Added Luis's English solution of 5.6.22 and 7.1.17

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+ <header>
+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <main>
+ <article>
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h4>Statement</h4>
+ <p>
+ $7.1.17^*$
+ There are two diametrically opposed holes in a non-conductive thin-walled sphere with an uniformly distributed charge $Q$ which has mass $M$. At the initial istant, the spehere is in rest. Certain particle of mass $m$ and charge $q$ (same sign of $Q$) is moved from infinite with speed $v$ over line that join the holes. Determine the time which particle will be inside the sphere.
+ </p>
+ <h4>Solution</h4>
+ <p class="TxtSolutions">
+ Firstly, we consider the situation before $q$-particle penetrates $Q$-sphere. Since both bodies have equal-sign charges, $q$-particle will move with a deacceleration while $Q$-sphere will accelerate.
+ </p>
+ <p class="TxtSolutions">
+ Applying Energy Conservation Law (from the beginning until $q$-particle reaches $Q$-sphere surface)
+ </p>
+ <center>
+ $\frac{mv^2}{2} = \frac{mv_q^2}{2}+\frac{kQq}{R}+\frac{Mv_Q^2}{2} \;(1)$
+ </center>
+ <p class="TxtSolutions">
+ From Momentum Conservation Law
+ </p>
+ <center>
+ $mv = mv_q + Mv_Q \;(2)$
+ </center>
+ <p class="TxtSolutions">
+ Inside of $Q$-sphere, there is no electric field, then both bodies have constant velocities while $q$-particle be inside $Q$-sphere (see Figure). The displacement of $q$-particle is
+ </p>
+ <center>
+ <figure>
+ <img src="7.1.17.png"
+ loading="lazy" alt="7.1.17" width="250" />
+ <figcaption>
+ Particle motion inside the sphere
+ </figcaption>
+ </figure>
+ </center>
+ <center>
+ $\Delta S_q = \Delta S_Q+2R = v_q~T \;(3)$
+ </center>
+ <p class="TxtSolutions">
+ And the sphere's displacement is
+ </p>
+ <center>
+ $\Delta S_Q = v_Q~T \;(4)$
+ </center>
+ <p class="TxtSolutions">
+ From $(4)$ in $(3)$, and separating $T$
+ </p>
+ <center>
+ $T = \frac{2R}{v_q-v_Q} \;(5)$
+ </center>
+ <p class="TxtSolutions">
+ Separating $v_q$ from $(2)$
+ </p>
+ <center>
+ $v_q = v - \frac{M}{m}v_Q \;(6)$
+ </center>
+ <p class="TxtSolutions">
+ Putting $(6)$ into $(1)$ and developing algebraically
+ </p>
+ <center>
+ $0 = \frac{M}{2}\left(1+\frac{M}{m}\right)v_Q^2-Mv~v_Q + \frac{kQq}{R} \;(7)$
+ </center>
+ <p class="TxtSolutions">
+ $(7)$ is a second-degree equation or quadratic equation in $v_Q$ ($AX^2+BX+C=0$). We can solve this by Discriminant method or General-Solution formula for quadratic equations. The Discriminant is
+ </p>
+ <center>
+ $D = B^2-4AC$
+ </center>
+ <center>
+ $D = M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}$
+ </center>
+ <p class="TxtSolutions">
+ The solutions are
+ </p>
+ <center>
+ $X = \frac{-B\pm\sqrt{D}}{2A}$
+ </center>
+ <center>
+ $v_Q = \frac{Mv\pm\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}{M\left(1+\frac{M}{m}\right)} \;(8)$
+ </center>
+ <p class="TxtSolutions">
+ Substituting $(6)$ into $(5)$
+ </p>
+ <center>
+ $T = \frac{2R}{v-v_Q\left(1+\frac{M}{m}\right)} \;(9)$
+ </center>
+ <p class="TxtSolutions">
+ Putting $(8)$ into $(9)$
+ </p>
+ <center>
+ $T = \frac{2MR}{\mp\sqrt{M^2v^2-2M\left(1+\frac{M}{m}\right)\frac{kQq}{R}}}$
+ </center>
+ <p class="TxtSolutions">
+ Choosing ``plus'' (+) sign, because time is non-negative, taking $k = 1/(4\pi\epsilon_0)$ and developing
+ </p>
+ <center>
+ $\boxed{T = \frac{2R}{v}\left[1-\frac{qQ(M+m)}{2\pi\epsilon_0mMv^2}\right]^{-\frac{1}{2}}}$
+ </center>
+ <p class="TxtSolutions">
+ This is valid for $\frac{\mu v^2}{2}>\frac{qQ}{4\pi\epsilon_0R}$, where $\mu = \frac{mM}{M+m}$ is the reduced mass.
+ </p>
+ <p class="TxtSolutions" 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>
+ </article>
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