A block of mass M = 40 kg is released on a smooth incline from point A. After travelling through a length of 30 cm it strikes an ideal spring of force constant `K = 1000 N//m` . It compresses the spring and then gets pushed back. How much time after its release, the block will be back to point A?

Options

A. `4 sec`.
B. `2.5 sec.`
C. `1.54 sec`.
D. `3.26 sec`.

Answer

C

Explanation

Time to travel from `A` to `B (t_(1))` is given by -
`x = ut + (1)/(2)at^(2) , [:. a = gsintheta = 5m//s^(2)]`
`0.3 = (1)/(2) xx 5 xx t_(1)^(2) rArr t_(1) = 0.35s`
Speed of the block when it hits the spring is given by -
`V^(2) = 0^(2) + 2ax = 2 xx 5 xx 0.3`
`V = sqrt(3) m//s`
The motion of the block in contact with the spring can be regarded as a part of `SHM`. In equilibrium, compression in the spring is given by
`kx_(0) = mg sin theta`
`x_(0) = (40 xx 10 xx (1)/(2))/(1000) = 0.2 m`
Let the block compress the spring by a length `x_(1)`.
Energy conservation gives -
`(1)/(2) kx_(1)^(2) = (1)/(2)MV^(2) + Mgx_(1) sintheta`
`(1)/(2) xx 1000. x_(1)^(2) = (1)/(2) xx 40 xx (sqrt(3))^(2) xx (40 xx 10 xx (1)/(2))x_(1)`
`5x_(1)^(2) - 2x_(1) - 0.6 m`
Solving `x_(1) = 0.5 m`
Hence, amplitude of `SHM` is `A = x_(1) - x_(0) = 0.4 m`

The motion from `B` to `C` and back to `B` can be regarded as motion of a particle performing `SHM` [from `x = + (A)/(2)` to negative extrem e and back to `x = + (A)/(2)`]
Time for this motion can be obtained from fig given below.

Particle on circle moves from `P_(1)` to `C` to `P_(2)`. Time required
for completing this two third circle `= (2T)/(3)`
Hence, desired time `t_(2) = (2T)/(3) = (2)/(3)2pisqrt((m)/(k))`
`= (4pi)/(3)sqrt((40)/(1000))l = (8pi)/(30)s = 0.84 s`
`:.` Time required for the block to come back to `A` is `t = 2t_(1) + t_(0)`
`= 2 xx 0.35 + 0.84 = 1.54 s`