A heavy particle of mass m_1 collides with a resting light particle of mass m_2. What is the greatest angle that a heavy particle can deflect as a result of an elastic impact?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#2.5">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.5.17^*.$ A heavy particle of mass $m_1$ collides with a resting light particle of mass $m_2$. What is the greatest angle that a heavy particle can deflect as a result of an elastic impact?
    </p>

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

Following from the law of conservation of energy $$\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}=\frac{p_0^2}{2m_1}$$ From where $$p_2^2=\frac{m_2}{m_1}\left(p_0^2-p_1^2\right)\quad\text{(1)}$$ From the cosine theorem $$p_2^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$ Substituting into $\text{(1)}$: $$\frac{m_2}{m_1}p_0^2-\frac{m_2}{m_1}p_1^2=p_1^2+p_0^2-2p_1p_0\cos\alpha$$ Let's regroup this into a quadratic equation

Solving relative to $p_1$ Since $\Delta = 0$ $$\cos^2\alpha=1-\frac{m_2^2}{m_1^2}$$ From where we find the required ratio $$\frac{m_2^2}{m_1^2}=1-\cos^2\alpha=\sin^2\alpha$$ $$\boxed{\sin\alpha=\frac{m_2}{m_1}}$$

    <h4>Answer</h4>
    <p>
        $$\sin\alpha=m_2/m_1$$
    </p>

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


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