Question-132102

Tinku Tara June 3, 2023 Mechanics 0 Comments

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Question Number 132102 by mr W last updated on 14/Feb/21

Commented by mr W last updated on 16/Feb/21

a ball is thrown from point A with speed u and strikes at a point B on the semispherical surface with radius R and returns back to point A. if the restitution coefficient of the collision is e, find the minimum speed u the ball must have and at which angle 𝛉 the ball should be thrown.

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

o n t h e s e m i s p h e r i c a l s u r f a c e w i t h

r a d i u s R a n d r e t u r n s b a c k t o p o i n t A .

i f 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 m i n i m u m

s p e e d u t h e b a l l m u s t h a v e a n d a t

w h i c h a n g l e θ t h e b a l l s h o u l d b e

t h r o w n .

Commented by otchereabdullai@gmail.com last updated on 14/Feb/21

You are the world best no one can challenge you!

You are the world best no one can

challenge you!

Commented by ajfour last updated on 15/Feb/21

For u_(min) I got φ=30° t_1 =(√((2(√3)R)/g)){1±(√((1/e^2 )−1))} u_(min) ^2 =(((R(√3))/(2t))+((gt)/4))^2 +((3(√3))/2)gR((1/e^2 )−1)

F o r u_{m i n} I g o t ϕ = 30 °

t_{1} = \sqrt{\frac{2 \sqrt{3} R}{g}} {1 \pm \sqrt{\frac{1}{e^{2}} - 1}}

u_{m i n}^{2} = {(\frac{R \sqrt{3}}{2 t} + \frac{g t}{4})}^{2} + \frac{3 \sqrt{3}}{2} g R (\frac{1}{e^{2}} - 1)

Commented by mr W last updated on 15/Feb/21

do you mean that u_(min) is always at φ=30° indepentently from e? i can′t confirm this sir.

d o y o u m e a n t h a t u_{m i n} i s a l w a y s a t

ϕ = 30 ° i n d e p e n t e n t l y f r o m e ?

i {c a n}^{'} t c o n f i r m t h i s s i r .

