Physics

posted by kt

A proton of mass m undergoes a head on collision with a stationary atom of mass 13m. If the initial speed of the proton is 485 m/s, find the speed of the proton after the collision.

  1. Anonymous

    485 m = m u + 13 m v
    now you have to know if the collision is elastic or if they proton sticks to the atom
    If they stick
    then u = v
    if elastic
    then (1/2) m(485^2) = (1/2) m u^2 + (1/2)(13 m) v^2

  2. Henry

    Given:
    M1 = m, V1 = 485 m/s.
    M2 = 13m, V2 = 0.
    V3 = Velocity of proton(M1) after collision.

    Momentum before = Momentum after.
    M1*V1 + M2*V2 = M1*V3 + M2*V4.
    m*485 + 13m*0 = m*V3 + 13m*V4,
    485m + 0 = m*V3 + 13m*V4,
    Divide both sides by m:
    485 = V3 + 13V4,

    V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
    V3 = (485(m-13m) + 26m*0)/14m = -5820m /14m = -416 m/s.

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