A particle of mass m_1 flew at a velocity v on a stationary particle of mass m_2, which, after an elastic impact, flew at an angle lpha to the initial direction of movement of the incoming particle. Determine the velocity of a particle of mass m_2 after impact.

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    <h3> Statement </h3>
    <p>
        $2.5.18^*.$ A particle of mass $m_1$ flew at a velocity $v$ on a stationary particle of mass $m_2$, which, after an elastic impact, flew at an angle $\alpha$ to the initial direction of movement of the incoming particle. Determine the velocity of a particle of mass $m_2$ after impact.
    </p>

    <h3>Solution</h3>
    <p>
        </p>

Following from the law of conservation of energy $$\frac{p_0^2}{2m_1}=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}$$ From where $$p_1^2=p_0^2-p_2^2\frac{m_1}{m_2}$$ From the cosine theorem $$p_1^2=p_2^2+p_0^2-2p_2p_0\cos\alpha$$ From where $$p_2\left(1+\frac{m_1}{m_2}\right)=2p_0\cos\alpha\quad (1)$$ By definition of momentum $$\left\{\begin{matrix} p_0=m_1v \\ p_2=m_2u \end{matrix}\right.$$ Substituting into $(1)$ expression $$m_2u\left(1+\frac{m_1}{m_2}\right)=2m_1v\cos\alpha$$ $$u(m_1+m_2)=2m_1v\cos\alpha$$ $$\boxed{u=v\frac{2m_1\cos\alpha}{m_1+m_2s}}$$

    <h4>Answer</h4>
    <p>
        $$u=v\frac{2m_1\cos\alpha}{m_1+m_2}$$
    </p>

    <p style="text-align: right; font-style: italic; font-size: 14;">
      Almaskhan Arsen<br>
    </p>

    <h3>Alternative solution</h3>
          <p>
        This phenomenon is also knowned as Compton scattering

Let's write the law of conservation of momentum:
$$\left{\begin{matrix} X\text{: }m_1v=m_1v_1\cos\beta+m_2v_2\cos\alpha \ Y\text{: }m_1v_1\sin\beta=m_2v_2\sin\alpha \ \end{matrix}\right.$$
Let's
$$\xi=\frac{m_2}{m_1}$$

From where:
$$v_1\sin\beta=\xi v_2\sin\alpha$$
$$v=v_1\cos\beta+\xi v_2\cos\alpha$$
We could find $\cos\beta$ from first expression:
$$\cos\beta=\sqrt{1-\sin^2\beta}$$
$$\cos\beta=\sqrt{1-(\xi \frac{v_2}{v_1}\sin\alpha)^2}$$
The law of conservation of energy:
$$\frac{m_1v^2}{2}=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}$$
From where
$$v_1^2=v^2-\xi v_2^2$$
From the other hand:

$$(v-\xi v_2\cos\alpha)^2=v_1^2\cos^2\beta$$
$$(v-\xi v_2\cos\alpha)^2=v_1^2\cdot(1-(\xi \frac{v_2}{v_1}\sin\alpha)^2)$$
$$(v-\xi v_2\cos\alpha)^2=v_1^2-(\xi v_2\sin\alpha)^2$$
$$v^2 - 2\xi vv_2\cos\alpha=v_1^2-\xi^2 v_2^2$$
From where
$$v_2^2\xi(1+\xi)=2\xi vv_2\cos\alpha$$
$$v_2=v\frac{2\cos\alpha}{1+\xi}$$
And, finally
$$\boxed{v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$

    <h4>Answer</h4>
    <p>
        $$v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}$$
    </p>

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