Answered by mr W last updated on 25/Feb/21

Commented by mr W last updated on 12/Feb/21

Commented by mr W last updated on 16/Feb/21

motion from A to B: t=((R(1+cos φ))/(u cos θ)) R sin φ=−u sin θ×((R(1+cos φ))/(u cos θ))+(g/2)×((R^2 (1+cos φ)^2 )/(u^2 cos^2 θ)) ((sin φ)/(1+cos φ))=((gR)/(2u^2 ))×(1+cos φ)(1+tan^2 θ)−tan θ let λ=((gR)/(2u^2 )) ((sin φ)/(1+cos φ))=λ(1+cos φ)(1+tan^2 θ)−tan θ λ(1+cos φ)tan^2 θ−tan θ+((λ(1+cos φ)^2 −sin φ)/(1+cos φ))=0 tan θ=((1±(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ))) ⇒θ=tan^(−1) [((1±(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] at point B: U_x =u cos θ U_y =gt−u sin θ=((gR(1+cos φ))/(u cos θ))−u sin θ U_⊥ =U_x cos φ+U_y sin φ U_⊥ =u cos θcos φ+[((gR(1+cos φ))/(u cos θ))−u sin θ]sin φ ⇒U_⊥ =u cos (φ+θ)+((gR(1+cos φ)sin φ)/(u cos θ)) U_∥ =−U_x sin φ+U_y cos φ U_∥ =−u cos θ sin φ+[((gR(1+cos φ))/(u cos θ))−u sin θ]cos φ ⇒U_∥ =−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ)) V_⊥ =eU_⊥ ⇒V_⊥ =eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ)) V_∥ =U_∥ ⇒V_∥ =−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ)) V_x =V_⊥ cos φ+V_∥ sin φ V_x =[eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))]cos φ+[−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))]sin φ V_x =eu cos (φ+θ)cos φ+((egR(1+cos φ)sin φ cos φ)/(u cos θ))−u sin (φ+θ)sin φ+((gR(1+cos φ)sin φcos φ)/(u cos θ)) ⇒V_x =u[(1+e)cos (φ+θ)cos φ−cos θ+((λ(1+e)(1+cos φ) sin 2φ)/(cos θ))] V_y =V_⊥ sin φ−V_∥ cos φ V_y =[eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))]sin φ−[−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))]cos φ ⇒V_y =u[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)((1+e) sin^2 φ−1))/(cos θ))] return from B to A: t=((R(1+cos φ))/((1+e)cos (φ+θ)cos φ−cos θ+((λ(1+cos φ)(1+e) sin 2φ)/(cos θ))))×(1/u) ⇒((tu)/R)=ξ=((1+cos φ)/((1+e)cos (φ+θ)cos φ−cos θ+((λ(1+e)(1+cos φ) sin 2φ)/(cos θ)))) R sin φ=t{u[(1+e)cos (φ+θ)sin φ+sin θ]+((2λ(1+cos φ)(e sin^2 φ−cos^2 φ))/(cos θ))−((gt)/2)} sin φ=((tu)/R)[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)(e sin^2 φ−cos^2 φ))/(cos θ))−((λtu)/R)] sin φ=ξ[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)((1+e)sin^2 φ−1))/(cos θ))−λξ] or sin φ=ξ[e′cos (φ+θ) sin φ+sin θ+((2λ(1+cos φ)(e′sin^2 φ−1))/(cos θ))−λξ] ...(I) with e′=1+e λ=((gR)/(2u^2 )) θ=tan^(−1) [((1+(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] ∗) ξ=((1+cos φ)/(e′cos (φ+θ) cos φ−cos θ+((λe′(1+cos φ) sin 2φ)/(cos θ)))) for a given e ∈(0,1] we can find solution(s) for φ in (I) for some values of parameter λ. the maximum value of λ such that a solution for φ exists can be determined numerically. this corresponds to the minimum speed the ball must have: u_(min) =(√((gR)/(2λ_(max) ))) examples: e=0.5 ⇒u_(min) ≈2.7682(√(gR)) e=0.75 ⇒u_(min) ≈1.6082(√(gR)) e=1 ⇒u_(min) ≈0.9098(√(gR)) (= Q65589) ∗) with the other solution θ=tan^(−1) [((1−(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] we can get a solution for the case that the ball strikes the surface upwards, i.e. counterclockwise. but in this case more energy is needed, i.e. u_(min) is larger than with θ=tan^(−1) [((1+(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] see examples with e=0.5 and 0.75.

\underset{―}{m o t i o n f r o m A t o B :}

t = \frac{R (1 + \cos ϕ)}{u \cos θ}

R \sin ϕ = - u \sin θ \times \frac{R (1 + \cos ϕ)}{u \cos θ} + \frac{g}{2} \times \frac{R^{2} {(1 + \cos ϕ)}^{2}}{u^{2} \cos^{2} θ}

\frac{\sin ϕ}{1 + \cos ϕ} = \frac{g R}{2 u^{2}} \times (1 + \cos ϕ) (1 + \tan^{2} θ) - \tan θ

l e t λ = \frac{g R}{2 u^{2}}

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

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

\tan θ = \frac{1 \pm \sqrt{1 + 4 λ \sin ϕ - 4 λ^{2} {(1 + \cos ϕ)}^{2}}}{2 λ (1 + \cos ϕ)}

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

\underset{―}{a t p o i n t B :}

U_{x} = u \cos θ

U_{y} = g t - u \sin θ = \frac{g R (1 + \cos ϕ)}{u \cos θ} - u \sin θ

U_{⊥} = U_{x} \cos ϕ + U_{y} \sin ϕ

U_{⊥} = u \cos θ \cos ϕ + [\frac{g R (1 + \cos ϕ)}{u \cos θ} - u \sin θ] \sin ϕ

\Rightarrow U_{⊥} = u \cos (ϕ + θ) + \frac{g R (1 + \cos ϕ) \sin ϕ}{u \cos θ}

U_{∥} = - U_{x} \sin ϕ + U_{y} \cos ϕ

U_{∥} = - u \cos θ \sin ϕ + [\frac{g R (1 + \cos ϕ)}{u \cos θ} - u \sin θ] \cos ϕ

