time for collision

Question

Two particles $A$ and $B$ move anticlockwise with the same speed $v$ in a circle of radius $R$ and are diametrically opposite to each other. At $t=0$, $A$ is given a constant acceleration (tangential) $a_t = \frac{72v^2}{25\pi R}$. Calculate the time in which $A$ collides with $B$, the angle traced by $A$, its angular velocity and radial acceleration at the time of collision.

In the first part of question I thought to equate the time for both particles when they collide. But I am not able to find the time taken by the first particle $A$ .

Please help me in this.

Comments

What would it mean for the particles to collide? The final times would be the same. True. But would anything else be equal besides the time?

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Public Service Announcement: I edited the question, but please, please *please* don't just plop in an image of the problem! Type it out and format the math equations. There should only be pictures if there's a diagram, for instance.

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Why   , we cannot post image of a question

Answer 1

Work in a rotating frame in which the particles are initially at rest. To catch up with B, particle A has to cover a distance $s=\pi R$. Use $s=ut+\frac12at^2$ where $u=0$ is the initial relative velocity. Solve to find $t$.

When A catches B, it has traced an angle of $\pi$ radians in the rotating frame. If the time to catch up was $T$ then during this time the rotating frame traced out arc length $vT$ which subtends angle $vT/R$ radians.

At the time of collision A's speed is $V=v+aT$ and its angular velocity is $V/R$. Its radial acceleration at this time is $V^2/R$.

Comments

Good answer. I though v was variable.

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No, I think $v$ must be a constant. Tangential acceleration is stated to be constant and proportional to $v^2$.