$2.5.17^*.$ 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
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
Since $\Delta = 0$
$$\cos^2\alpha=1-\frac{m_2^2}{m_1^2}$$
From where we find the required ratio
$$\frac{m_2^2}{m_1^2}=1-\cos^2\alpha=\sin^2\alpha$$
$$\boxed{\sin\alpha=\frac{m_2}{m_1}}$$