Two blocks of masses `3kg` and `6kg` respectivley are placed on a smooth horizontal surface. They are connected by a light spring of force constant `k=200N//m`. Initially the spring is unstretched. The indicated velocities are imparted to the blocks. Find the maximum extension of the spring.

Answer

C

Explanation

At maximum extension their velocities are same.
This common velocity is given by
`v=("Total momentum")/("Total mass")`
`=(2xx6-1xx3)/(3+6)=1m//s`
Now, `E_i=E_f`
`:. 1/2xx6xx(2)^2+1/2xx3xx(1)^2=1/2xx9(1)^2+1/2xx200xxx_m^2`
Solving we get, `x_m=0.3m`
`=30cm`