For the arrangement shown in figure, the spring is initially compressed by `3 cm`. When the spring is released the block collides with the wall and rebounds to compress the spring again.

(a) If the coefficient of restitution is `(1)/sqrt(2)` , find the maximum compression in the spring after collision.
(b) If the time starts at the instant when spring is released, find the minimum time after which the block becomes stationary.

Answer

17

Explanation

The motion starts from position `A` the time taken from `A` to `W_(2) (t_(1)) = (T)/(4) + (T)/(12)`
Before collision the energy of the block just before collision
`K_(i) = (1)/(2)K(2A)^(2) - (1)/(2) KA^(2)` & just after collision `K_(1) = (K_(f))/(3)` (given) `= (1)/(2)KA^(2)`
Now during motion after collision, the energy is again conserved
Hence, `K_(1) + (1)/(2) KA^(2) = (1)/(2)KA^('2)`
`A' =` maximum compresion after collision `rArr = Asqrt(2)`
ie. Now motion has amlitude `Asqrt(2)`
Now time taken by block from
`W_(2)` to positon `B = (T)/(4) + (T)/(8)`
`:.` total time taken `= t_(1) + t_(2) = (T)/(4) + (T)/(12) + (T)/(4) + (T)/(8) = (17)/(24)T = (17pi)/(12)sqrt((m)/(K)) = 17sec {sqrt((m)/(k)) = (12)/(pi)}`