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4 November, 04:57

A projectile proton with a speed of 1100 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 45° from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?

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  1. 4 November, 05:06
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    So, the target proton's speed is 777.82 m/s

    And, the projectile proton's speed is 777.82 m/s

    Explanation:

    as per the system, it conserves the linear momentum,

    so along x axis:

    Mp V1 (i) = Mp V1 (f) cos θ1 + Mp V2 (f) cos θ2

    along y axis:

    0 = - Mp V1 (f) sin θ1 + Mp V2 (f) sin θ2

    let us assume before collision it was moving on positive x axis, hence target angle will be θ2 = 45° from x axis

    V2 (f) = V1 (i) sin θ1 / (cosθ2 sin θ1 + cos θ1 sin θ2)

    = 1100 * sin 45 / (cos 45 sin 45 + cos 45 sin 45)

    = 1100 * 0.7071 / (0.7071 * 0.7071 + 0.7071 * 0.7071)

    = 777.82 / (0.5 + 0.5)

    = 777.82 m/s

    (b)

    the speed of projectile, V1 (f) = sinθ2 * V2 (f) / sinθ1

    = sin 45 * 777.82 / sin 45

    = 777.82 m/s

    So, the target proton's speed is 777.82 m/s

    And, the projectile proton's speed is 777.82 m/s
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