\Rightarrow U_{∥} = - u \sin (ϕ + θ) + \frac{g R (1 + \cos ϕ) \cos ϕ}{u \cos θ}

V_{⊥} = {e U}_{⊥}

\Rightarrow V_{⊥} = e u \cos (ϕ + θ) + \frac{e g R (1 + \cos ϕ) \sin ϕ}{u \cos θ}

V_{∥} = U_{∥}

\Rightarrow V_{∥} = - u \sin (ϕ + θ) + \frac{g R (1 + \cos ϕ) \cos ϕ}{u \cos θ}

V_{x} = V_{⊥} \cos ϕ + V_{∥} \sin ϕ

V_{x} = [e u \cos (ϕ + θ) + \frac{e g R (1 + \cos ϕ) \sin ϕ}{u \cos θ}] \cos ϕ + [- u \sin (ϕ + θ) + \frac{g R (1 + \cos ϕ) \cos ϕ}{u \cos θ}] \sin ϕ

V_{x} = e u \cos (ϕ + θ) \cos ϕ + \frac{e g R (1 + \cos ϕ) \sin ϕ \cos ϕ}{u \cos θ} - u \sin (ϕ + θ) \sin ϕ + \frac{g R (1 + \cos ϕ) \sin ϕ \cos ϕ}{u \cos θ}

\Rightarrow V_{x} = u [(1 + e) \cos (ϕ + θ) \cos ϕ - \cos θ + \frac{λ (1 + e) (1 + \cos ϕ) \sin 2 ϕ}{\cos θ}]

V_{y} = V_{⊥} \sin ϕ - V_{∥} \cos ϕ

V_{y} = [e u \cos (ϕ + θ) + \frac{e g R (1 + \cos ϕ) \sin ϕ}{u \cos θ}] \sin ϕ - [- u \sin (ϕ + θ) + \frac{g R (1 + \cos ϕ) \cos ϕ}{u \cos θ}] \cos ϕ

\Rightarrow V_{y} = u [(1 + e) \cos (ϕ + θ) \sin ϕ + \sin θ + \frac{2 λ (1 + \cos ϕ) ((1 + e) \sin^{2} ϕ - 1)}{\cos θ}]

\underset{―}{r e t u r n f r o m B t o A :}

t = \frac{R (1 + \cos ϕ)}{(1 + e) \cos (ϕ + θ) \cos ϕ - \cos θ + \frac{λ (1 + \cos ϕ) (1 + e) \sin 2 ϕ}{\cos θ}} \times \frac{1}{u}

\Rightarrow \frac{t u}{R} = ξ = \frac{1 + \cos ϕ}{(1 + e) \cos (ϕ + θ) \cos ϕ - \cos θ + \frac{λ (1 + e) (1 + \cos ϕ) \sin 2 ϕ}{\cos θ}}

R \sin ϕ = t {u [(1 + e) \cos (ϕ + θ) \sin ϕ + \sin θ] + \frac{2 λ (1 + \cos ϕ) (e \sin^{2} ϕ - \cos^{2} ϕ)}{\cos θ} - \frac{g t}{2}}

\sin ϕ = \frac{t u}{R} [(1 + e) \cos (ϕ + θ) \sin ϕ + \sin θ + \frac{2 λ (1 + \cos ϕ) (e \sin^{2} ϕ - \cos^{2} ϕ)}{\cos θ} - \frac{λ t u}{R}]

\sin ϕ = ξ [(1 + e) \cos (ϕ + θ) \sin ϕ + \sin θ + \frac{2 λ (1 + \cos ϕ) ((1 + e) \sin^{2} ϕ - 1)}{\cos θ} - λ ξ]

o r

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

w i t h

e^{'} = 1 + e

λ = \frac{g R}{2 u^{2}}

θ = \tan^{- 1} [\frac{1 + \sqrt{1 + 4 λ \sin ϕ - 4 λ^{2} {(1 + \cos ϕ)}^{2}}}{2 λ (1 + \cos ϕ)}] *)

