At a time when the particle velocity is 10^6 \frac{m}{s}, its acceleration is 10^4 \frac{m}{s^2} and is directed at an angle of 30^\circ to the velocity. How much will the speed increase in 10^{-2} seconds? At what angle will the speed direction change? What is the angular velocity of rotation of the velocity vector at this moment?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#1.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.3.24.$ At a time when the particle velocity is $10^6$ $\frac{m}{s}$, its acceleration is $10^4$ $\frac{m}{s^2}$ and is directed at an angle of $30^\circ$ to the velocity. How much will the speed increase in $10^{-2}$ seconds? At what angle will the speed direction change? What is the angular velocity of rotation of the velocity vector at this moment?
    </p>

    <h3>Solution</h3>
    <p>
        <p>We decompose the acceleration into two components, tangential $a_\parallel $ and normal $a_\perp$:</p>

$$a_\parallel = a \cdot \cos 30^{\circ} = a \frac{\sqrt{3}}{2}$$

$$a_\perp = a \cdot \sin 30^{\circ}= a \frac{1}{2}$$

During time $\Delta t$ the speed will increase by $\Delta v_\perp = a_\perp \Delta t$ and $\Delta v _\parallel = a_\parallel \Delta t$, in the transverse and longitudinal directions, respectively.

New speed values:

$$v_\parallel = \frac{\sqrt{3}}{2}a \Delta t +v_0$$

$$v_\perp = \frac{1}{2}a \Delta t$$

Thus, the desired angle:

$$\tan \alpha = \frac{v_\perp}{v_\parallel}$$

Angular velocity

$$\omega = \frac{\alpha}{t}$$

    <h4>Answer</h4>
    <p>
        $\text{With}(\sqrt{3}/2)\cdot10^2\text{ m/s;}$ на $5\cdot10^{-5}\text{ rad;}$ $\omega=5\cdot10^{-3}\text{ s}^{-1}.$
    </p>


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