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A particle of mass m is placed at rest on the top of a smooth wedge of mass M,which in turn is placed at rest on a smooth horizontal surface.If hbe the height of wedge and θ is the inclination,then the distance moved by the wedge as the particle reaches the foot of the wedge is ______width=151________

sindhuja P , 16 Years ago
Grade 11
anser 1 Answers
Badiuddin askIITians.ismu Expert

Dear sindhuja

there is no external horizontal force is acting on the system so center of mass will not move in x direction.

base of the wedge = hcotΘ

let the left lower  corner of wedge is origin and wedge move by x' distance ,and center of mass of wedge is at x1 distance from origin

so (mhcotΘ  +Mx1)/(m+M)  ={m(x'-hcotΘ) +M(x1+x')}/(m+M)

or  mhcotΘ  +Mx1  =m(x'-hcotΘ) +M(x1+x')

or 2mhcotΘ =(m+M)x'

 x'= 2mhcot/(m+M)

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