ξ = \frac{1 + \cos ϕ}{e^{'} \cos (ϕ + θ) \cos ϕ - \cos θ + \frac{λ e^{'} (1 + \cos ϕ) \sin 2 ϕ}{\cos θ}}

f o r a g i v e n e \in (0, 1] w e c a n f i n d

s o l u t i o n (s) f o r ϕ i n (I) f o r s o m e

v a l u e s o f p a r a m e t e r λ . t h e m a x i m u m

v a l u e o f λ s u c h t h a t a s o l u t i o n f o r ϕ

e x i s t s c a n b e d e t e r m i n e d n u m e r i c a l l y .

t h i s c o r r e s p o n d s t o t h e m i n i m u m

s p e e d t h e b a l l m u s t h a v e :

u_{m i n} = \sqrt{\frac{g R}{2 λ_{m a x}}}

e x a m p l e s :

e = 0.5 \Rightarrow u_{m i n} \approx 2.7682 \sqrt{g R}

e = 0.75 \Rightarrow u_{m i n} \approx 1.6082 \sqrt{g R}

e = 1 \Rightarrow u_{m i n} \approx 0.9098 \sqrt{g R} (= Q 65589)

*)

w i t h t h e o t h e r s o l u t i o n

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

w e c a n g e t a s o l u t i o n f o r t h e c a s e

t h a t t h e b a l l s t r i k e s t h e s u r f a c e

u p w a r d s, i . e . c o u n t e r c l o c k w i s e .

b u t i n t h i s c a s e m o r e e n e r g y i s

n e e d e d, i . e . u_{m i n} i s l a r g e r t h a n w i t h

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

s e e e x a m p l e s w i t h e = 0.5 a n d 0.75 .

Commented by mr W last updated on 15/Feb/21

Commented by mr W last updated on 13/Feb/21

Commented by mr W last updated on 13/Feb/21

Commented by mr W last updated on 12/Feb/21

Commented by mr W last updated on 12/Feb/21

Commented by mr W last updated on 14/Feb/21

following diagrams show different situations for e=0.75. generally: for a given e∈(0,1] there exists an u_(min) . for any u>u_(min) there are four possible ways how the ball returns back to its starting position: two ways clockwise and two counter− clockwise.

f o l l o w i n g d i a g r a m s s h o w d i f f e r e n t

s i t u a t i o n s f o r e = 0.75 .

g e n e r a l l y :

f o r a g i v e n e \in (0, 1] t h e r e e x i s t s a n

u_{m i n} . f o r a n y u > u_{m i n} t h e r e a r e f o u r

p o s s i b l e w a y s h o w t h e b a l l r e t u r n s

b a c k t o i t s s t a r t i n g p o s i t i o n : t w o

w a y s c l o c k w i s e a n d t w o c o u n t e r -

c l o c k w i s e .

Commented by mr W last updated on 15/Feb/21

Commented by mr W last updated on 13/Feb/21

Commented by mr W last updated on 14/Feb/21

Commented by mr W last updated on 13/Feb/21

Commented by mr W last updated on 14/Feb/21

Answered by mr W last updated on 14/Feb/21

Commented by mr W last updated on 14/Feb/21

background knowledge what happens when a ball strikes on a wall and the restitution coefficient of the collision is e? U=coming speed V=leaving speed U_⊥ =U cos δ_1 U_∥ =U sin δ_1 V_⊥ =V cos δ_2 V_∥ =V sin δ_2 V_⊥ =e×U_⊥ ⇒V cos δ_2 =e×U cos δ_1 V_∥ =U_∥ ⇒ V sin δ_2 =U sin δ_1 tan δ_2 =((tan δ_1 )/e) δ_2 ≥δ_1 KE_1 =(1/2)mU^2 KE_2 =(1/2)mV^2 =(1/2)mU^2 (sin^2 δ_1 +e^2 cos^2 δ_1 ) ⇒KE_2 =(sin^2 δ_1 +e^2 cos^2 δ_1 )KE_1 we see KE_2 =KE_1 only when e=1, otherwise KE_2 <KE_1 . the collision is called elastic if e=1.

b a c k g r o u n d k n o w l e d g e

w h a t h a p p e n s w h e n a b a l l s t r i k e s o n

a w a l l a n d 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 ?

