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| <div id="logo"> | | <div id="logo"> |
| <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> |
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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> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.4.16.$ Two beads of mass $m$ each, connected to each other by a spring of stiffness $k$, are held on smooth rods rigidly fixed in the wall. The spring is stretched and its length is $l$. The distance between the free ends of the rods is equal to the length of the undeformed spring. The beads are released. How fast will the spring move in the $x$-direction after the beads jump off the rod? What will be the greatest compression strain of the spring? | | $2.4.16.$ Two beads of mass $m$ each, connected to each other by a spring of stiffness $k$, are held on smooth rods rigidly fixed in the wall. The spring is stretched and its length is $l$. The distance between the free ends of the rods is equal to the length of the undeformed spring. The beads are released. How fast will the spring move in the $x$-direction after the beads jump off the rod? What will be the greatest compression strain of the spring? |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.4.16.png" | | <img src="2.4.16.png" |
| loading="lazy" width="200" /> | | loading="lazy" width="200" /> |
| <figcaption> | | <figcaption> |
| For problem $2.4.16$ | | For problem $2.4.16$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Preservation of energy at the initial stage | | Preservation of energy at the initial stage |
| $$\frac{k\left(l-l_0\right)^2}{2} = 2\frac{mv^2}{2}$$ | | $$\frac{k\left(l-l_0\right)^2}{2} = 2\frac{mv^2}{2}$$ |
| From where | | From where |
| $$ v = \sqrt{\frac{k}{2m}}\left(l-l_0\right)$$ | | $$ v = \sqrt{\frac{k}{2m}}\left(l-l_0\right)$$ |
| The greatest compression will be when it has nowhere to compress $(v_y=0)$ | | The greatest compression will be when it has nowhere to compress $(v_y=0)$ |
| Alternative for horizontal projection | | Alternative for horizontal projection |
| $$v_x=v\cos\alpha;v_x=\text{const}$$ | | $$v_x=v\cos\alpha;v_x=\text{const}$$ |
| After substituting | | After substituting |
| $$ \boxed{v_x = \sqrt{\frac{k}{2m}}\left(l-l_0\right)\cos\alpha} $$ | | $$ \boxed{v_x = \sqrt{\frac{k}{2m}}\left(l-l_0\right)\cos\alpha} $$ |
