Question-133056

Tinku Tara June 3, 2023 Mechanics 0 Comments

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Question Number 133056 by mr W last updated on 19/Feb/21

Commented by mr W last updated on 18/Feb/21

a ball is thrown from point A with speed u at an angle 𝛉 to horizontal and strikes at a point on the inclined plane and returns back to point A. the restitution coefficient of the collision is e. find the speed u in terms of 𝛉 and determine the minimum speed u_(min) the ball must have.

a b a l l i s t h r o w n f r o m p o i n t A w i t h

s p e e d u a t a n a n g l e θ t o h o r i z o n t a l

a n d s t r i k e s a t a p o i n t o n t h e

i n c l i n e d p l a n e a n d r e t u r n s b a c k t o

p o i n t A . t h e r e s t i t u t i o n c o e f f i c i e n t

o f t h e c o l l i s i o n i s e .

f i n d t h e s p e e d u i n t e r m s o f θ a n d

d e t e r m i n e t h e m i n i m u m s p e e d u_{m i n}

t h e b a l l m u s t h a v e .

Commented by liberty last updated on 18/Feb/21

Commented by liberty last updated on 18/Feb/21

exactly the same. hahaha

exactly the same . hahaha

Commented by mr W last updated on 18/Feb/21

hahaha, if you mean they are exactly the same, then you seem not to have read both questions exactly.

h a h a h a, i f y o u m e a n t h e y a r e e x a c t l y

t h e s a m e, t h e n y o u s e e m n o t t o h a v e

r e a d b o t h q u e s t i o n s e x a c t l y .

Commented by liberty last updated on 18/Feb/21

yes maybe the question is different. the exact same is the picture

yes maybe the question is

different . the exact same is the

picture

Commented by liberty last updated on 18/Feb/21

hahahaha

hahahaha

Answered by mr W last updated on 19/Feb/21

Commented by EDWIN88 last updated on 20/Feb/21

amazing

amazing

Commented by mr W last updated on 20/Feb/21

B(L+s cos α, s sin α) with s=ξL, ξ≥0 from A to B: t=((L+s cos α)/(u cos θ))=((L(1+ξ cos α))/(u cos θ)) s sin α=((L(1+ξ cos α))/(u cos θ))[u sin θ−(g/2)×((L(1+ξ cos α))/(u cos θ))] ξ sin α=(1+ξ cos α)[tan θ−((gL(1+ξ cos α)(1+tan^2 θ))/(2u^2 ))] let λ=((gL)/(2u^2 )) ξ sin α=(1+ξ cos α)[tan θ−λ(1+ξ cos α)(1+tan^2 θ)] λ(1+ξ cos α)tan^2 θ−tan θ+((ξ sin α+λ(1+ξ cos α)^2 )/((1+ξ cos α)))=0 ⇒tan θ=((1±(√(1−4λ[ξ sin α+λ(1+ξ cos α)^2 ])))/(2λ(1+ξ cos α))) at point B: (see Q132102 for more detail) U=coming speed V=leaving speed (→) U_x =u cos θ U_y =−u sin θ+g×((L(1+ξ cos α))/(u cos θ)) (↓) U_y =u[−sin θ+((2λ(1+ξ cos α))/(cos θ))] let e′=1+e V_x =(e′ sin^2 α−1)U_x +e′ sin α cos α U_y (←) V_x =u[−e′ sin α sin (θ−α)−cos θ+((λe′(1+ξ cos α) sin 2α)/(cos θ))] V_y =e′ sin α cos α U_x +(e′ cos^2 α−1)U_y (↑) V_y =u[−e′ cos α sin (θ−α)+sin θ+((2λ(e′ cos^2 α−1)(1+ξ cos α))/(cos θ))] from B to A: t=((L(1+ξ cos α))/(u[−e′ sin α sin (θ−α)−cos θ+((λe′(1+ξ cos α) sin 2α)/(cos θ))])) ((tu)/L)=η=((1+ξ cos α)/(−e′ sin α sin (θ−α)−cos θ+((λe′(1+ξ cos α) sin 2α)/(cos θ)))) s sin α=t{−u[−e′ cos α sin (θ−α)+sin θ+((2λ(e′ cos^2 α−1)(1+ξ cos α))/(cos θ))]+(1/2)gt} ξ sin α=((tu)/L)[e′ cos α sin (θ−α)−sin θ−((2λ(e′ cos^2 α−1)(1+ξ cos α))/(cos θ))+((gL)/(2u^2 ))×((tu)/L)] ⇒ξ sin α=η[e′ cos α sin (θ−α)−sin θ−((2λ(e′ cos^2 α−1)(1+ξ cos α))/(cos θ))+λη] ...(I) for a given λ we can find ξ through (I).

