A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered at an angle of 85.0 0 to the initial dire

Question

A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered at an angle of 85.0 0 to the initial direction of the bowling ball and with a speed of 15.0 m/s. (a) Calculate the final velocity (magnitude and direction) of the bowling ball.

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Nguyệt Ánh 4 years 2021-08-31T16:08:56+00:00 1 Answers 71 views 0

Answers ( )

  1. Dulcie
    0
    2021-08-31T16:10:04+00:00
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    Answer:

    9.05 m/s ,   -14.72°  (respect to x axis)

    Explanation:

    To find the final velocity of the bowling ball you take into account the conservation of the momentum for both x and y component of the total momentum. Then, you have:

    p_{xi}=p_{xf}\\\\p_{yi}=p_{yf}\\\\

    m_1v_{1xi}+m_2v_{2xi}=m_1v_1cos\theta+m_2v_{2}cos\phi\\\\0=m_1v_1sin\theta-m_2v_2sin\phi

    m1: mass of the bowling ball = 5.50 kg

    m2: mass of the bowling pin = 0.850 kg

    v1xi: initial velocity of the bowling ball = 9.0 m/s

    v2xi: initial velocity of bowling pin = 0m/s

    v1: final velocity of bowling ball = ?

    v2: final velocity of bowling pin = 15.0 m/s

    θ: angle of the scattered bowling pin = ?

    Φ: angle of the scattered bowling ball = 85.0°

    Where you have used that before the bowling ball hits the pin, the y component of the total momentum is zero.

    First you solve for v1cosθ in the equation for the x component of the momentum:

    v_1cos\theta=\frac{m_1v_{1xi}-m_2v_2cos\phi}{m_1}\\\\v_1cos\theta=\frac{(5.50kg)(9.0m/s)-(0.850kg)(15.0m/s)cos85.0\°}{5.50kg}\\\\v_1cos\theta=8.79m/s

    and also you solve for v1sinθ in the equation for the y component of the momentum:

    v_1sin\theta=\frac{(0.850kg)(15.0m/s)sin(85.0\°)}{5.50kg}\\\\v_1sin\theta=2.3m/s

    Next, you divide v1cosθ and v1sinθ:

    \frac{v_1sin\theta}{v_1cos\theta}=tan\theta=\frac{2.3}{8.79}=0.26\\\\\theta=tan^{-1}(0.26)=14.72

    the direction of the bawling ball is -14.72° respect to the x axis

    The final velocity of the bawling ball is:

    v_1=\frac{2.3m/s}{sin\theta}=\frac{2.3}{sin(14.72\°)}=9.05\frac{m}{s}

    hence, the final velocity of the bawling ball is 9.05 m/s

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