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Question

# A plane mirror is moving with velocity (4^i+5^j+7^k) m/s. A point object in front of the mirror moves with a velocity (3^i+4^j+5^k) m/s. Here ^k is along the normal to plane mirror and facing towards the object. The velocity of image w.r.t ground is (in SI units)

A
3^i+4^j+9^k
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B
3^i+4^j+11^k
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C
3^i4^j+11^k
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D
3^i4^j+9^k
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Solution

## The correct option is A 3^i+4^j+9^kWe know that for fixed plane mirror, (1) Perpendicular to the mirror (x - axis), vo=−vi (2) Parallel to the mirror (y - axis), vo=vi Since ^k is along the normal to mirror. Applying (→vo,m)⊥=−(→vi,m)⊥ ⇒(vo,m)z=−(vi,m)z ⇒(vo)z−(vm)z=−[(vi)z−(vm)z] ⇒5^k−7^k=−(vi)z+(vm)z or −2^k=−(vi)z+7^k ∴(vi)z=9^k m/s Normal direction for mirror is ^k (z - axis), thus mirror is in (xy - plane). Hence, x and y component of velocity of the image will be same as that of the object. ⇒(vi)y=4^j & (vi)x=3^i ∴→vi=(3^i+4^j+9^k) m/s

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