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