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.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <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;

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.

    <h3>Solution</h3>
    <p>
        <p>$a)$ Since the collision is elastic, then according to the Law of Conservation of Momentum:</p>

$$v \sin \alpha = u \sin \alpha$$

$$\fbox{$v = u $}$$

$б)$ Further, this problem is a little reminiscent of 1.4.8.

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 $\vec{u} = \vec{v} - 2\vec{w}$.

Working with vector quantities is clearly demonstrated below

Let's find the projections of the vector $\vec{u}$ on the horizontal and vertical axes:

$$u_y = v \sin \alpha$$

$$u_x = v \cos \alpha + 2w$$

Using the Pythagorean theorem, we find the modulus of the vector $\vec{u}$

$$u = \sqrt{u_x^2+u_y^2}$$

$$u = \sqrt{(v \sin \alpha)^2 + (v \cos \alpha + 2w)^2}$$

$$\fbox{$u=\sqrt{v^{2}+4vw\cos\alpha+4w^{2}}$}$$

$в)$ Similar to the previous subparagraph

$$\vec{u} = \vec{v} - 2\vec{w}$$

We will show these vectors in the figure

We will find the projections of the vector $\vec{u}$ on the horizontal and vertical axis:

$$u_y = v \sin \alpha - 2w \sin \beta$$ $$u_x = v \cos \alpha + 2w \cos \beta$$

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>
        $\text{a) } u=v.\quad\text{b) } u=\sqrt{v^{2}+4vw\cos\alpha+4w^{2}}.\quad\text{c) } u=\sqrt{v^{2}+4vw\cos\alpha\cos\beta+4w^{2}\cos^{2}\beta}.$
    </p>


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