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    				| 1. | 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)A. `sqrt(14) m`B. `sqrt(2.8) m`C. `sqrt(14) m`D. `sqrt(0.7) m` | 
| Answer» Correct Answer - B 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 | |