Two loads of mass $m_1$ and $m_2$ ($m_1 > m_2$) are connected by a thread thrown over a fixed block. At the initial moment, the load of mass $m_1$ is held at a height $h$ above the floor. Then it is released without a push. How much heat will be released when the load hits the floor? The impact is absolutely inelastic.

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Solutions of Savchenko Problems in Physics
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    <h3 id="back-link"><a href="/#2.4">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
      $2.4.34$
      Two loads of mass $m_1$ and $m_2$ ($m_1 \geq m_2$) are connected by a thread thrown over a fixed block. At the initial moment, the load of mass $m_1$  is held at a height $h$ above the floor. Then it is released without a push. How much heat will be released when the load hits the floor? The impact is absolutely inelastic.
    </p>

    <center>
        <figure>
          <img src="statement.png"
            loading="lazy"  width="250" />
          <figcaption>
            For problem $2.4.34$
          </figcaption>
        </figure>
      </center>

    <h3>Solution</h3>
    <p>
        Applying Energy Conservation Law:
        $$m_1 g h = Q + m_2 g H\\;(1)$$
        where $Q$ is lost energy by heat and $H$ is the height achieved by body of mass $m_2$ after the impact of body of mass $m_1$, and is given by,
        $$H = h + \frac{v^2}{2g}\\;(2)$$
        Why? Potential energy of body 1 is converted in kinetic energy of body 1 and kinetic energy of body 2 with same velocity due to the constraint ($v_1=v_2=v$), after totally inelastic collision with ground, part of this energy is transformed in heat and the other one in potential energy for body 2, which behaves as a projectile launched vertically upwards from height $h$. For determining this velocity $v$, it's necessary to know what acceleration did the system have before impact. Applying Newton Second Law, for body 1:
        $$m_1 g - T = m_1 a \\;(3)$$
        and for body 2,
        $$T - m_2 g = m_2 a\\;(4)$$
        Suming up equations (3) and (4) side by side,
        $$a = \frac{m_1-m_2}{m_1+m_2}g\\;(5)$$
        As $h = \frac{v^2}{2a}$, so,
        $$v = \sqrt{2ah}\\;(6)$$
        Substituting (5) into (6),
        $$v = \sqrt{2gh\left(\frac{m_1-m_2}{m_1+m_2}\right)}\\;(7)$$
        Putting (7) into (2)
        $$H = \frac{2m_1}{m_1+m_2}h\\;(8)$$
        Finally, substituting (8) into (1) and separating $Q$,
    </p>
    <h4>Answer</h4>
    <p>
        $$Q = m_1gh \frac{m_1-m_2}{m_1+m_2}$$
    </p>


    <p style="text-align: right; font-style: italic; font-size: 14;">   
      BSc. Luis Daniel Fernández Quintana<br>
      Physics Department (FCNE)<br>
      Universidad de Oriente, Cuba<br>
    </p>



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