied by vi


(b) the direction of the velocity vectors after the collision
initially moving proton __° relative to the +x direction
initially at rest proton __° relative to the +x direction

  • Physics -

    before collision
    V1x is x velocity of proton 1 (your Vihat)
    V1y = 0 is y velocity
    V2x = V2y = 0 = velocities of proton 2

    after collision
    W1x = x velocity of proton 1
    W1y = y velocity of proton 1
    W2x = x velocity of proton 2
    W2y = y velocity of proton 2

    Now try

    now constraints (m = mass of proton)
    momentum same before and after
    m V1x = m W1x + m W2x
    m V1y = 0 = m W1y + m W2y
    energy(ke) the same before and after
    .5 m V1x^2 = .5m (W1x^2+W1y^2 +W2x^2+W2y^2)
    special 2 times constraint
    W1x^2+W1y^2 = 2 (W2x^2+W2y^2)

  • Typo -

    special 2 times constraint
    W1x^2+W1y^2 = 4 (W2x^2+W2y^2)

    must be 4 not two because it is the speed twice, therefore the square of the speed 4 times

  • Physics -

    (a) They tell you that is is an elastic collision. Let's call the velocities
    V1i = vi = speed of initially moving proton
    V2i = 0 (initially at rest proton)
    Vif (final speed of initially moving proton)
    V2f = (final speed of initially moving proton)
    They tell you that V2f/V2f = 2

    From energy conservation
    Mp vi^2 = Mp V2f^2 + Mp V2f^2/4
    = (5/4)Mp Vf2
    V1f = sqrt(4/5) * vi = 0.8944 vi
    V2f = 0.4472 vi

    (b) Now you have to consider the momentum equation to determine the angles the protons leave. Assume +x is the initial direction of motion.

    Let the angles be A1 and A2
    Total momentum in the y direction must remain zero
    V1f sin A1 = -V2f sin A2
    sin A2/sin A1 = -V1f/V2f = -2
    Mp V1f cos A1 + Mp V2f cos A2 = Mp vi
    Vif cos A1 + (Vif/2) cos A2 = vi
    0.8944 cos A1 + 0.4472 cos A2 = 1

    You have two equations in the two unknown angles A1 and A2. The remaining steps involve some trig which I will leave up to you. If it gets too messy, solve graphically. Remember A1 and A2 must be on opposite sides of the x axis.

  • Physics -

    What is MP?

  • Physics -

    MP = Moving Proton

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