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| <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | | <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> |
| </div> | | </div> |
| <p class="author"> | | <p class="author"> |
| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
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| <h3 id="back-link"><a href="../../#2.4">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.4">$\leftarrow$Back</a></h3> |
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| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="250" /> | | loading="lazy" width="250" /> |
| <figcaption> | | <figcaption> |
| For problem $2.4.34$ | | For problem $2.4.34$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Applying Energy Conservation Law: | | Applying Energy Conservation Law: |
| $$m_1 g h = Q + m_2 g H\;(1)$$ | | $$m_1 g h = Q + m_2 g H\;(1)$$ |
| 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, | | 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, |
| $$H = h + \frac{v^2}{2g}\;(2)$$ | | $$H = h + \frac{v^2}{2g}\;(2)$$ |
| 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: | | 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: |
| $$m_1 g - T = m_1 a \;(3)$$ | | $$m_1 g - T = m_1 a \;(3)$$ |
| and for body 2, | | and for body 2, |
| $$T - m_2 g = m_2 a\;(4)$$ | | $$T - m_2 g = m_2 a\;(4)$$ |
| Suming up equations (3) and (4) side by side, | | Suming up equations (3) and (4) side by side, |
| $$a = \frac{m_1-m_2}{m_1+m_2}g\;(5)$$ | | $$a = \frac{m_1-m_2}{m_1+m_2}g\;(5)$$ |
| As $h = \frac{v^2}{2a}$, so, | | As $h = \frac{v^2}{2a}$, so, |
| $$v = \sqrt{2ah}\;(6)$$ | | $$v = \sqrt{2ah}\;(6)$$ |
| Substituting (5) into (6), | | Substituting (5) into (6), |
| $$v = \sqrt{2gh\left(\frac{m_1-m_2}{m_1+m_2}\right)}\;(7)$$ | | $$v = \sqrt{2gh\left(\frac{m_1-m_2}{m_1+m_2}\right)}\;(7)$$ |
| Putting (7) into (2) | | Putting (7) into (2) |
| $$H = \frac{2m_1}{m_1+m_2}h\;(8)$$ | | $$H = \frac{2m_1}{m_1+m_2}h\;(8)$$ |
| Finally, substituting (8) into (1) and separating $Q$, | | Finally, substituting (8) into (1) and separating $Q$, |
| </p> | | </p> |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$Q = m_1gh \frac{m_1-m_2}{m_1+m_2}$$ | | $$Q = m_1gh \frac{m_1-m_2}{m_1+m_2}$$ |
| </p> | | </p> |
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| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
| </p> | | </p> |
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