A 0.5 kg block slides down a frictionless semicircular track from a height, h, and collides with a second 1.5 kg block at the bottom. The tw

Question

A 0.5 kg block slides down a frictionless semicircular track from a height, h, and collides with a second 1.5 kg block at the bottom. The two blocks stick together via some Velcro and slide up the other side of the track. What height do the two blocks reach?

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Hồng Cúc 4 years 2021-07-22T15:30:52+00:00 1 Answers 21 views 0

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  1. trucchi
    0
    2021-07-22T15:31:58+00:00
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    Answer:

    h' = 0.062\cdot h

    Explanation:

    The speed of the 0.5 kg block just before the collision is found by the Principle of Energy Conservation:

    U_{g} = K

    m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}

    v = \sqrt{2\cdot g \cdot h}

    v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot h}

    v \approx 4.429\cdot \sqrt{h}

    Knowing that collision is inelastic, the speed just after the collision is determined with the help of the Principle of Momentum Conservation:

    (0.5\,kg) \cdot (4.429\cdot \sqrt{h}) = (2\,kg)\cdot v

    v = 1.107\cdot \sqrt{h}

    Lastly, the height reached by the two blocks is:

    K = U_{g}

    \frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot h'

    h' = \frac{v^{2}}{2\cdot g}

    h' = \frac{(1.107\cdot \sqrt{h})^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

    h' = 0.062\cdot h

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