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

Commented by ajfour last updated on 13/Jul/20

$Q . 103072 (A r e v i s i t)$
	Answered by ajfour last updated on 13/Jul/20

	$l e t s t h e f r i c t i o n a l f o r c e b e f (θ) .$ $\underline{w o r k - e n e r g y e q}$ $m g r \sin θ = \frac{1}{2} {M V}^{2} + \int f d x$ $+ \frac{1}{2} m [{(u \sin θ - V)}^{2} + u^{2} \cos^{2} θ]$ $\underline{I m p u l s e - m o m e n t u m e q .}$ $- M V + m (u \sin θ - V) = \int f d t$ $\underline{f r i c t i o n f o r c e, n o r m a l N, a c c . A}$ $N + m A \cos θ - m g \sin θ = \frac{{m u}^{2}}{r}$ $f = μ (N \sin θ + M g)$ $N \cos θ - f = M A$ $(e v e r y t h i n g m u c h i n t e r d e p e n d e n t$ $s o n o t e a s y t o r e s o l v e e v e n i f w e$ $d e s i r e t o k n o w j u s t f i n a l V, u . . . .)$
	Commented by Dwaipayan Shikari last updated on 13/Jul/20
	$Is the circular path (from where the small box slides) frictionless? If friction is absent so there is no need to calculate i think sir! As μ→0 Mv+mV=0 { Sir can i apply momentum conservaion here? v=−((mV)/M) (v=velocity of the wedge ,V=velocity of th block v=−((m(√(2gR)))/M) { (1/2)mV^2 =mgR⇒V=(√(2gR)) I am not sure If i am wrong then kindly rectify me$
	$I s t h e c i r c u l a r p a t h (f r o m w h e r e t h e s m a l l b o x s l i d e s)$ $f r i c t i o n l e s s ? I f f r i c t i o n i s a b s e n t s o t h e r e i s n o n e e d$ $t o c a l c u l a t e i t h i n k s i r!$ $A s$ $μ \to 0 M v + m V = 0 {S i r c a n i a p p l y m o m e n t u m c o n s e r v a i o n h e r e ?$ $v = - \frac{m V}{M} (v = v e l o c i t y o f t h e w e d g e, V = v e l o c i t y o f t h b l o c k$ $v = - \frac{m \sqrt{2 g R}}{M} {\frac{1}{2} {m V}^{2} = m g R \Rightarrow V = \sqrt{2 g R}$ $I a m n o t s u r e$ $I f i a m w r o n g t h e n k i n d l y r e c t i f y m e$
	Commented by ajfour last updated on 13/Jul/20

	$w e a r e n o t s o l v i n g f o r μ = 0$ $t h a n k y o u S i r .$
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