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.