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

U_{⊥} = U \cos δ_{1}

U_{∥} = U \sin δ_{1}

V_{⊥} = V \cos δ_{2}

V_{∥} = V \sin δ_{2}

V_{⊥} = e \times U_{⊥} \Rightarrow V \cos δ_{2} = e \times U \cos δ_{1}

V_{∥} = U_{∥} \Rightarrow V \sin δ_{2} = U \sin δ_{1}

\tan δ_{2} = \frac{\tan δ_{1}}{e}

δ_{2} ⩾ δ_{1}

{K E}_{1} = \frac{1}{2} {m U}^{2}

{K E}_{2} = \frac{1}{2} {m V}^{2} = \frac{1}{2} {m U}^{2} (\sin^{2} δ_{1} + e^{2} \cos^{2} δ_{1})

\Rightarrow {K E}_{2} = (\sin^{2} δ_{1} + e^{2} \cos^{2} δ_{1}) {K E}_{1}

w e s e e {K E}_{2} = {K E}_{1} o n l y w h e n e = 1,

o t h e r w i s e {K E}_{2} < {K E}_{1} .

t h e c o l l i s i o n i s c a l l e d e l a s t i c i f e = 1 .

Commented by Ar Brandon last updated on 13/Feb/21

Wow ! 😃 Where did you learn your geometry Sir ? I feel you have this unique skill here.

Wow !

Where did you learn your geometry Sir ? I feel you have this unique skill here.

Commented by Ar Brandon last updated on 14/Feb/21

Commented by mr W last updated on 14/Feb/21

the best teacher is the own interest...

t h e b e s t t e a c h e r i s t h e o w n i n t e r e s t \dots

Answered by ajfour last updated on 15/Feb/21

Commented by ajfour last updated on 15/Feb/21

First just assuming (without proof) that u is u_(min) when least energy is lost upon collision. Let just before collision velocity v_1 ^� =−qi^� −pj^� and just after v_2 ^� =−qi^� +epj^� −△K =(p^2 /2)(1−e^2 ) This is minimum if p is minimum. But ball has to reach back ⇒ (e^2 p^2 )_(min) =2(gsin 2φ)R(1+cos 2φ) say 2φ=θ , and f(θ)=sin θ(1+cos θ) f ′(θ)=cos θ+cos^2 θ−sin^2 θ =2cos^2 θ+cos θ−1 =0 ⇒ cos 2φ=cos θ=−(1/4)±(√((1/(16))+(1/2))) cos 2φ=cos θ=(1/2) ⇒ φ=30° Now from (e^2 p^2 )_(min) =2(gsin 2φ)R(1+cos 2φ) ⇒ p_(min) ^2 =((2/e^2 ))(((g(√3))/2))(((3R)/2)) p_(min) =(√((3(√3)gR)/(2e^2 ))) u_(y,min) ^2 =p_(min) ^2 −2(gsin 2φ)R(1+cos 2φ) =((3(√3)gR)/(2e^2 ))−((3(√3)gR)/2) u_(y,min) =(√((3(√3)gR(1−e^2 ))/(2e^2 ))) Rsin 2φ=u_x t_1 −(((gcos 2φ)/2))t_1 ^2 u_x =((R(√3))/(2t_1 ))+((gt_1 )/4) t_1 =((p_(min) −u_(y,min) )/(gsin 2φ))=2(((p_(min) −u_(y,min) )/(g(√3)))) t_1 =(2/(g(√3)))((√((3(√3)gR)/(2e^2 ))) )(1±(√(1−e^2 )) ) u_(min) ^2 =(u_x ^2 +u_y ^2 ) with φ=30° =(((R(√3))/(2t_1 ))+((gt_1 )/4))^2 +((3(√3)gR(1−e^2 ))/(2e^2 ))

F i r s t j u s t a s s u m i n g (w i t h o u t

p r o o f) t h a t u i s u_{m i n} w h e n

l e a s t e n e r g y i s l o s t u p o n c o l l i s i o n .

