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+ <meta name="description" content="The body hits the wall with velocity v and angle \alpha to the line perpendicular to the wall. Determine the velocity of the body after an elastic impact if the wall is: a) stationary; b) moving perpendicular to itself at a speed w towards the body; c) moving at an angle \beta to the line perpendicular to it at a speed w towards the body.">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
+ <meta property="og:title" content="The body hits the wall with velocity v and angle \alpha to the line perpendicular to the wall. Determine the velocity of the body after an elastic impact if the wall is: a) stationary; b) moving perpendicular to itself at a speed w towards the body; c) moving at an angle \beta to the line perpendicular to it at a speed w towards the body.">
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+ <title>The body hits the wall with velocity v and angle \alpha to the line perpendicular to the wall. Determine the velocity of the body after an elastic impact if the wall is: a) stationary; b) moving perpendicular to itself at a speed w towards the body; c) moving at an angle \beta to the line perpendicular to it at a speed w towards the body.</title>
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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#1.4">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $1.4.9.$ The body hits the wall with velocity $v$ and angle $\alpha$ to the line perpendicular to the wall. Determine the velocity of the body after an elastic impact if the wall is: </p><p>
+a) stationary; </p><p>
+b) moving perpendicular to itself at a speed $w$ towards the body; </p><p>
+c) moving at an angle $\beta$ to the line perpendicular to it at a speed $w$ towards the body.
+
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/1/1.4.9/statement.png"
+ loading="lazy" width="150" />
+ <figcaption>
+ For problem $1.4.9$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <p>$a)$ Since the collision is elastic, then according to the Law of Conservation of Momentum:</p>
+$$v \sin \alpha = u \sin \alpha $$
+
+$$\fbox{$v = u $}$$
+
+<p>$б)$ Further, this problem is a little reminiscent of <a href="../1.4.8" target="_blank">1.4.8</a>. </p>
+<p>In the frame of reference associated with the wall, the relative velocity of the ball $\vec{v_{rel}} = \vec{v} - \vec{w}$. During elastic reflection, passing into the earth's frame of reference, the velocity is equal to $\vec{u} = \vec{v} - 2\vec{w}$. </p>
+<p>Working with vector quantities is clearly demonstrated below</p>
+<center>
+<figure>
+<img src="https://savchenkosolutions.com/1/1.4.9/draw.png"
+loading="lazy" width="250" />
+<figcaption>
+Illustration of the ball's velocities
+</figcaption>
+</figure>
+</center>
+<p>Let's find the projections of the vector $\vec{u}$ on the horizontal and vertical axes:</p>
+$$u_y = v \sin \alpha $$
+
+$$u_x = v \cos \alpha + 2w$$
+
+
+<p>Using the Pythagorean theorem, we find the modulus of the vector $\vec{u}$</p>
+
+$$u = \sqrt{u_x^2+u_y^2}$$
+
+$$u = \sqrt{(v \sin \alpha)^2 + (v \cos \alpha + 2w)^2}$$
+
+$$\fbox{$u=\sqrt{v^{2}+4vw\cos\alpha+4w^{2}}$}$$
+
+<p>$в)$ Similar to the previous subparagraph</p>
+$$\vec{u} = \vec{v} - 2\vec{w}$$
+
+<p>We will show these vectors in the figure</p>
+<center>
+<figure>
+<img src="https://savchenkosolutions.com/1/1.4.9/draw1.png"
+loading="lazy" width="250" />
+<figcaption>
+Illustration of the ball's velocities
+</figcaption>
+</figure>
+</center>
+<p>We will find the projections of the vector $\vec{u}$ on the horizontal and vertical axis:</p>
+$$u_y = v \sin \alpha - 2w \sin \beta$$
+ $$u_x = v \cos \alpha + 2w \cos \beta$$
+ <p>Using the Pythagorean equation, we find the modulus of the vector $\vec{u}$
+ $$u = \sqrt{u_x^2+u_y^2}$$
+ $$u = \sqrt{(v \sin \alpha - 2w \sin \beta)^2 + (v \cos \alpha + 2w \cos \beta)^2}$$
+ $$\fbox{$u=\sqrt{v^{2}+4vw\cos\alpha\cos\beta+4w^{2}\cos^{2}\beta}$}$$
+
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $\text{a) } u=v.\quad\text{b) } u=\sqrt{v^{2}+4vw\cos\alpha+4w^{2}}.\quad\text{c) } u=\sqrt{v^{2}+4vw\cos\alpha\cos\beta+4w^{2}\cos^{2}\beta}.$
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