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Question Number 195870 by mr W last updated on 12/Aug/23

Commented by mr W last updated on 12/Aug/23

find the minimum speed with which  the ball is launced from point A such  that it can return to this point again  after rebounded at point B.  assume the coefficient of restitution  is e.

findtheminimumspeedwithwhichtheballislauncedfrompointAsuchthatitcanreturntothispointagainafterreboundedatpointB.assumethecoefficientofrestitutionise.

Commented by liuxinnan last updated on 12/Aug/23

it as if  has not the min v_0   but it has a assured angle   between the v_0   and the incline

itasifhasnottheminv0butithasaassuredanglebetweenthev0andtheincline

Commented by mahdipoor last updated on 13/Aug/23

It seems that the question does not haven  minimum speed ,instead ,i find relationship   between tangential and vertical speed   (relative to the surface):  (V_t /V_v )=(v/u)=((e^2 +2e+1)/(e+1))tanθ=(e+1)tanθ

Itseemsthatthequestiondoesnothavenminimumspeed,instead,ifindrelationshipbetweentangentialandverticalspeed(relativetothesurface):VtVv=vu=e2+2e+1e+1tanθ=(e+1)tanθ

Commented by mr W last updated on 13/Aug/23

it′s correct. there is no minimum for  speed u, since the rebound point B  is not fixed.  but i got tan ϕ=(1/((e+1)tan θ))

itscorrect.thereisnominimumforspeedu,sincethereboundpointBisnotfixed.butigottanφ=1(e+1)tanθ

Answered by mr W last updated on 13/Aug/23

Commented by mr W last updated on 13/Aug/23

motion from A to B:  0=u sin ϕ t−((g cos θ)/2)t^2   ⇒t=((2u sin ϕ)/(g cos θ))  s=u cos ϕ t−((g sin θ)/2)t^2   s=u cos ϕ×((2u sin ϕ)/(g cos θ))−((g sin θ)/2)×(((2u sin ϕ)/(g cos θ)))^2   s=(u^2 /(g cos θ))(sin 2ϕ−2 sin^2  ϕ tan θ)  U_x =u cos ϕ−g sin θ ×((2u sin ϕ)/(g cos θ))       =u(cos ϕ−2 tan θ sin ϕ)   U_y =u sin ϕ−g cos θ×((2u sin ϕ)/(g cos θ))       =−u sin ϕ  V_x =U_x =u(cos ϕ−2 tan θ sin ϕ)   V_y =−eU_y =eu sin ϕ  motion from B to A:  t=((2eu sin ϕ)/(g cos θ))  −s=u(cos ϕ−2 tan θ sin ϕ)t−((g sin θ)/2)t^2   −s=u(cos ϕ−2 tan θ sin ϕ)×((2eu sin ϕ)/(g cos θ))−((g sin θ)/2)×(((2eu sin ϕ)/(g cos θ)))^2   −(u^2 /(g cos θ))[sin 2ϕ−2 sin^2  ϕ tan θ]=u(cos ϕ−2 tan θ sin ϕ)×((2eu sin ϕ)/(g cos θ))−((g sin θ)/2)×(((2eu sin ϕ)/(g cos θ)))^2   −sin 2ϕ+2 sin^2  ϕ tan θ=2e sin ϕ (cos ϕ−2 tan θ sin ϕ)−2e^2  tan θ sin^2  ϕ  (e+1) sin ϕ cos ϕ=(e+1)^2  tan θ sin^2  ϕ  ⇒tan ϕ=(1/((e+1) tan θ))  s=(u^2 /(g cos θ))(sin 2ϕ−2 sin^2  ϕ tan θ)  ⇒(u^2 /(gs))=[(e+1)^2 +(1/(tan^2  θ))]((sin θ)/(2e))

motionfromAtoB:0=usinφtgcosθ2t2t=2usinφgcosθs=ucosφtgsinθ2t2s=ucosφ×2usinφgcosθgsinθ2×(2usinφgcosθ)2s=u2gcosθ(sin2φ2sin2φtanθ)Ux=ucosφgsinθ×2usinφgcosθ=u(cosφ2tanθsinφ)Uy=usinφgcosθ×2usinφgcosθ=usinφVx=Ux=u(cosφ2tanθsinφ)Vy=eUy=eusinφmotionfromBtoA:t=2eusinφgcosθs=u(cosφ2tanθsinφ)tgsinθ2t2s=u(cosφ2tanθsinφ)×2eusinφgcosθgsinθ2×(2eusinφgcosθ)2u2gcosθ[sin2φ2sin2φtanθ]=u(cosφ2tanθsinφ)×2eusinφgcosθgsinθ2×(2eusinφgcosθ)2sin2φ+2sin2φtanθ=2esinφ(cosφ2tanθsinφ)2e2tanθsin2φ(e+1)sinφcosφ=(e+1)2tanθsin2φtanφ=1(e+1)tanθs=u2gcosθ(sin2φ2sin2φtanθ)u2gs=[(e+1)2+1tan2θ]sinθ2e

Commented by mr W last updated on 13/Aug/23

Commented by mr W last updated on 13/Aug/23

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