A solid cylinder of mass `3 kg` is rolling on a horizontal surface with velocity `4 ms^(-1)`. It collides with a horizontal spring of force constant `200 Nm^(-1)`. The maximum compression produced in the spring will be `:`

Options

A. 0.7 m
B. 0.2 m
C. 0.5 m
D. 0.6 m

Answer

D

Explanation

`(1)/(2) mv^(2)[1+(K^(2))/(R^(2))]=(1)/(2)kx_("max")^(2)`
Putting `m=3kg, v=4 m//s, (K^(2))/(R^(2))=1//2` (For solid cylinder), `k=200 Nm^(-1)` we get `x_("max")=0.6 m`