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Question

An impulse J is applied on a ring of mass m along a line passing through its centre O. The ring is placed on a rough horizontal surface. The linear velocity of centre of ring, once it starts rolling without slipping, is

160459_9a81aa82fbff467aa4fdf3c24dc8637a.png

A
J/m
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B
J/2m
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C
J/4m
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D
J/3m
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Solution

The correct option is A J/2m
Let v and vo be the velocity of CM after pure rolling and before pure rolling.
J=Δp=mvo
Also, v=Rw
Now f=maa=fm
fR=Iα
fR=(mR2)αmRα=f
v=voatv=vofmt ............(1)
w=0+αt=fmRt ............(2)
Thus v=vowR
v=vovv=vo2=J2m

446156_160459_ans_2391719ccfc4495dbf9ad95069e8f8b5.png

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