L e t j u s t b e f o r e c o l l i s i o n

v e l o c i t y {\bar{v}}_{1} = - q \hat{i} - p \hat{j}

a n d j u s t a f t e r {\bar{v}}_{2} = - q \hat{i} + e p \hat{j}

- △ K = \frac{p^{2}}{2} (1 - e^{2})

T h i s i s m i n i m u m i f p i s

m i n i m u m .

B u t b a l l h a s t o r e a c h b a c k \Rightarrow

{(e^{2} p^{2})}_{m i n} = 2 (g \sin 2 ϕ) R (1 + \cos 2 ϕ)

s a y 2 ϕ = θ, a n d

f (θ) = \sin θ (1 + \cos θ)

f^{'} (θ) = \cos θ + \cos^{2} θ - \sin^{2} θ

= 2 \cos^{2} θ + \cos θ - 1 = 0

\Rightarrow

\cos 2 ϕ = \cos θ = - \frac{1}{4} \pm \sqrt{\frac{1}{16} + \frac{1}{2}}

\cos 2 ϕ = \cos θ = \frac{1}{2}

\Rightarrow ϕ = 30 ° N o w f r o m

{(e^{2} p^{2})}_{m i n} = 2 (g \sin 2 ϕ) R (1 + \cos 2 ϕ)

\Rightarrow p_{m i n}^{2} = (\frac{2}{e^{2}}) (\frac{g \sqrt{3}}{2}) (\frac{3 R}{2})

p_{m i n} = \sqrt{\frac{3 \sqrt{3} g R}{2 e^{2}}}

u_{y, m i n}^{2} = p_{m i n}^{2} - 2 (g \sin 2 ϕ) R (1 + \cos 2 ϕ)

= \frac{3 \sqrt{3} g R}{2 e^{2}} - \frac{3 \sqrt{3} g R}{2}

u_{y, m i n} = \sqrt{\frac{3 \sqrt{3} g R (1 - e^{2})}{2 e^{2}}}

R \sin 2 ϕ = u_{x} t_{1} - (\frac{g \cos 2 ϕ}{2}) t_{1}^{2}

u_{x} = \frac{R \sqrt{3}}{2 t_{1}} + \frac{{g t}_{1}}{4}

t_{1} = \frac{p_{m i n} - u_{y, m i n}}{g \sin 2 ϕ} = 2 (\frac{p_{m i n} - u_{y, m i n}}{g \sqrt{3}})

t_{1} = \frac{2}{g \sqrt{3}} (\sqrt{\frac{3 \sqrt{3} g R}{2 e^{2}}}) (1 \pm \sqrt{1 - e^{2}})

u_{m i n}^{2} = (u_{x}^{2} + u_{y}^{2}) w i t h ϕ = 30 °

= {(\frac{R \sqrt{3}}{2 t_{1}} + \frac{{g t}_{1}}{4})}^{2} + \frac{3 \sqrt{3} g R (1 - e^{2})}{2 e^{2}}

Commented by ajfour last updated on 15/Feb/21

Please review it Sir, it dont seem to taly with your solution and hence with the original question, but yet i think it is a nice question, and its answer as i have derived, did have a chance (i suspected) to yield the same answer...

