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

Options

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

Answer

B

Explanation

According to law of conservation of energy-Loss in KE of cylinder=Gain in PE of spring
`implies (1)/(2)mv^(2)=(1)/(2) Iomega^(2)=(1)/(2) kx^(2)`
`=(1)/(2)mv^(2)=(1)/(2)(mR^(2))/(2)((v)/(R))^(2)=(1)/(2)kx^(2)`
`implies (3)/(4)mv^(2)=(1)/(2)kx^(2)`
`implies (3)/(4)xx3xx(4)^(2)=(1)/(2)xx200xx x^(2)`
`x=0.6`m.