SEE I GOT THE ANSWER ITS VERY COMPLEX AND THIS IS NOT AN IIT SUM THIS IS AN  PHYSICS OYMAPAID SUM

SEE FIRST CALCULATE THE INERTIA OF THE COMBINED MASS (THIS IS THE TOUGHEST PART OF THE WHOLE SUM)

SEE FIRST WE NEED TO FIND THE POSITION OF THE COM IF THE SYSTEM AND THEN WE CAN FIND OUT THE THE INERTIA OF EACH SPHERE ABT THIS PT AND ADD IT GIVING THE INERTIA OF THE COMBINATION

THE Y AXIS OF THE COM WILL BE TAKING THE ORIGION AT THE GROUND  COM=M1R1+M2R2/M1+M2

SIGNIFYING THIS AS Y1

TAKING THE RADIUS OF THE FIRST SPHERE AS THE ORIGION THE COM  IS M2(R1+R2) /M1+M2    

WE KNOW THE INERTIA ABT THE COM OF THE FIRST SPHERE (THAT IS THROUGH ITS RADIUS)  SO D= M2(R1+R2) /M1+M2   SO D^2 CAN BE EASILY CALCULATED AND SO IST INERTIA IS EASILY CACULATED

INERTIA OF THE 2ND SPHERE ABT THE COM OF THE SYSTEM  

D=R1+R2-M2(R1+R2) /M1+M2  SO D^2 CAN BE CALCULATED AND SO ANOTHE INERTIA IS ALSO CALCULATED

NOW LETS TAKE LINEAR MOMENTUM THIS CAN BE CONSERVED SINCE BEFORE COLLISION FRICION IS NOT ACTING AND IMMEDIATELY AFTER COLLISION IT IS ALSO NOT ACTING

SO  M1V1+M2V2=(M1+M2)V3           (HERE V3 IS THE NEW VELOCITY)

V3=M1V1+M2V2/(M1+M2)

NOW CONSERVING ANGULAR MOMENTUM ABT THE BOTTOM MOST PT ANYWHERE U CAN TAKE SINCE INITIALLY NO FRICTION IS ACTING

V1=W1R1  W1=V1/R1  SIMILARLY   W2=V2/R2

INERTIA OF MASS M1=2/5M1R1^2 AND INERTIA IF 2ND MASS =2/5M2R2^2

SO CONSERVING ANGUAR MOMENTUM

I1W1+M1VIR1+I2W2+M2V2R2=INERTIA CALCULATED ERALIER W3 +(M1+M2)V3AND R IS THE POSITION OF THE COM ON THE ORIGION

SO WE ET W

NOW SINCE THE DIRECTION OF ROLLING IS NOT CHANGING THE VEOCITY OF THE BOTTOMMOST PT OF CONTACT =V-W3Y1

AND SO WE FIND OUT THE VEOCITY OF PT OF CONTACT  

BUT NOW WE KNOW THE MASS OF THE SYSTEM N=2MG

SO WE KNOW THE VALUE OF uN

TIME TAKEN TO STOP=   U/A

AND SO THIS IS THE ANSWER

PLS RATE ME FOR MY EFFORTS

!!!!!!!CHEERS!!!!!!!!!!