The system described in previous question is kept on a frictionless horizontal floor with spring at its natural length. Now a sharp impulse is applied on the block m2 to impart it a speed of v. Calculate velocity acquired by the centre of mass just after the impulse is applied on the block m2. Also calculate maximum possible elongation suffered by the spring.

Explanation

Velocity of the centre of mass can be written as follows:
`v_(cm)=(m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2))=(m_(1)xx0+m_(2)xxv)/(m_(1)xxm_(2))=(m_(2)v)/(m_(1)xxm_(2))`
When block my starts moving then it elongates the spring which in turn applies force on blockm, to move it along. Spring force opposes the motion of m, and so its velocity decreases and spring force accelerates the block m, and its velocity increases. Elongation of the spring increases till the time both the blocks acquire common velocity. And this common velocity will be equal to velocity of centre of mass. At this instant elongation of the spring will be maximum.
Let x be the maximum elongation, then using conservation of mechanical energy we can write the following:
Loss in kinetic energy = gain in spring potential energy
`(1)/(2)m_(2)v^(2)-(1)/(2)(m_(1)+m_(2))[(m_(2)v)/(m_(1)+m_(2))]^(2)=(1)/(2) kx^(2)`
`rArr kx^(2)= m_(2)v^(2)-(m_(2)^(2)v^(2))/(m_(1)+m_(2))`
`rArr kx^(2)=(m_(1)m_(2)^(2)v^(2))/(m_(1)+m_(2))`
`rArr x=v sqrt((m_(1)m_(2))/(k(m_(1)+m_(2))))`