Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

Statement

$2.5.17^*.$ 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?

Solution

Direction of balls momentum

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

$$p_1^2\left(1+\frac{m_2}{m_1}\right)-2p_0p_1\cos\alpha+p_0^2\left(1-\frac{m_2}{m_1}\right)=0$$
Solving relative to $p_1$
$$p_1=\frac{2p_0\cos\alpha\pm \sqrt{4p_0^2\cos^2\alpha-4p_0^2\left(1-\frac{m_2^2}{m_1^2}\right)}}{2\left(1+\frac{m_2}{m_1}\right)}$$
Since $\Delta = 0$ $$\cos^2\alpha=1-\frac{m_2^2}{m_1^2}$$ From where we find the required ratio $$\frac{m_2^2}{m_1^2}=1-\cos^2\alpha=\sin^2\alpha$$ $$\boxed{\sin\alpha=\frac{m_2}{m_1}}$$

Answer

$$\sin\alpha=m_2/m_1$$

Almaskhan Arsen