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

A, B, C, D

Explanation

(a) Velocity of the block just before collision,
`(1)/(2)mv_(0)^(2) + (1)/(2)kx^(2) = (1)/(2)kx_(0)^(2)`
or `v_(0) = sqrt((k)/(m)(x_(0)^(2) - x^(2))`
Here, `x_(0) = 0.03m, x = 0.01m, k = 10^(4)N//m,m = 1kg`
`:. v_(0) = 2sqrt(2)m//s`
After collision, `v = ev_(0) = (1)/sqrt(2) 2sqrt(2) = 2m//s`
Maximum compression is the spring is
`(1)/(2)kx_(m)^(2) = (1)/(2)kx^(2) + (1)/(2)mv^(2)`
or `= x_(m) = sqrt(x^(2) + (m)/(k)v^(2))`
`= sqrt((0.01)^(2) + ((1)(2)^(2))/(10^(4))m = 2.23 cm`
(b) In the case of spring - mass system, since the time period is independent of the amplitude of oscillation.
Time `= t_(AB) + t_(BC) + t_(CB) + t_(BD)`
` = (T_(0))/(4) + ((T_(0))/(2pi))sin^(-1) ((1)/(3)) + (T_(0)/(2pi))sin^(-1)((1)/(2.23)) + (T_(0))/(4)`
Here, `T_(0) = 2pi sqrt((m)/(k))`

Substituting the values, we get
Total time `= sqrt((m)/(k))[pi + sin^(-1)((1)/(3)) + sin^(-1)((1)/(2.23))]`.