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Question Number 139167 by Dwaipayan Shikari last updated on 23/Apr/21

Answered by mr W last updated on 23/Apr/21

Commented by mr W last updated on 23/Apr/21

$$y=\frac{x^2}{2a}$$$$\tan \theta =y^{\prime }=\frac{x}{a}=\lambda ,say$$$$y^{″}=\frac{1}{a}$$$$r=\frac{{(1+y^{\prime 2})}^{3/2}}{y^{″}}=a{(1+\frac{x^2}{a^2})}^{3/2}=a{(1+{\lambda }^2)}^{\frac{3}{2}}$$$$u=velocityat(x,y)$$$$\frac{1}{2}{mv}^2=\frac{1}{2}{mu}^2+mgy$$$$\Rightarrow u^2=v^2-\frac{{gx}^2}{a}=v^2-ag{\lambda }^2$$$$athighestpositionu=0:$$$$v^2-ag{\lambda }_{max}^2=0$$$$\Rightarrow {\lambda }_{max}=\frac{v}{\sqrt{ag}}$$$$R-mg\cos \theta =\frac{{mu}^2}{r}=\frac{m(v^2-ag{\lambda }^2)}{a{(1+{\lambda }^2)}^{\frac{3}{2}}}$$$$R=\frac{mg}{\sqrt{1+t^2}}+\frac{mg(\frac{v^2}{ag}-{\lambda }^2)}{{(1+{\lambda }^2)}^{\frac{3}{2}}}=\frac{mg(1+\frac{v^2}{ag})}{{(1+{\lambda }^2)}^{\frac{3}{2}}}$$$$suchthattheblock{doesn}^{\prime }tslip,$$$$R\sin \theta ⩽\mu (Mg+R\cos \theta )$$$$R(\sin \theta -\mu \cos \theta )⩽\mu Mg$$$$\frac{m(1+\frac{v^2}{ag})(\lambda -\mu )}{{(1+{\lambda }^2)}^2}⩽\mu M$$$$\frac{\mu M}{m(1+\frac{v^2}{ag})}⩾\frac{\lambda -\mu }{{(1+{\lambda }^2)}^2}=f\left(\lambda \right)say$$$$formaximumf\left(\lambda \right):$$$$\frac{df\left(\lambda \right)}{d\lambda }=\frac{1}{{(1+{\lambda }^2)}^2}-\frac{4(\lambda -\mu )\lambda }{{(1+{\lambda }^2)}^3}=0$$$$\frac{4(\lambda -\mu )\lambda }{1+{\lambda }^2}=1$$$$3{\lambda }^2-4\mu \lambda -1=0$$$$\Rightarrow \lambda =\frac{2\mu +\sqrt{4{\mu }^2+3}}{3}$$$$with\mu =\frac{1}{2}$$$$\Rightarrow \lambda =\frac{2\times \frac{1}{2}+\sqrt{4\times \frac{1}{4}+3}}{3}=1$$$$f{\left(\lambda \right)}_{max}=\frac{1-\frac{1}{2}}{{(1+1)}^2}=\frac{1}{8}$$$$\frac{\mu M}{m(1+\frac{v^2}{ag})}⩾\frac{1}{8}$$$$\left(\frac{4M-m}{m}\right)ag⩾\frac{v^2}{ag}$$$$v⩽\sqrt{\frac{4M-m}{m}ag}$$$${\lambda }_{max}=\frac{v}{\sqrt{ag}}=\sqrt{4\times \frac{M}{m}-1}>\sqrt{4\times \frac{1}{2}-1}=1$$$$thatmeans\lambda =1isreachedbeforethe$$$$highestpositionisreached.$$$$i.e.v_{max}=\sqrt{\frac{4M-m}{m}ag}isvalid.$$$$\Rightarrow \alpha =4,\beta =1,\gamma =1$$$$\Rightarrow \alpha +\beta +\gamma =6$$

Commented by Dwaipayan Shikari last updated on 23/Apr/21

$$Greatsir!Butineedasuggestion.Howdoyouthinkthiskind$$$$solutionseasily?$$

Commented by mr W last updated on 24/Apr/21

$${it}^{\prime }snoteasytogiveageneral$$$$sugestion.actuallythisisnotan$$$$easyquestion,becauseoneneedsto$$$$applyalotofthingsbothinphysics$$$$andincalculus.$$

Commented by peter frank last updated on 24/Apr/21

$$GODhasblesshimwith$$$$greatabilitymathsandphysics.$$

Commented by otchereabdullai@gmail.com last updated on 24/Apr/21

$$\mathrm{Is}\mathrm{very}\mathrm{true}\mathrm{sir}\mathrm{Frank}\mathrm{i}\mathrm{really}\mathrm{love}\mathrm{this}$$$$\mathrm{man}.\mathrm{For}\mathrm{me}\mathrm{he}\mathrm{is}\mathrm{the}\mathrm{world}\mathrm{best}!$$$$\mathrm{Because}\mathrm{he}\mathrm{looks}\mathrm{very}\mathrm{different}\mathrm{in}$$$$\mathrm{solving}\mathrm{mathematics}\mathrm{and}\mathrm{physics}$$$$\mathrm{questions}.\mathrm{And}\mathrm{he}\mathrm{has}\mathrm{idea}\mathrm{in}\mathrm{almost}$$$$\mathrm{all}\mathrm{questions}\mathrm{even}\mathrm{if}\mathrm{the}\mathrm{question}$$$$\mathrm{is}\mathrm{wrong}\mathrm{he}\mathrm{can}\mathrm{tell}\mathrm{you}\mathrm{the}\mathrm{quetion}\mathrm{is}$$$$\mathrm{wrong}\mathrm{and}\mathrm{tell}\mathrm{the}\mathrm{reason}\mathrm{and}\mathrm{how}\mathrm{the}$$$$\mathrm{question}\mathrm{should}\mathrm{look}\mathrm{like}\mathrm{even}\mathrm{if}\mathrm{is}$$$$\mathrm{a}\mathrm{very}\mathrm{difficult}\mathrm{question}!\mathbf{I}\mathbf{wish}\mathbf{he}\mathbf{is}$$$$\mathbf{my}\mathbf{father}!\mathrm{God}\mathrm{bless}\mathrm{you}\mathbf{prof}\mathbf{W}!\mathbf{long}$$$$\mathbf{life}\mathbf{is}\mathbf{what}\mathbf{we}\mathbf{pray}\mathbf{for}\mathbf{you}!$$

Commented by mr W last updated on 24/Apr/21

$$thankstoyoubothforyourheart-$$$$warmingwords!$$

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