At time t=0, two point objects A and B respectively are at pole and centre of curvature of a fixed concave mirror of focal length f; the velocity vectors of A and B are always V→A=ui^ and V→B=−ui^ respectively, where i^ is unit vector along principal axis directed from pole towards focus and u is a positive constant

A⇒OD=−ut ID=? 1−ut+1ID=1−f 1ID=1ut−1f ID=utff−ut=ffut−1 B⇒OD=−(2f−ut) ID=? 1−(2f−ut)+1ID=1−f 1ID=12f−ut−1f ID=(2f−ut)f−f+ut=(2f−ut)f−f+ut

At time t=0, two point objects A and B respectively are at pole and centre of curvature of a fixed concave mirror of focal length f; the velocity vectors of A and B are always V→A=ui^ and V→B=−ui^ respectively, where i^ is unit vector along principal axis directed from pole towards focus and u is a positive constant

A⇒OD=−ut ID=? 1−ut+1ID=1−f 1ID=1ut−1f ID=utff−ut=ffut−1 B⇒OD=−(2f−ut) ID=? 1−(2f−ut)+1ID=1−f 1ID=12f−ut−1f ID=(2f−ut)f−f+ut=(2f−ut)f−f+ut

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At time $t = 0$ , two point objects A and B respectively are at pole and centre of curvature of a fixed concave mirror of focal length f; the velocity vectors of A and B are always ${\vec{V}}_{A} = u \hat{i} a n d {\vec{V}}_{B} = - u \hat{i}$ respectively, where $\hat{i}$ is unit vector along principal axis directed from pole towards focus and u is a positive constant

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The distance between images of A and B will be 4f at time $t = \frac{f}{2 u}$

Magnitude of relative velocity of image of A and image of B at $t = 0$ is 2u

Starting from $t = 0$ and before the particles come in contact, distance between image of A and image of B increases

Starting from $t = 0$ and before the particles come in contact, distance between image of A and image of B first increases and then decreases

answer is A, B, C.

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Detailed Solution

$A \Rightarrow O D = - u t I D = ? \frac{1}{- u t} + \frac{1}{I D} = \frac{1}{- f} \frac{1}{I D} = \frac{1}{u t} - \frac{1}{f}$

$I D = \frac{u t f}{f - u t} = \frac{f}{\frac{f}{u t} - 1} B \Rightarrow O D = - (2 f - u t) I D = ? \frac{1}{- (2 f - u t)} + \frac{1}{I D} = \frac{1}{- f} \frac{1}{I D} = \frac{1}{2 f - u t} - \frac{1}{f} I D = \frac{(2 f - u t) f}{- f + u t} = \frac{(2 f - u t) f}{- f + u t}$