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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> |
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| <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> |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.1.9.$ To test the equipment in zero-gravity conditions, the container is thrown up by a pneumatic piston device located at the bottom of the evacuated shaft. The piston acts on the container for a time $\Delta t$ with a force $F = nmg$, where $m$ is the mass of the container with the equipment. How long will it take for the container to fall to the bottom of the mine? How long does the zero-gravity state last for the equipment, if $\Delta t = 0.04$ s, and $n = 125$? | | $2.1.9.$ To test the equipment in zero-gravity conditions, the container is thrown up by a pneumatic piston device located at the bottom of the evacuated shaft. The piston acts on the container for a time $\Delta t$ with a force $F = nmg$, where $m$ is the mass of the container with the equipment. How long will it take for the container to fall to the bottom of the mine? How long does the zero-gravity state last for the equipment, if $\Delta t = 0.04$ s, and $n = 125$? |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| <p> | | <p> |
| 1. The container's movement can be divided into three sections: in the acceleration section $OA$. the piston exerts a force $F = nmg$, which accelerates the container to the speed $v_0$; in the second section $AB$ the container moves like a body thrown vertically upwards, in the third section, after stopping, the container with the equipment will freely fall to the bottom of the shaft. | | 1. The container's movement can be divided into three sections: in the acceleration section $OA$. the piston exerts a force $F = nmg$, which accelerates the container to the speed $v_0$; in the second section $AB$ the container moves like a body thrown vertically upwards, in the third section, after stopping, the container with the equipment will freely fall to the bottom of the shaft. |
| </p> | | </p> |
| <p> | | <p> |
| 2. Let us write the equation of the fundamental law of dynamics for the acceleration section, which, in combination with the kinematic conditions of uniformly accelerated motion, allows us to determine the values $y_1$, $t_1$ and $v_0$ | | 2. Let us write the equation of the fundamental law of dynamics for the acceleration section, which, in combination with the kinematic conditions of uniformly accelerated motion, allows us to determine the values $y_1$, $t_1$ and $v_0$ |
| $$ nmg-mg = ma $$ | | $$ nmg-mg = ma $$ |
| | | |
| $$ a = g(n-1) = 1240 \,m/s^2 $$ | | $$ a = g(n-1) = 1240 \,m/s^2 $$ |
| | | |
| $$ V_0 = a \Delta t = g(n-1) \Delta t = 50 \,m/s $$ | | $$ V_0 = a \Delta t = g(n-1) \Delta t = 50 \,m/s $$ |
| | | |
| $$ y_1 = \frac{a\Delta t^2}{2} = \frac{g (n-1)\Delta t^2}{2} = 2 \,m $$ | | $$ y_1 = \frac{a\Delta t^2}{2} = \frac{g (n-1)\Delta t^2}{2} = 2 \,m $$ |
| 3. Let us determine the time of lifting the container from point $A$ to point $B$ and the value $y_2$ | | 3. Let us determine the time of lifting the container from point $A$ to point $B$ and the value $y_2$ |
| $$ t_2 = \frac{v_0}{g} = (n-1) \Delta t = 5\,c $$ | | $$ t_2 = \frac{v_0}{g} = (n-1) \Delta t = 5\,c $$ |
| | | |
| $$ y_2 = v_0 t_2 - \frac{gt_2^2}{2} $$ | | $$ y_2 = v_0 t_2 - \frac{gt_2^2}{2} $$ |
| | | |
| $$ y_2= g(n-1)^2 \Delta t^2 - \frac{g}{2}(n-1)^2 \Delta t^2 $$ | | $$ y_2= g(n-1)^2 \Delta t^2 - \frac{g}{2}(n-1)^2 \Delta t^2 $$ |
| 4. Thus, the container will stop when it reaches a height of: | | 4. Thus, the container will stop when it reaches a height of: |
| $$ y_3=y_2+y_1 = \frac{g(n-1) \Delta t^2}{2} +\frac{g(n-1)^2 \Delta t^2}{2} $$ | | $$ y_3=y_2+y_1 = \frac{g(n-1) \Delta t^2}{2} +\frac{g(n-1)^2 \Delta t^2}{2} $$ |
| 5. Time of container fall from height $y_3$ | | 5. Time of container fall from height $y_3$ |
| $$ t_3 = \sqrt{\frac{2 y_3}{g}} = \Delta t \sqrt{n(n-1)} = 5 \,s $$ | | $$ t_3 = \sqrt{\frac{2 y_3}{g}} = \Delta t \sqrt{n(n-1)} = 5 \,s $$ |
| 6. The time the container with the equipment stays in the “airless” space | | 6. The time the container with the equipment stays in the “airless” space |
| $$ t=\Delta t + t_2 + t_3 $$ | | $$ t=\Delta t + t_2 + t_3 $$ |
| | | |
| $$ t= \Delta t [1 + (n-1) + \sqrt{n(n-1)}] $$ | | $$ t= \Delta t [1 + (n-1) + \sqrt{n(n-1)}] $$ |
| | | |
| $$ t = \Delta t [n + \sqrt{n(n-1)}]=10 \,s $$ | | $$ t = \Delta t [n + \sqrt{n(n-1)}]=10 \,s $$ |
| 7. The equipment in the container will experience a state of weightlessness for a period of time | | 7. The equipment in the container will experience a state of weightlessness for a period of time |
| $$ t_н =10 \,s $$ | | $$ t_н =10 \,s $$ |
| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$t = n\Delta t(1 + \sqrt{1 − 1/n}) \text{; }t_н \approx 10 \,s$$ | | $$t = n\Delta t(1 + \sqrt{1 − 1/n}) \text{; }t_н \approx 10 \,s$$ |
