Question Number 20786 by Tinkutara last updated on 02/Sep/17
$$\mathrm{Two}\:\mathrm{particles}\:{A}\:\mathrm{and}\:{B}\:\mathrm{start}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{position}\:\mathrm{along}\:\mathrm{the}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{of} \\ $$$$\mathrm{radius}\:\mathrm{0}.\mathrm{5}\:\mathrm{m}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:{v}_{{A}} \:=\:\mathrm{1}\:\mathrm{ms}^{−\mathrm{1}} \\ $$$$\mathrm{and}\:{v}_{{B}} \:=\:\mathrm{1}.\mathrm{2}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{in}\:\mathrm{opposite}\:\mathrm{direction}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{time}\:\mathrm{before}\:\mathrm{they}\:\mathrm{collide}. \\ $$
Answered by dioph last updated on 03/Sep/17
$$\theta_{{A}} \:=\:\omega_{{A}} {t}\:=\:\frac{{v}_{{A}} {t}}{{r}}\:=\:\frac{\mathrm{1}×{t}}{\mathrm{0}.\mathrm{5}}\:=\:\mathrm{2}{t} \\ $$$$\theta_{{B}} \:=\:\mathrm{2}\pi\:−\:\omega_{{B}} {t}\:=\:\mathrm{2}\pi\:−\:\frac{{v}_{{B}} {t}}{{r}}\:=\:\mathrm{2}\pi\:−\frac{\mathrm{1}.\mathrm{2}×{t}}{\mathrm{0}.\mathrm{5}}\:=\:\mathrm{2}\pi−\mathrm{2}.\mathrm{4}{t} \\ $$$$\theta_{{A}} \:=\:\theta_{{B}} \:\Leftrightarrow\:\mathrm{2}{t}\:=\:\mathrm{2}\pi−\mathrm{2}.\mathrm{4}{t}\:\Leftrightarrow \\ $$$$\Leftrightarrow\:\mathrm{4}.\mathrm{4}{t}\:=\:\mathrm{2}\pi\:\Leftrightarrow\:{t}\:=\:\frac{\mathrm{2}\pi}{\mathrm{4}.\mathrm{4}}\:{s} \\ $$
Commented by Tinkutara last updated on 03/Sep/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Answered by mrW1 last updated on 03/Sep/17
$$\mathrm{t}=\frac{\mathrm{2}\pi\mathrm{r}}{\mathrm{v}_{\mathrm{A}} +\mathrm{v}_{\mathrm{B}} }=\frac{\mathrm{2}\pi×\mathrm{0}.\mathrm{5}}{\mathrm{1}+\mathrm{1}.\mathrm{2}}=\frac{\pi}{\mathrm{2}.\mathrm{2}}=\mathrm{1}.\mathrm{43}\:\mathrm{s} \\ $$
Commented by Tinkutara last updated on 03/Sep/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$