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

Answer

`(2)`

Explanation

At maximum compression the solid cylinder will stop. According to law of conservation of mechanical energy
Loss of kinetic energy = Gain in potential energy of cylinder of spring
`1/2(mv^(2)) + 1/2 (I omega ^(2)) = 1/2 (kx)^(2)`
`1/2 mv^(2) + 1/2((mR^(2))/(2))((v)/(R))^(2) = 1/2 kx^(2)z`
`(therefore v = R omega and for solid cylinder, I = 1/2 mR^(2))`
`1/2 mv^(2) + 1/4 mv^(2) = 1/2 kx^(2)`
`3/4 mv^(2) = 1/2 kx^(2) or x^(2) = 3/2 (mv)^(2)/k`
Here,` m = 3kg, v = 4ms^(-1), k = 200 Nm^(-1)`
Substituting the given values, we get
`x^(2) = (3xx 3 xx 4 xx4)/(2 xx 200)`
`x^(2) = 36/100 or x = 0.6 m`