P l e a s e r e v i e w i t S i r, i t d o n t

s e e m t o t a l y w i t h y o u r s o l u t i o n

a n d h e n c e w i t h t h e o r i g i n a l

q u e s t i o n, b u t y e t i t h i n k i t i s

a n i c e q u e s t i o n, a n d i t s a n s w e r

a s i h a v e d e r i v e d, d i d h a v e a

c h a n c e (i s u s p e c t e d) t o y i e l d

t h e s a m e a n s w e r \dots

Commented by mr W last updated on 15/Feb/21

let me some time to follow your working in detail. i think you interpreted the question differently as i did. the question should ask with which smallest speed one can throw the ball into the bowl so that it can return back to the start point. e.g. if e=0.75 the smallest speed is 1.6082(√(gR)) when the ball follows following track:

l e t m e s o m e t i m e t o f o l l o w y o u r

w o r k i n g i n d e t a i l . i t h i n k y o u

i n t e r p r e t e d t h e q u e s t i o n d i f f e r e n t l y

a s i d i d . t h e q u e s t i o n s h o u l d a s k

w i t h w h i c h s m a l l e s t s p e e d o n e c a n

t h r o w t h e b a l l i n t o t h e b o w l s o t h a t

i t c a n r e t u r n b a c k t o t h e s t a r t p o i n t .

e . g . i f e = 0.75 t h e s m a l l e s t s p e e d i s

1.6082 \sqrt{g R} w h e n t h e b a l l f o l l o w s

f o l l o w i n g t r a c k :

Commented by mr W last updated on 15/Feb/21

Commented by mr W last updated on 15/Feb/21

for any other cases, i think also your result, the ball needs more speed than 1.6082(√(gR)). if e=1, the smallest speed should be 0.9098(√(gR)).

f o r a n y o t h e r c a s e s, i t h i n k a l s o y o u r

r e s u l t, t h e b a l l n e e d s m o r e s p e e d

t h a n 1.6082 \sqrt{g R} .

i f e = 1, t h e s m a l l e s t s p e e d s h o u l d b e

0.9098 \sqrt{g R} .

Commented by mr W last updated on 15/Feb/21

acc. to your result, for e=1, t_1 =(√((2(√3)R)/g)) u_(min) =(((√3)/2)+((√(2(√3)))/4))(√(gR))≈1.3313(√(gR)) >0.9098(√(gR)) (what i got)

a c c . t o y o u r r e s u l t, f o r e = 1,

t_{1} = \sqrt{\frac{2 \sqrt{3} R}{g}}

u_{m i n} = (\frac{\sqrt{3}}{2} + \frac{\sqrt{2 \sqrt{3}}}{4}) \sqrt{g R} \approx 1.3313 \sqrt{g R}

> 0.9098 \sqrt{g R} (w h a t i g o t)

Commented by mr W last updated on 15/Feb/21

i think the least energy loss during collision doesn′t mean the minimum total energy the ball needs.

i t h i n k t h e l e a s t e n e r g y l o s s d u r i n g

c o l l i s i o n {d o e s n}^{'} t m e a n t h e m i n i m u m

t o t a l e n e r g y t h e b a l l n e e d s .

Commented by ajfour last updated on 16/Feb/21

I know you are right Sir.

I k n o w y o u a r e r i g h t S i r .

Commented by mr W last updated on 16/Feb/21

i think i have found the logic fault in your solution: with But ball has to reach back ⇒ (e^2 p^2 )_(min) =2(gsin 2φ)R(1+cos 2φ) we only ensure that the ball can reach back its height (position in y− direction), the ball may not come back to its start point (also in x− direction), so it is something like this:

i t h i n k i h a v e f o u n d t h e l o g i c f a u l t

i n y o u r s o l u t i o n :

w i t h

B u t b a l l h a s t o r e a c h b a c k \Rightarrow

{(e^{2} p^{2})}_{m i n} = 2 (g \sin 2 ϕ) R (1 + \cos 2 ϕ)

w e o n l y e n s u r e t h a t t h e b a l l c a n r e a c h

b a c k i t s h e i g h t (p o s i t i o n i n y -

d i r e c t i o n), t h e b a l l m a y n o t c o m e

b a c k t o i t s s t a r t p o i n t (a l s o i n x -

d i r e c t i o n), s o i t i s s o m e t h i n g l i k e

t h i s :

Commented by mr W last updated on 16/Feb/21

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