Two blocks of masses `m_1 and m_2` are connected by a spring of spring constant k figure. The block of mass `m_2` is given a shapr imulse so that it acquires a velocity `v_0` twoards right. Find a. the velocity of the cenre of mas b. the maximum elongation that teh spring will suffer.

Explanation

Velocity of centre of mass
`=(m_1v_1+m_2v_2)/(m_1+m_2)`
Hence `v_1=0,v_2=v_0`
`=(m_1xx0+m_2xxv_0)/(m_1+m-2)`
`=(m_2v_0)/(m_1+m_2)`
b. The spring wil attain maximum elongation when both velocities of two blocks wil attain are veolicy of centre of mass.
`xrarr maximum elongation of spring
change in kinetic energy
=potential energy stored in spring
`rarr 1/2 m_2v_2^2-1/2(m_1+m_2)((m_2v_0)/(m_1+m_2))^2`
`=1/2kx^2`
`rarr m_2v_0^2(1-m^2/(m_1+m_2)=kx^2`

`rarr x=[(m_1m_2)/(m_1+m_2)]^91/2)V_0`