Physics - NLM Question with Solution | TestHub

For the blocks A and B FBD as shown below
Equations of motion $$\begin{array}{l}{\mathrm{a}}_{\mathrm{A}}=\frac{\mathrm{F}}{\mathrm{M}}(\text{ in }-\mathrm{x}\text{ direction })\\ {\mathrm{a}}_{\mathrm{B}}=\frac{\mathrm{F}}{\mathrm{M}}(\text{ in }+\mathrm{x}\text{ direction) }\end{array}$$ Relative acceleration, of A w.r.t. B, $$\begin{array}{rl}{\mathrm{a}}_{\mathrm{A},\mathrm{B}}& ={\mathrm{a}}_{\mathrm{A}}-{\mathrm{a}}_{\mathrm{B}}=-\frac{\mathrm{F}}{\mathrm{m}}-\frac{\mathrm{F}}{\mathrm{M}}\\ & =-\mathrm{F}\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{Mm}}\right)(\text{ along }-\mathrm{x}\text{ direction })\end{array}$$ Initial relative velocity of A w.r.t. B, ${\mathrm{u}}_{\mathrm{AB}}={\mathrm{v}}_0$ using equation $v^2=u^2+2$ as $$0=v_0^2-\frac{2\mathrm{F}(\mathrm{m}+\mathrm{M})\mathrm{S}}{\mathrm{Mm}}\Rightarrow \mathrm{S}=\frac{{\mathrm{Mmv}}_0^2}{2\mathrm{F}(\mathrm{m}+\mathrm{M})}$$ i.e., Distance moved by A relative to B $${\mathrm{S}}_{\mathrm{AB}}=\frac{{\mathrm{Mmv}}_0^2}{2\mathrm{F}(\mathrm{m}+\mathrm{M})}$$