a block of mass M is placed on the top of a bigger block of mass 10M. (consider the bigger block 2 b a rt. triangular wedge whose base length is 2.2m)  all the surfaces are smooth. the system is released frm rest.

find the dist. moved by the bigger block at the instant the smaller block reaches the ground.

ans. 0.2m

LET INITIALY THE CENTRE OF MASS OF COMBINED SYSTEM WAS AT ORIGIN

SINCE NO NET HORIZONTAL FORCE IS ACTING ON THE SYSTEM AND ALSO CENTRE OF MASS WAS AT REST INITYALY SINCE BOTH BLOCKS WERE AT REST HENCE CENTRE OF MASS WILL REMAIN ON Y AXIS i.E IT WILL NOT MOVE HORIZONTALY

LET DISTANCE MOVED BY BIGGER BLOCK=X1

THEN DISTANCE MOVED BY SMALLER BLOCK WILL BE =2.2-X1

M*(2.2-X1) =10MX1

2.2-X1=10X1

11X1=2.2

X1=2m