Question Number 103143 by ajfour last updated on 13/Jul/20

Commented byajfour last updated on 13/Jul/20

Q.103072    (A revisit)

Answered by ajfour last updated on 13/Jul/20

lets  the frictional force be f(θ).  work-energy eq_(−)   mgrsin θ=(1/2)MV^2 +∫fdx      +(1/2)m[(usin θ−V)^2 +u^2 cos ^2 θ]  Impulse-momentum eq._(−)   −MV+m(usin θ−V)=∫fdt    friction force, normal N, acc. A_(−)   N+mAcos θ−mgsin θ=((mu^2 )/r)    f=μ(Nsin θ+Mg)     Ncos θ−f=MA  (everything much interdependent   so not easy to resolve even if we  desire to know just final V, u ....)

Commented byDwaipayan Shikari last updated on 13/Jul/20

Is the circular path (from where the small box slides)  frictionless? If friction is absent so there is no need  to calculate i think sir!  As   μ→0      Mv+mV=0   {  Sir can i apply momentum conservaion here?          v=−((mV)/M)  (v=velocity of the wedge  ,V=velocity of th block      v=−((m(√(2gR)))/M)         { (1/2)mV^2 =mgR⇒V=(√(2gR))  I  am not sure  If i am wrong then kindly rectify me

Commented byajfour last updated on 13/Jul/20

we are not solving for μ=0  thank you Sir.