B (L + s \cos α, s \sin α)

w i t h s = ξ L, ξ ⩾ 0

f r o m A t o B :

t = \frac{L + s \cos α}{u \cos θ} = \frac{L (1 + ξ \cos α)}{u \cos θ}

s \sin α = \frac{L (1 + ξ \cos α)}{u \cos θ} [u \sin θ - \frac{g}{2} \times \frac{L (1 + ξ \cos α)}{u \cos θ}]

ξ \sin α = (1 + ξ \cos α) [\tan θ - \frac{g L (1 + ξ \cos α) (1 + \tan^{2} θ)}{2 u^{2}}]

l e t λ = \frac{g L}{2 u^{2}}

ξ \sin α = (1 + ξ \cos α) [\tan θ - λ (1 + ξ \cos α) (1 + \tan^{2} θ)]

λ (1 + ξ \cos α) \tan^{2} θ - \tan θ + \frac{ξ \sin α + λ {(1 + ξ \cos α)}^{2}}{(1 + ξ \cos α)} = 0

\Rightarrow \tan θ = \frac{1 \pm \sqrt{1 - 4 λ [ξ \sin α + λ {(1 + ξ \cos α)}^{2}]}}{2 λ (1 + ξ \cos α)}

a t p o i n t B :

(s e e Q 132102 f o r m o r e d e t a i l)

U = c o m i n g s p e e d

V = l e a v i n g s p e e d

(\to) U_{x} = u \cos θ

U_{y} = - u \sin θ + g \times \frac{L (1 + ξ \cos α)}{u \cos θ}

(↓) U_{y} = u [- \sin θ + \frac{2 λ (1 + ξ \cos α)}{\cos θ}]

l e t e^{'} = 1 + e

V_{x} = (e^{'} \sin^{2} α - 1) U_{x} + e^{'} \sin α \cos α U_{y}

(\leftarrow) V_{x} = u [- e^{'} \sin α \sin (θ - α) - \cos θ + \frac{λ e^{'} (1 + ξ \cos α) \sin 2 α}{\cos θ}]

V_{y} = e^{'} \sin α \cos α U_{x} + (e^{'} \cos^{2} α - 1) U_{y}

(↑) V_{y} = u [- e^{'} \cos α \sin (θ - α) + \sin θ + \frac{2 λ (e^{'} \cos^{2} α - 1) (1 + ξ \cos α)}{\cos θ}]

f r o m B t o A :

t = \frac{L (1 + ξ \cos α)}{u [- e^{'} \sin α \sin (θ - α) - \cos θ + \frac{λ e^{'} (1 + ξ \cos α) \sin 2 α}{\cos θ}]}

\frac{t u}{L} = η = \frac{1 + ξ \cos α}{- e^{'} \sin α \sin (θ - α) - \cos θ + \frac{λ e^{'} (1 + ξ \cos α) \sin 2 α}{\cos θ}}

s \sin α = t {- u [- e^{'} \cos α \sin (θ - α) + \sin θ + \frac{2 λ (e^{'} \cos^{2} α - 1) (1 + ξ \cos α)}{\cos θ}] + \frac{1}{2} g t}

ξ \sin α = \frac{t u}{L} [e^{'} \cos α \sin (θ - α) - \sin θ - \frac{2 λ (e^{'} \cos^{2} α - 1) (1 + ξ \cos α)}{\cos θ} + \frac{g L}{2 u^{2}} \times \frac{t u}{L}]

\Rightarrow ξ \sin α = η [e^{'} \cos α \sin (θ - α) - \sin θ - \frac{2 λ (e^{'} \cos^{2} α - 1) (1 + ξ \cos α)}{\cos θ} + λ η] \dots (I)

f o r a g i v e n λ w e c a n f i n d ξ t h r o u g h (I) .

Commented by mr W last updated on 20/Feb/21

Commented by mr W last updated on 20/Feb/21

Commented by mr W last updated on 19/Feb/21

Commented by mr W last updated on 19/Feb/21

Commented by mr W last updated on 19/Feb/21

Commented by mr W last updated on 19/Feb/21

for large angle θ there are more than one possibilities for the same speed.

f o r l a r g e a n g l e θ t h e r e a r e m o r e t h a n

o n e p o s s i b i l i t i e s f o r t h e s a m e s p e e d .

Commented by mr W last updated on 20/Feb/21

Commented by mr W last updated on 20/Feb/21

Commented by mr W last updated on 20/Feb/21

Commented by mr W last updated on 19/Feb/21

Commented by mr W last updated on 20/Feb/21

Commented by mr W last updated on 20/Feb/21

Commented by otchereabdullai@gmail.com last updated on 23/Feb/21

world best !

world best!

world best!

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