A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless pla

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A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless plane before the collision, what is the velocity of the block after the bullet passes through

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Thu Nguyệt 5 years 2021-08-11T23:11:05+00:00 1 Answers 57 views 0

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  1. ThanhThu
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    2021-08-11T23:12:20+00:00
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    Answer:

    The final velocity of the block after the bullet passes through is 0.66 meters per second.

    Explanation:

    The interaction between the bullet and the block of woods is a clear example of a perfectly inelastic collision, which can be modelled after the Principle of Momentum Conservation. There are no external forces exerted on the bullet-block system. The equation describing the collision is described below:

    m_{B}\cdot v_{B,o} + m_{W}\cdot v_{W,o} = m_{B}\cdot v_{B,f} + m_{W}\cdot v_{W,f}

    Where:

    m_{B}, m_{W}– Masses of the bullet and the block of wood, measured in kilograms.

    v_{B,o}, v_{W,o} – Initial speeds of the bullet and the block of wood, measured in meters per second.

    v_{B,f}, v_{W,f}– Final speeds of the bullet and the block of wood, measured in meters per second.

    The final speed of the block is cleared:

    v_{W,f} = \frac{m_{B}\cdot (v_{B,o}-v_{B,f})+m_{W}\cdot v_{W,o}}{m_{W}}

    v_{W,f} = v_{W,o} + \frac{m_{B}}{m_{W}} \cdot (v_{B,o}-v_{B,f})

    If v_{W,o} = 0\,\frac{m}{s}, m_{B} = 0.022\,kg, m_{W} = 2\,kg, v_{B,o} = 210\,\frac{m}{s} and v_{B,f} = 150\,\frac{m}{s}, then the final velocity of the block after the bullet passes through is:

    v_{W,f} = 0\,\frac{m}{s}+\left(\frac{0.022\,kg}{2\,kg}\right)\cdot \left(210\,\frac{m}{s}-150\,\frac{m}{s} \right)

    v_{W,f} = 0.66\,\frac{m}{s}

    The final velocity of the block after the bullet passes through is 0.66 meters per second.

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