Two identical blocks A and B, each of mass `m=3kg`, are connected with the help of an ideal spring and placed on a smooth horizontal surface as shown in Fig. Another identical blocks C moving velocity `v_0=0.6(m)/(s)` collides with A and sticks to it, as a result, the motion of system takes place in some way
Based on this information answer the following questions:
Q. Oscillation energy of the system i.e., part of the energy which is oscillation (changing) between potention and kinetic forms is

Options

A. `0.27 J`
B. `0.09J`
C. `0.18 J`
D. `0.45 J`

Answer

A

Explanation

`B` accelerates and combined mass decelerates. The centre of mass of the system continues to move rightward with constant speed. While both the blocks oscillate about centre of mass.
The velocity just after collision is: `v = (v_(0))/(2)` The velocity of centre of mass is:
`V_(c.m) = (m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2)), T = 2pi sqrt((mu)/(k))`: reduced mass concept:
Oscillation energy of the system is: `E = (1)/(2) (2m) v^(2)`
Translational kinetic energy of the centre of mass of the system `= (1)/(2) (3m) v_(cm)^(2)`
The remaining energy is oscillating between kinetic and potential energy during the motion of blocks: Oscillating energy `=` maximum elastic potential energy