A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob

Question

A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length script l and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? (Use any variable or symbol stated above along with the following as necessary: g. Be sure to use script l from physPad.)
v =

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Thanh Thu 3 years 2021-09-04T19:42:42+00:00 1 Answers 2 views 0

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  1. lethu
    0
    2021-09-04T19:43:42+00:00
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    Answer:

    v = \sqrt{16\cdot g \cdot L}

    Explanation:

    The physical phenomenon is described by the Principles of Momentum Conservation and Energy Conservation:

    Momentum

    m \cdot v = M\cdot \frac{v}{2} + m \cdot v'

    Energy

    \frac{1}{2}\cdot m \cdot v^{2} = \frac{1}{8}\cdot M \cdot v^{2} + \frac{1}{2}\cdot m \cdot v'^{2}

    \frac{1}{8}\cdot M\cdot v^{2} = 2\cdot M\cdot g \cdot L

    The minimum speed of the pendulum bob so that it could barely swing through a complete vertical cycle is:

    \frac{1}{8}\cdot v^{2} = 2\cdot g\cdot L

    v^{2} = 16\cdot g\cdot L

    v = \sqrt{16\cdot g \cdot L}

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