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Question Number 133275 by mr W last updated on 20/Feb/21

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

A hall with horizontal smooth floor has a parabola shaped wall as shown. A ball is projected along the floor from point A and returns back to this point after two impacts with the wall. If the impacts are elastic, find the coordinates of the impact points B and C.

Ahallwithhorizontalsmoothfloorhasaparabolashapedwallasshown.AballisprojectedalongthefloorfrompointAandreturnsbacktothispointaftertwoimpactswiththewall.Iftheimpactsareelastic,findthecoordinatesoftheimpactpointsBandC.

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

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

since the impacts are elastic, the path of the ball is the same as of a light ray. A′ =image of A in tangent at B A” =image of A in tangent at C then A′, B, C and A” are collinear. let η=(h/b), ξ=(a/b), λ=(p/b), μ=(q/b) equation of parabola: y=h−((hx^2 )/b^2 ) (dy/dx)=−((2hx)/b^2 ) say B(p, y_B ) with y_B =h−((hp^2 )/b^2 ) say C(q, y_B ) with y_C =h−((hq^2 )/b^2 ) tan θ=((2hp)/b^2 )=2ηλ tan ϕ=−((2hq)/b^2 )=−2ημ tan α=(y_B /(p−a))=((h−((hp^2 )/b^2 ))/(p−a))=((η(1−λ^2 ))/(λ−ξ)) tan β=(y_C /(a−q))=((h−((hq^2 )/b^2 ))/(a−q))=((η(1−μ^2 ))/(ξ−μ)) tan φ=((y_C −y_B )/(p−q))=((−((hq^2 )/b^2 )+((hp^2 )/b^2 ))/(p−q))=η(λ+μ) γ=π−α−θ δ=π−β−ϕ γ+φ=θ π−(α−φ)=2θ tan 2θ=−tan (α−φ) ((2 tan θ)/(1−tan^2 θ))=−((tan α−tan φ)/(1+tan α tan φ)) ((2×2ηλ)/(1−(2ηλ)^2 ))=−((((η(1−λ^2 ))/(λ−ξ))−η(λ+μ))/(1+((η(1−λ^2 ))/(λ−ξ))×η(λ+μ))) ((4λ)/(1−4η^2 λ^2 ))+((1−λ^2 −(λ+μ)(λ−ξ))/(λ−ξ+η^2 (1−λ^2 )(λ+μ)))=0 ...(i) δ−φ=ϕ π−(β+φ)=2ϕ tan 2ϕ=−tan (β+φ) ((2 tan ϕ)/(1−tan^2 ϕ))=−((tan β+tan φ)/(1−tan β tan φ)) ((2(−2ημ))/(1−(−2ημ)^2 ))=−((((η(1−μ^2 ))/(ξ−μ))+η(λ+μ))/(1−((η(1−μ^2 ))/(ξ−μ))×η(λ+μ))) ((4μ)/(1−4η^2 μ^2 ))+((1−μ^2 +(λ+μ)(ξ−μ))/(μ−ξ+η^2 (1−μ^2 )(λ+μ)))=0 ...(ii) from (i) and (ii) we get λ and μ. examples:

sincetheimpactsareelastic,thepathoftheballisthesameasofalightray.A′=imageofAintangentatBA″=imageofAintangentatCthenA′,B,CandA″arecollinear.letη=hb,ξ=ab,λ=pb,μ=qbequationofparabola:y=h−hx2b2dydx=−2hxb2sayB(p,yB)withyB=h−hp2b2sayC(q,yB)withyC=h−hq2b2tanθ=2hpb2=2ηλtanφ=−2hqb2=−2ημtanα=yBp−a=h−hp2b2p−a=η(1−λ2)λ−ξtanβ=yCa−q=h−hq2b2a−q=η(1−μ2)ξ−μtanϕ=yC−yBp−q=−hq2b2+hp2b2p−q=η(λ+μ)γ=π−α−θδ=π−β−φγ+ϕ=θπ−(α−ϕ)=2θtan2θ=−tan(α−ϕ)2tanθ1−tan2θ=−tanα−tanϕ1+tanαtanϕ2×2ηλ1−(2ηλ)2=−η(1−λ2)λ−ξ−η(λ+μ)1+η(1−λ2)λ−ξ×η(λ+μ)4λ1−4η2λ2+1−λ2−(λ+μ)(λ−ξ)λ−ξ+η2(1−λ2)(λ+μ)=0...(i)δ−ϕ=φπ−(β+ϕ)=2φtan2φ=−tan(β+ϕ)2tanφ1−tan2φ=−tanβ+tanϕ1−tanβtanϕ2(−2ημ)1−(−2ημ)2=−η(1−μ2)ξ−μ+η(λ+μ)1−η(1−μ2)ξ−μ×η(λ+μ)4μ1−4η2μ2+1−μ2+(λ+μ)(ξ−μ)μ−ξ+η2(1−μ2)(λ+μ)=0...(ii)from(i)and(ii)wegetλandμ.examples:

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

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

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

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

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

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

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

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