Question-144496

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} \dots (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 \dots (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 \dots (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

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}}})

\dots (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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