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A system of two blocks A and B and a wedge C are released from rest as shown in figure. Masses of the blocks and the wedge are m, 2m and 2mrespectively. The displacement of the wedge C when the block slides down the plane a distance 10 cm is (neglect friction)
Just do cmponents of the 10 cm displacement in x- and y- directionfor both m and 2m masses adn do Momentum Conservation in x and y direction.In x direcn.,0 + 0 + 0 = 2m*10cos45 + m*10cos45 + 2mx we get,x = -30/2root2 i cap In y direcn.,2msin45*y(-j) + msin45*y(j) + 2my = 0 + 0 + 0we get,y = 5/root2jcap.Resultant displac,ent = root((5/root2)^2 + (15/root2)^2) = > root125 = > 11.18 cm.
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