Two blocks of mass 3 kg and 6 kg resp are placed on a smooth horizontal surface. They are connected by a light spring of force constant k=200 N/m.Initially the spring is unstretched.Velocity of 1 m/s is imparted to the block of mass 3 kg and vel of 2 m/s is imparted to the 6 kg block.The velocities are in opposite directions(such that they are moving away from each other).Find the maximum extension in the spring.

SINCE CHANGE IN K.E + CHANGE IN P.E = WORK DONE BY NONCONSERVATIVE FORCE+WORK DONE BY EXTERNAL FORCE

SINCE THERE IS NO NON CONSERVATIVE FORCE OR EXTERNAL FORCE THEREFORE CHANGE IN K.E + CHANGE IN P.E=0

 SINCE THE MAXIMUM COMPRESSION WILL OCCUR WHEN BOTH THE BLOCK STOPS OR FINAL K.E =0

 CONSIDER EXTENSION ON LEFT SIDE IS ''X'' AND RIGHT SIDE IS ''X*'' THEREFORE TOTAL EXTENSION IS X+X*

 NOW 3X(-)+6X*(+)=0 THEREFORE X=2X*..........................(1)

SINCE POTENTIAL ENERGY OF A SPRING IS (1/2)Kx²

  THEREFORE FROM CHANGE IN K.E + CHANGE IN P.E =0

 WE GET FINALK.E - INITIALK.E + FINAL P.E -INITIALP.E=0

           (0)-[(1/2 * 3 * 1²)+(1/2 * 6 * 2²)]+[(-1/2 * K *X²)+(1/2 *K * (X*)²)]-(0)]=0............................(2)    NOTE: ALSO MENTION THE SIGN OF THE DIRECTION 

     SOLVE EQUATION (1) AND (2) BY SUBSTITUTING (1) IN (2) AND GET YOUR ANSWER WHICH WILL BE X+X* 

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