How would I go about solving a 1D two body problem as follows:
Consider two blocks A (mass $m_1$) and B (mass $m_2$) placed on a horizontal smooth surface with a spring of constant $k$ between them. A constant force of $F_1$ and $F_2$ is acting on A and B respectively in the opposite directions.
How would I go about converting it into a equivalent 1 body problem using the reduced mass idea?
It's easier to think this situation in terms of centre of mass frame.
Let $F_2>F_1$. Hence $F_{net}=F_2-F_1$ $$F_{net}=(m_1+m_2) a_{CM}$$ $$a_{CM}=\frac{F_2-F_1}{m_1+m_2}$$
Where $a_{CM}$ is acceleration of center of mass of our system.
Since whole system moves with some constant acceleration, our individual blocks $m_1,\,\,m_2$ experience the pseudo force $m_1a_{CM}\,\,,m_2a_{CM}$ respectively.
Finding $F_{net}$ on individual blocks:
$$F_1-m_1a_{CM}=F_1'$$ $$F_1'=\frac{m_1F_2+m_2F_1}{m_1+m_2}$$ And: $$F_2+m_2a_{CM}=F_2'$$ $$F_2'=\frac{m_1F_2+m_2F_1}{m_1+m_2}$$
So now you can predict further variables regarding situation, having find $F_{net}$ on each blocks.
Hope it helped you out!
