Question

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Answer 1

$\huge{\pink{\underline{\mathfrak{your\:answer}}}}$

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$\bold{(C)\:2}$

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step-by-step explanation:

Let the radius of the circle be 'r'.

Then,

the two distance travelled by the two particles before the first collision is 2πr.

therefore,

(2v × t) + (v × t) = 2πr

where,

t is the time taken to occur first collision.

.°. $t\:=\:\frac{2πr}{3v}$

.°. Distance travelled by particle with velocity 'v' is equal to

$v\:×\:\frac{2πr}{3v}=\frac{2πr}{3}$

Therefore,

the collision occur at B.

(see the attachment)

Now,

as the collision is elastic and the particles have equal masses, the velocities will interchange as shown in the figure.

(Refer to the attachments)

Now,

According to the same reasoning as above, 2nd collision will take place at C and the velocities will again interchange.

Also,

With the same reasoning, 3rd collision will occur at the point A.

Hence,

there will be two elastic collisions before the particles again reach at A.

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