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26 February, 12:44

A proton, moving with a velocity of vii⁄, collides elastically with another proton that is initially at rest. Assuming that the two protons have equal speeds after the collision, find (a) the speed of each proton after the collision in terms of vi and (b) the direction of the velocity vectors after the collision.

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  1. 26 February, 12:50
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    Given that,

    Mass of proton

    Mp=m=1.627*10^-27kg

    One of the proton is at rest then it velocity is 0

    Let the other proton be moving at vi

    Since the after collision the two proton moves together with the same velocity (i. e inelastic collision)

    Then, using conservation of energy

    Kinetic energy before collision = kinetic energy after collision

    Given that, K. E=½mv²

    Before collision = after collision

    ½mvi²+½m (0) ²=½ (m+m) vf²

    ½mvi²=½ (2m) vf²

    ½mvi²=mvf²

    Divide through by m

    ½vi²=vf²

    vi²=2vf²

    Take square root of both sides

    √vi²=√ (2vf²).

    vi=√2 * vf

    Then, the final velocity is

    vf = vi / √2

    b. Direction of the velocity vectors after collision

    Let vf1 be the final velocity for the incident proton

    vf2 be the final velocity for the proton initially at rest.

    Conserving momentum in the ˆj direction

    Piy = 0 = Pfy = mp•vf1y + mp•vf2y

    vf1y = - vf2y

    So the protons have equal magnitude speeds in the ˆj direction. Because the speed of the particles are equal, the magnitude of their speeds in the ˆi direction should also be equal |vf1x| = |vf2x|.

    Conserving momentum in the ˆi direction.

    Pix = mp•vi = Pfx = mp•vf1x + mp•vf2x = 2mp•vfx

    Then, vfx = vi/2

    Using the Pythagorean theorem to solve for the magnitude of vfy

    vf²=vi²/2 = vfx² + vfy² = vi²/4+vfy²

    vfy = vi√ (½-¼) = vi/2=vfx

    So because the ˆi and ˆj components of vf are the same, both protons are deflected away at an angle of θ = 45° from the ˆi direction, with opposite ˆj components (so the angle between vf1 and vf2 is 90°).
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