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25 December, 22:30

A billiards ball B rests on a horizontal surface and is struck by another billiards ball A of the same mass m = 0.2 kg. Ball A is initially moving at 18 m/s and is deflected 28° from its original direction. Assume the collision is elastic. a) find the final velocity vector of each ball, b) find the mechanical energy both before and after the collision.

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  1. S
    25 December, 23:47
    0
    v1 = 15.90 m/s

    v2 = 8.46 m/s

    mechanical energy before collision = 32.4 J

    mechanical energy after collision = 32.433 J

    Explanation:

    given data

    mass m = 0.2 kg

    speed = 18 m/s

    angle = 28°

    to find out

    final velocity and mechanical energy both before and after the collision

    solution

    we know that conservation of momentum remain same so in x direction

    mv = mv1 cosθ + mv2cosθ

    put here value

    0.2 (18) = 0.2 v1 cos (28) + 0.2 v2 cos (90-28)

    3.6 = 0.1765 V1 + 0.09389 v2 ... 1

    and

    in y axis

    mv = mv1 sinθ - mv2sinθ

    0 = 0.2 v1 sin28 - 0.2 v2 sin (90-28)

    0 = 0.09389 v1 - 0.1768 v2 ... 2

    from equation 1 and 2

    v1 = 15.90 m/s

    v2 = 8.46 m/s

    so

    mechanical energy before collision = 1/2 mv1² + 1/2 mv2²

    mechanical energy before collision = 1/2 (0.2) (18) ² + 0

    mechanical energy before collision = 32.4 J

    and

    mechanical energy after collision = 1/2 (0.2) (15.90) ² + 1/2 (0.2) (8.46) ²

    mechanical energy after collision = 32.433 J
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