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18 May, 09:44

In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s [forward], and after the impact, rolls away at an angle of 35.0° counterclockwise from its initial direction with a velocity of 3.2 m/s. What is the velocity of the eight ball after the collision?

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  1. 18 May, 10:33
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    another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

    Explanation:

    given data

    mass m1 = 0.165 kg

    mass m2 = 0.155 kg

    before collision velocity v1 = 5.8 m/s

    before collision velocity v2 = 0

    angle = 35.0° from initial direction

    after collision 1st ball velocity v3 = 3.2 m/s

    to find out

    after collision another ball velocity v4

    solution

    we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

    so from conservation of momentum we say

    m1v1 = m1v3cos35 + m2v4cosθ with x axis ... 1

    m1v3sin35 = m2v4sinθ with y axis ... 2

    so from 1 equation

    0.165 * 5.8 = 0.165 (3.2) cos35 + 0.155 (v4) cosθ

    v4 cosθ = 3.38 ... 3

    form 2 equation

    0.165 (3.2) sin35 = 0.155 (v4) sinθ

    v4 sinθ = 1.95 ... 4

    so magnitude of another ball velocity is square and adding equation 3 and 4

    another ball velocity = √ (3.39²+1.96²)

    another ball velocity = 3.92 m/s

    and direction is tanθ = 1.96/3.39

    θ = 30° clockwise from initial direction
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