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Question Number 144496 by ajfour last updated on 25/Jun/21

Commented by ajfour last updated on 25/Jun/21

$A r o d i s t o b e r e l e a s e d$ $h o r i z o n t a l l y a t h e i g h t y$ $s u c h t h a t i t h i t s g r o u n d$ $i n v e r t i c a l o r i e n t a t i o n .$ $F i n d y i n t e r m s o f a, b, h .$ $R o d s u f f e r s a n e l a s t i c$ $c o l l i s i o n w i t h t h e c o r n e r$ $o f t a b l e .$
	Answered by mr W last updated on 27/Jun/21

	Commented by mr W last updated on 27/Jun/21

	$u = \sqrt{2 g y}$ $\frac{m}{2} (u^{2} - v^{2}) = \frac{1}{2} \times \frac{{m b}^{2}}{12} \times ω^{2}$ $\Rightarrow 12 (u^{2} - v^{2}) = b^{2} ω^{2} . . . (i)$ $m u - P = m v$ $P (\frac{b}{2} - a) = \frac{{m b}^{2}}{12} ω$ $m (u - v) (\frac{b}{2} - a) = \frac{{m b}^{2}}{12} ω$ $\Rightarrow 6 (u - v) (b - 2 a) = b^{2} ω . . . (i i)$ $(i) / (i i) :$ $ω b = \frac{2 (u + v)}{1 - \frac{2 a}{b}}$ $p u t i n t o (i i) :$ $6 (u - v) (b - 2 a) = b \frac{2 (u + v)}{1 - \frac{2 a}{b}}$ $3 {(1 - \frac{2 a}{b})}^{2} (u - v) = u + v$ $\Rightarrow v = \frac{3 {(1 - \frac{2 a}{b})}^{2} - 1}{3 {(1 - \frac{2 a}{b})}^{2} + 1} u = β u$ $v + u = \frac{6 {(1 - \frac{2 a}{b})}^{2}}{3 {(1 - \frac{2 a}{b})}^{2} + 1} u$ $\Rightarrow b ω = \frac{12 (1 - \frac{2 a}{b})}{3 {(1 - \frac{2 a}{b})}^{2} + 1} u = α u$ $v t + \frac{1}{2} {g t}^{2} = h - \frac{b}{2}$ $2 v t + {g t}^{2} = 2 h - b$ $t = \frac{- v + \sqrt{v^{2} + g (2 h - b)}}{g}$ $ω t = \frac{π}{2}$ $2 ω [\sqrt{v^{2} + g (2 h - b)} - v] = π g . . . (i i i)$ $2 α [u \sqrt{β^{2} u^{2} + g (2 h - b)} - β u^{2}] = π g b$ $u \sqrt{β^{2} u^{2} + g (2 h - b)} = β u^{2} + \frac{π g b}{2 α}$ $(2 h - b - \frac{β π b}{α}) u^{2} = \frac{π^{2} {g b}^{2}}{4 α^{2}}$ $(2 h - b - \frac{β π b}{α}) 2 g y = \frac{π^{2} {g b}^{2}}{4 α^{2}}$ $\Rightarrow y = \frac{π^{2} b}{8 α^{2} (\frac{2 h}{b} - 1 - \frac{β π}{α})}$ $w i t h$ $α = \frac{12 (1 - \frac{2 a}{b})}{3 {(1 - \frac{2 a}{b})}^{2} + 1}$ $β = \frac{3 {(1 - \frac{2 a}{b})}^{2} - 1}{3 {(1 - \frac{2 a}{b})}^{2} + 1}$
	Commented by ajfour last updated on 27/Jun/21
	Thanks sir, our approach are the same, but you could bring it to completion; excellent soln.
	Answered by ajfour last updated on 27/Jun/21
	$Let impulse received at corner be J. u=(√(2gy)) mv=mu−J −h=vt−((gt^2 )/2) gt^2 −2vt−2h=0 t=(v/g)+(√((v^2 /g^2 )+((2h)/g))) ωt=(π/2) I_0 ω=J((b/2)−a) ⇒ I_0 ((π/(2t)))=m(u−v)((b/2)−a) I_0 ((π/2))=((mv(u−v))/g)((b/2)−a)(1+(√(1+((2gh)/v^2 )))) & mg(y+h−(b/2))=(1/2)mv^2 +(1/2)I_0 ω^2 and as y=(u^2 /(2g)) ⇒ mg((u^2 /(2g))+h−(b/2))=(1/2)mv^2 +(1/(2I_0 )){m(u−v)((b/2)−a)}^2 ..(I) I_0 ((π/2))= ((mv(u−v))/g)((b/2)−a)(1+(√(1+((2gh)/v^2 )))) ...(II) we solve for u and v from eqs. (I)& (II) and then y=(u^2 /(2g))$
	$L e t i m p u l s e r e c e i v e d a t$ $c o r n e r b e J .$ $u = \sqrt{2 g y}$ $m v = m u - J$ $- h = v t - \frac{{g t}^{2}}{2}$ ${g t}^{2} - 2 v t - 2 h = 0$ $t = \frac{v}{g} + \sqrt{\frac{v^{2}}{g^{2}} + \frac{2 h}{g}}$ $ω t = \frac{π}{2}$ $I_{0} ω = J (\frac{b}{2} - a)$ $\Rightarrow I_{0} (\frac{π}{2 t}) = m (u - v) (\frac{b}{2} - a)$ $I_{0} (\frac{π}{2}) = \frac{m v (u - v)}{g} (\frac{b}{2} - a) (1 + \sqrt{1 + \frac{2 g h}{v^{2}}})$ $& m g (y + h - \frac{b}{2}) = \frac{1}{2} {m v}^{2}$ $+ \frac{1}{2} I_{0} ω^{2}$ $a n d a s y = \frac{u^{2}}{2 g}$ $\Rightarrow m g (\frac{u^{2}}{2 g} + h - \frac{b}{2}) = \frac{1}{2} {m v}^{2}$ $+ \frac{1}{2 I_{0}} {m (u - v) (\frac{b}{2} - a)}^{2} . . (I)$ $I_{0} (\frac{π}{2}) =$ $\frac{m v (u - v)}{g} (\frac{b}{2} - a) (1 + \sqrt{1 + \frac{2 g h}{v^{2}}})$ $. . . (I I)$ $w e s o l v e f o r u a n d v f r o m$ $e q s . (I) & (I I) a n d t h e n$ $y = \frac{u^{2}}{2 g}$
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