A solid sphere of mass 2 kg is rolling on a frictionless horizontal surface with velocity `6 m//s`. It collides on the free and of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be
(Force constant of the spring = 36 N/m)

Options

A. `sqrt(14) m`
B. `sqrt(2.8) m`
C. `sqrt(14) m`
D. `sqrt(0.7) m`

Answer

B

Explanation

Given, `m = 2` kg, `v = 6` m/s and `k = 36` N/m
`= 1/2 mv^(2) + 1/2 lomega^(2)`
`= 1/2 mv^(2) + 1/2 xx 2/5 mr^(2) omega^(2)`
`=1/2 mv^(2) + 1/5 mv^(2) = 7/10 mv^(2) `
The potential energy of the spring on maximum compression x
`:. 1/2 kx^(2) = 7/10 mv^(2)`
`rArr x^(2) = 14/10 (mv^(2))/(k)`
Substituting the given values, we get
`= 14/10 xx (2 xx (16)^(2))/(36) = 2.8`
`:. x = sqrt(2.8) ` m