A cue ball of mass ml 0.365 kg is shot at another billiard ball, with mass m2 055 kg, which is at rest. The cue ball has an initial speed of

Question

A cue ball of mass ml 0.365 kg is shot at another billiard ball, with mass m2 055 kg, which is at rest. The cue ball has an initial speed of v = 8.5 m/s in the positive direction. Assume that the collision is elastic and exactly head-on. ? 25% Part (a) write an expression for the horizontal component of the billiard ball’s velocity, vr after the collision, in terms of the other variables of the problem. -& 25% Part (b) what is this velocity, in meters per second? 25% Part (c) Write an expression for the horizontal component of the cue ball’s velocity, vr, after the collision. 25% Part (d) what is the horizontal component of the cue ball’s final velocity, in meters per second?

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Bình An 5 years 2021-08-10T02:16:38+00:00 1 Answers 23 views 0

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  1. giabao
    0
    2021-08-10T02:18:10+00:00
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    Answer:

    Part a)

    Expression of speed of billiard ball is given as

    v_r = \frac{2m_1 v}{m_1 + m_2}

    Part b)

    Velocity of billiard ball is given as

    v_r = 6.78 m/s

    Part c)

    Expression for the speed of cue ball is given as

    v_c = \frac{(m_1 - m_2) v}{m_1 + m_2}

    Part d)

    Speed of cue ball is given as

    v_c = -1.72 m/s

    Explanation:

    Part a)

    As we know that there is no external force on the system of two balls

    so momentum is conserved here

    so we have

    m_1 v + m_2 (0) = m_1 v_c + m_2 v_r

    also we know that the collision is elastic collision so we have

    v_r - v_c = v - 0

    so we have

    v_c = v_r - v

    so we have

    m_1 v = m_1(v_r - v) + m_2 v_r

    v_r = \frac{2m_1 v}{m_1 + m_2}

    Part b)

    Velocity of billiard ball is given as

    v_r = \frac{2(0.365)(8.5)}{0.365 + 0.55}

    v_r = 6.78 m/s

    Part c)

    Now the speed of cue ball is given as

    v_c = v_r - v

    v_c = \frac{2m_1 v}{m_1 + m_2} - v

    v_c = \frac{(m_1 - m_2) v}{m_1 + m_2}

    Part d)

    Now by above formula the speed of the cue ball is given as

    v_c = \frac{0.365 - 0.55}{0.365 + 0.55}(8.5)

    v_c = -1.72 m/s

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