A skater of mass m standing on ice throws a stone of mass M with a velocity of v in a horizontal direction. The distance over which the skater will move back (the coefficient of friction between the skater and the ice is μ ) :
A
B
C
D
Text Solution
Verified by ExpertsThe correct answer is:
C
P i = 0 ....(i)
P f = MV – mV 1 ....(ii)
MV – mV 1 = 0 ⇒ v 1 = 
V.
using 0 2 = v 1 2 – 2ax
⇒ v 1 2 = 2 μ gx
⇒ 
= 2 μ g x.
∴ x = 
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