A block of mass `(10 alpha)`, where `alpha` is a constant is moving with velocity 3 m/s to the right collides inelastically with the block on the right with mass 10g and sticks to it. The right block is connected to three springs as shown in the figure. The spring constant of each spring is 2N/m. If the amplitude of the resulting simple harmonic motion is `(1)/(2 sqrt2)m`, then the value of `alpha` is

Options

A. 5
B. 2.5
C. 7
D. 10

Answer

A

Explanation

Equivalent spring constant of system of springs `= (3k)/(2) =3` N/m
Momentum is conserved in collision. Let blocks moves with velocity v just after collision, then
`m_(1) v_(1) = (m_(1) + m_(2))v`
`10 alpha xx 10^(-3) xx 3 = (10 alpha + 10) xx 10^(-3) xx v`
`rArr v = (30 alpha)/(10 (alpha + 1)) = (3alpha)/(alpha + 1) ms^(-1)`
Displacement of blocks is `(1)/(2 sqrt2)m`
So, energy conservation gives
`(1)/(2) (m_(1) + m_(2))v^(2) = (1)/(2) k_(eq)x^(2)` [friction is absent]
`rArr ((10 alpha + 10) xx 10^(-3) xx 9 alpha^(2))/((alpha + 1)^(2)) = 3 xx (1)/(8)`
`rArr (9 alpha^(2))/(100(alpha + 1)) = (3)/(8)`
`24 alpha^(2) = 100 alpha + 100 rArr alpha = 5`