A bead of mass m kept at the top of a smooth hemispherical wedge of mass M and radius R, is gently pushed towards right. As a result, the wedge slides due left Find the a. speed of the wedge b. magnitude of velocity of the beard relative to the wedge.

A bead of mass m kept at the top of a smooth hemispherical wedge of ma

1.	A bead of mass m kept at the top of a smooth hemispherical wedge of mass M and radius R, is gently pushed towards right. As a result, the wedge slides due left Find the a. speed of the wedge b. magnitude of velocity of the beard relative to the wedge.
Answer» Solution :Applying work energy theorem `W_("gravity")=/_\K'+/_\K_C`…………..i `/_\K'=` change in `KE` w.r.t centre of mass from and `/_\K_C=` change in kinetic energy of centre of mass w.r.t ground. Let `v` is the velocities of ball wilth respect to wedge and `V` is the velocity of wedge. Linear mometum will be conserved in the horizontal DIRECTION . Hence `MV=(vcostheta-V)` `implies V=(mvcostheta)/((m+M))` ...............ii The velocity of centre ofmas will be zero in horizontal direction but the velocity of centre of mass will change in vertical direction . `(v_(cm))_y=(mvsintheta)/((m+M))` .............iii Now applying WE theorem `-mgR(1-costheta)=1/2muv^(2)+1/2(m+M)v_(cm,y)^(2)` Here `mu=(mM)/((m+M))` and `v_(cm,y)=(mvsintheta)/((m+M))` which GIVES `v=sqrt((2(M+m)gR(1-costheta))/((M+Msin^(2)THETA)))` and from eqn ii `V=sqrt((2m^(2)gR(1-costheta)cos^(2)theta)/((M+m)(M+Msin^(2)theta)))`