Question

Two particles $\text{A}$ and $\text{B}$ are moving in opposite direction on a circle. Initially particle $\text{A}$ and $\text{B}$ are diagonally opposite to each other. Particle $\text{A}$ move with angular velocity $\pi \text{rad/s}$, angular acceleration $\frac{\pi }{2}{\text{rad/s}}^2$ and particle $\text{B}$ moves with constant angular velocity $2\pi \text{rad/s}$. Find the time after which both the particles $\text{A}$ and $\text{B}$ will collide.

• A. $3.2\text{sec}$
• B. $2\text{sec}$
• C. $3.6\text{sec}$
• D. $4.5\text{sec}$

Answer 1

The correct option is B $2\text{sec}$


Let us consider the motion of $\text{B}$ w.r.t $\text{A}$

${\omega }_{rel}=2\pi -\pi =\pi \text{rad/s}$

Similarly, angular acceleration of $\text{B}$ w.r.t $\text{A}$ will be

${\alpha }_{rel}=0-\frac{\pi }{2}=-\frac{\pi }{2}{\text{rad/s}}^2$

When the two bodies collide, the relative angular displacement will be,

$\Delta \theta ={180}^{\circ }=\pi$

Applying kinetic equation in relative terms:

$(\Delta \theta {)}_{rel}={\omega }_{rel}t+\frac{1}{2}{\alpha }_{rel}{t}^2$

$\Rightarrow \pi =\pi t+\frac{1}{2}\left(\frac{-\pi }{2}\right)t^2$

$\Rightarrow \pi =\pi t-\frac{1}{4}\pi t^2$

$\Rightarrow t^2-4t+4=0\Rightarrow (t-2{)}^2=0$

$\therefore t=2\text{sec}$

Hence, option $\left(b\right)$ is right.

Why this question? Tip: When two bodies are in motion simultaneously, then it's always advisable to solve from relative frame. Use of relative motion technique will make one body come to rest and problem will become simple.