1.

The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) = ((x^(4))/(4) - x^(2)/(2)) J` The total mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) isA. `(3)/sqrt(2))`B. `sqrt(2)`C. `(1)/(sqrt(2))`D. `2`

Answer» Correct Answer - A
(a) Velocity is muximum when K. E. is maximum for minimum P.E.
` (dV)/(dx) = 0 rArr x^(2) - x = 0 rArr x = +- 1`
`rArr Min P.E. = (1)/(4) - (1)/(2) = - (1)/(4) J`
`K. E _(max) + P.E_(min) = 2(Given)`
`:. K.E _(max) = 2 + (1)/(4) = (9)/(4) = (1) /(2) m nu_(max) ^(2)`
`rArr (1)/(2) xx 1 xx nu_(max)^(2) . = (9)/(4) rArr nu_(max) . = (3)/(sqrt(2))`


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