Question

Two particles $A$ and $B$ start at the origin $O$ and travel in opposite directions along the circular path at constant speeds $V_A=0.7\text{m/s}$ and $V_B=1.5\text{m/s}$ respectively. Determine the time when they collide and the magnitude of the acceleration of $B$ just before the collision happens.

• A. Time when they collide is $14.3\text{s}$.
• B. Time when they collide is $13.4\text{s}$.
• C. Acceleration of $B$ is $0.45{\text{m/s}}^2$.
• D. Acceleration of $A$ is $9.8{\text{m/s}}^2$.

Answer 1

Answer: (A), (C)

The correct options are
A Time when they collide is $14.3\text{s}$.
C Acceleration of $B$ is $0.45{\text{m/s}}^2$.


$t_A=\frac{\theta }{{\omega }_A}=\frac{\theta R}{V_A}{t}_B=\frac{2\pi -\theta }{{\omega }_B}=\frac{(2\pi -\theta )R}{V_B}$

$\frac{\theta R}{V_A}=\frac{(2\pi -\theta )R}{V_B}$

$\frac{\theta }{0.7}=\frac{2\pi -\theta }{1.5}$
$\frac{\theta }{0.7}=\frac{2\pi -\theta }{1.5}$
$\frac{\theta }{7}=\frac{2\pi -\theta }{15}$
$15\theta =14\pi -7\theta$
$22\theta =14\pi$
$\theta =\frac{14\pi }{22}$

$t=\theta /{\omega }_A=\theta R/V_A$
putting the values of $\theta ,R$ and $V_A$ and
On solving you get t = 14.3 sec

Centripetal Acceleration $=v^2/R$
$=(1.5{)}^2$
$=0.45m/s^2$