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Give me answers. I have a problem of this question Fast . Physics . Centre of Mass
From momenta conservatio in x direction:Mv2 = mv1cos(thetha)..........(1) From energy conservation,mgh = 0.5mv1^2 + 0.5Mv2^2...........(2)Solving both the eqns. we get,v2 = sqroot(2mgh/((M/cos(thetha))^2 + M)and v1 = (M*sqroot(2mgh/((M/cos(thetha))^2 + M))/(m*cos(thetha)) time for block m to get down to the height h would be:v1 = 0 + gtt = v1/gDisplacement of wedge will be in x direction only.Therfore,x = v2*t + 0x = v2*v1/g => ((M*sqroot(2mgh/((M/cos(thetha))^2 + M))/(m*cos(thetha))*sqroot(2mgh/((M/cos(thetha))^2 + M))/g afer solving ,x = 2h/(M/(m*cos(thetha)) + costhetha) m.
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