Question

A particle 'P' moves with velocity 10m/sec towards the intersection point'O' of the 2 plane mirror kept at right angle to each other. I1&I2; are the images formed due to direct reflection from mirror M1 and mirror M2 respectively. Then the relative speed of I1 with respect to I2 will be
(1)20m/sec
(2)12m/sec
(3)10√2m/sec
(4)16m/sec

Answer 1

Answer on Question#52005 - Physics - Optics

A particle 'P' moves with velocity $10\mathrm{m / sec}$ towards the intersection point 'O' of the 2 plane mirror kept at right angle to each other. I1&I2 are the images formed due to direct reflection from mirror M1 and mirror M2 respectively. Then the relative speed of I1 with respect to I2 will be

(1) $20\mathrm{m / sec}$

(2) $12\mathrm{m / sec}$

(3) $10 \mathrm{v} 2 \mathrm{~m} / \mathrm{sec}$

(4) $16 \mathrm{~m} / \mathrm{sec}$

Solution:

If the velocity of particle $P$ is

$$\boldsymbol {v} = \left(- v _ {x}, - v _ {y}\right),$$

then velocities of images $l_{1}$ and $l_{2}$ are

$$\boldsymbol {v} ^ {l _ {1}} = \left(- v _ {x}, v _ {y}\right)$$ $$\boldsymbol {v} ^ {l _ {2}} = \left(v _ {x}, - v _ {y}\right)$$

The relative velocity of I1 with respect to I2 is

$$\boldsymbol {v} ^ {l _ {1} l _ {2}} = \boldsymbol {v} ^ {1} - \boldsymbol {v} ^ {l _ {2}} = (- v _ {x}, v _ {y}) - (v _ {x}, - v _ {y}) = (- 2 v _ {x}, 2 v _ {y})$$

Then the relative speed of I1 with respect to I2 is

$$\left| \boldsymbol {v} ^ {l _ {1} l _ {2}} \right| = \sqrt {(- 2 v _ {x}) ^ {2} + (2 v _ {y}) ^ {2}} = 2 \sqrt {v _ {x} ^ {2} + v _ {y} ^ {2}} = 2 | \boldsymbol {v} |$$

Since $|\pmb{v}| = 10\frac{\mathrm{m}}{\mathrm{s}}$ , we obtain

$$\left| \boldsymbol {v} ^ {l _ {1} l _ {2}} \right| = 2 | \boldsymbol {v} | = 2 \cdot 1 0 \frac {\mathrm {m}}{\mathrm {s}} = 2 0 \frac {\mathrm {m}}{\mathrm {s}}$$

So the correct answer is (1).

Answer: (1).

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