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Question

Find the velocity of image of a moving particle O in the situation as shown in the figure.


A
164 m/s, at tan1(4/5) with horizontal
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B
10 m/s, at 53 with horizontal
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C
68 m/s, at tan1(12) with vertical
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D
104 m/s, at tan1(15) with vertical.
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Solution

The correct option is A 164 m/s, at tan1(4/5) with horizontal
Let,
Vo= velocity of object,
Vi= velocity of image,
Vm= velocity of mirror

Given:
Vox=10cos53=6 m/s
Voy=10sin53=8 m/s
Vm=2 m/s

Using the formula of moving plane mirror and object along the optical axis

Vix=Vox±2Vm

take positive sign when mirror moving towards object

Vix=Vox+2Vm

Vix=6+2(2)=10 ms1

And in vertical direction of moving mirror

Viy=Voy

Viy=8 ms1

So, Vi=V2ix+V2iy=(10)2+82

Vi=164 m/s

And, tanα=ViyVix=810=45

α=tan1(45) with horizontal.

Hence, option (b) is correct.

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