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Question Number 20786 by Tinkutara last updated on 02/Sep/17

Two particles A and B start from the  same position along the circular path of  radius 0.5 m with a speed v_A  = 1 ms^(−1)   and v_B  = 1.2 ms^(−1)  in opposite direction.  Determine the time before they collide.

$$\mathrm{Two}\:\mathrm{particles}\:{A}\:\mathrm{and}\:{B}\:\mathrm{start}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{position}\:\mathrm{along}\:\mathrm{the}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{of} \\ $$$$\mathrm{radius}\:\mathrm{0}.\mathrm{5}\:\mathrm{m}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:{v}_{{A}} \:=\:\mathrm{1}\:\mathrm{ms}^{−\mathrm{1}} \\ $$$$\mathrm{and}\:{v}_{{B}} \:=\:\mathrm{1}.\mathrm{2}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{in}\:\mathrm{opposite}\:\mathrm{direction}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{time}\:\mathrm{before}\:\mathrm{they}\:\mathrm{collide}. \\ $$

Answered by dioph last updated on 03/Sep/17

θ_A  = ω_A t = ((v_A t)/r) = ((1×t)/(0.5)) = 2t  θ_B  = 2π − ω_B t = 2π − ((v_B t)/r) = 2π −((1.2×t)/(0.5)) = 2π−2.4t  θ_A  = θ_B  ⇔ 2t = 2π−2.4t ⇔  ⇔ 4.4t = 2π ⇔ t = ((2π)/(4.4)) s

$$\theta_{{A}} \:=\:\omega_{{A}} {t}\:=\:\frac{{v}_{{A}} {t}}{{r}}\:=\:\frac{\mathrm{1}×{t}}{\mathrm{0}.\mathrm{5}}\:=\:\mathrm{2}{t} \\ $$$$\theta_{{B}} \:=\:\mathrm{2}\pi\:−\:\omega_{{B}} {t}\:=\:\mathrm{2}\pi\:−\:\frac{{v}_{{B}} {t}}{{r}}\:=\:\mathrm{2}\pi\:−\frac{\mathrm{1}.\mathrm{2}×{t}}{\mathrm{0}.\mathrm{5}}\:=\:\mathrm{2}\pi−\mathrm{2}.\mathrm{4}{t} \\ $$$$\theta_{{A}} \:=\:\theta_{{B}} \:\Leftrightarrow\:\mathrm{2}{t}\:=\:\mathrm{2}\pi−\mathrm{2}.\mathrm{4}{t}\:\Leftrightarrow \\ $$$$\Leftrightarrow\:\mathrm{4}.\mathrm{4}{t}\:=\:\mathrm{2}\pi\:\Leftrightarrow\:{t}\:=\:\frac{\mathrm{2}\pi}{\mathrm{4}.\mathrm{4}}\:{s} \\ $$

Commented by Tinkutara last updated on 03/Sep/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Answered by mrW1 last updated on 03/Sep/17

t=((2πr)/(v_A +v_B ))=((2π×0.5)/(1+1.2))=(π/(2.2))=1.43 s

$$\mathrm{t}=\frac{\mathrm{2}\pi\mathrm{r}}{\mathrm{v}_{\mathrm{A}} +\mathrm{v}_{\mathrm{B}} }=\frac{\mathrm{2}\pi×\mathrm{0}.\mathrm{5}}{\mathrm{1}+\mathrm{1}.\mathrm{2}}=\frac{\pi}{\mathrm{2}.\mathrm{2}}=\mathrm{1}.\mathrm{43}\:\mathrm{s} \\ $$

Commented by Tinkutara last updated on 03/Sep/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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