A skater of mass m standing on ice throws a stone of mass M with a velocity of v in a horizontal direction. Assuming coefficient of friction between the skater and the ice be μ, what will be the distance across which the skater will slide back ?
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a
M2v22mμg
b
Mv22m2μg
c
M2v22m2μg
d
M2v22m2μ2g
answer is C.
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Detailed Solution
Pi=0Pf=MV−mV1…………..(ii)MV−mV1=0⇒v1=MmV using 02=v12−2ax⇒ v12=2μgx⇒ MVm2=2μgx. ∴x=MzV22m2μg