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Question

A man of mass m starts moving w.r.t. a platform of mass 2m with a velocity u =913m/s as shown in the figure. The platform is fitted with a concave mirror of focal length f. The velocity of image (in m/s) at the initial moment is
75498.JPG

A
3
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B
2.5
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C
2
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D
3.5
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Solution

The correct option is A 3
Using the relation m=ffu we get

Magnification m = ff+3f2 m= -2

Taking the direction in the right hand side Vimage/mirror=m2Vobject/mirror

Vimage/mirror is velocity of image w.r.t. mirror

Vobject/mirror is velocity of object w.r.t mirror

Vimage/mirror=4u

assume velocity of platform is v towards left

as the man is moving with velocity u towards right w.r.t. platform

vman=vman/platform+vplatform

vman=uv

writing momentum balance equation
m(uv)2mv=0v=u/3

the velocity of platform = velocity of mirror
Vimage/mirror=VimageVmirror

Vimage is velocity of image w.r.t ground

Vimage=Vimage/mirror+Vmirror=4uu3=133u

Vimage=3m/s

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