Two block `A` and `B` of masses `m_(1) = 3kg` and `m_(2) = 6kg` respectively are connected with each other by a spring of force constant `k = 200 N//m` as shown in figure. Blocks are pulled away from each other by `x_(0) = 3cm` and then released. When spring is in its natural length and block are moving towards each other, another block `C` of mass `m = 3kg` moving with velocity `v_(0) = 0.4m//s` (towards right) collides with `A` and gets stuck to it. Neglecting friction, calculate

(a) velocity `v_(1)` and `v_(2)` of the blocks `A` and `B` respectively just before collision and their angular frequency .
(b) velocity of centre of mass of the system, after collision,
( c) amplitude of oscillation of combined body,
(d) loss of energy during collision.

Answer

A, B, C, D

Explanation

(a) `(1)/(2) kx_(0)^(2) = (1)/(2)mu v_(r)^(2)`
Here, `mu =` reduced mass `= (6 xx 3)/(6 + 3)`
`= 2kg`
`:. v_(r) = sqrt((k)/(mu))x_(0) = (sqrt((200)/(2)))(3 xx10^(-2))`
`= 0.3m//s = 2v + v`
`:. v = 0.1m//s`
`:. v_(1) = 0.2m//s`
and `v_(2) = 0.1m//s`
Angular frequency
`omega = sqrt((k)/(mu)) = sqrt((200)/(2))`
`= 10rad//s`

(b) `v_(cm) = (m_(1)v_(1) + m_(2)v_(2) + m_(3)v_(3))/(m_(1) + m_(2) + m_(3))`
` = ((3)(0.2) - (6)(0.1) + 3(0.4))/(3 + 6 + 3)`
`= 0.1m//s` (towards right)
( c) After collision velocity of combined blocks `(A + C)`
`v_(0) = ((3 xx 0.2) + (3)(0.4))/(3 + 3) = 0.3m//s`
and velocity of block `B` is `v_(2) = 0.1m//s`

The spring will compress till velocity of all the blocks equal to the centre of mass. Applying conservation of mechanical energy, `(1)/(2)(3 + 3)(0.3)^(2) + (1)/(2)(6)(0.1)^(2) = (1)/(2)(3 + 3 + 6)(0.1)^(2) + (1)/(2)kA^(2)`
Solving this we get ,`A = 0.048m`
or `A = 4.8cm`
(d) `Delta E = (1)/(2)(3)(0.4)^(2) + (1)/(2)(3)(0.2)^(2) - (1)/(2)(3 + 3)(0.3)^(2)`
`= 0.24 + 0.06 - 0.27 = 0.03 J`