$2.5.18^*.$ 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.
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 $$m_2u\left(1+\frac{m_1}{m_2}\right)=2m_1v\cos\alpha$$ $$u(m_1+m_2)=2m_1v\cos\alpha$$ $$\boxed{u=v\frac{2m_1\cos\alpha}{m_1+m_2s}}$$
$$u=v\frac{2m_1\cos\alpha}{m_1+m_2}$$
Almaskhan Arsen
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 $$\boxed{v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}}$$
$$v_2=v\frac{2m_1\cos\alpha}{m_1+m_2}$$