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. After the collision of C and A, the combined body and block B would

Options

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

Answer

B

Explanation

The velocity of the combined mass just after collision is `mv_(0)=2mv`
`v=(v_(0))/(2)=0.3m//s`
Velocity of centre of mass of system,
`v_(CM)=(2mv+0)/(3m)=(2v)/(3)=(v_(0))/(3)=0.2m//s`.
Time period with which block B and the combined mass oscillate about centre of mass could be computed by using the reduced mass concept.
`T=2pisqrt((2m)/(3k))=(pi)/(5sqrt10)s`
Oscillation energy of the system is
`E=(2mv^(2))/(2)=0.27J`
The translational kinetic energy of the centre of mass of system is
`E_(cm)=(3mv^(2)CM)/(2)=0.18J`
The remaining energy is oscillating between kinetic and potential energy during the motion of blocks.