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9 April, 18:06

A uniform thin rod of length 0.60 m and mass 3.5 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 14 rad/s immediately after the collision, what is the bullet's speed just before impact? (Answer is in m/s)

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  1. 9 April, 20:50
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    The moment of inertia for the rod is

    mr*L^2/12

    After collision

    I=mr*l^2/12+mb*r^2

    I=0.073

    The momentum of the rod after collision is

    I*w

    =1.022

    The momentum of the bullet that is radial

    to the rod is

    vr=v*sin (60)

    vr/r=w

    so the radial momentum of the bullet is

    3*vr / (0.25*1000)

    vr*0.012

    applying conservation of momentum

    vr*0.012=1.022

    vr=85 m/s

    and

    v=85/sin (60)

    v=98 m,/s

    j
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