Question Number 23294 by ajfour last updated on 28/Oct/17
Commented by ajfour last updated on 28/Oct/17
$${The}\:{player}\:{failed}\:{to}\:{pocket}\:{the} \\ $$$${queen}\:{and}\:{it}\:{could}\:{reach}\:{only} \\ $$$${till}\:{P}.\:{Find}\:\boldsymbol{{u}},\:\boldsymbol{\theta}\:. \\ $$$${Striker}\:{has}\:{mass}\:\boldsymbol{{M}},\:{radius}\:\boldsymbol{{R}}. \\ $$$${queen}\:{has}\:{mass}\:\boldsymbol{{m}},\:{radius}\:\boldsymbol{{r}}. \\ $$$${friction}\:{coefficient}\:{between} \\ $$$${caromboard}\:{surface}\:{and}\:{queen}, \\ $$$${or}\:{striker}\:{is}\:\boldsymbol{\mu}.\:{Coefficient}\:{of} \\ $$$${restitution}\:{for}\:{the}\:{hit}\:{is}\:\boldsymbol{{e}}. \\ $$$$\left({Let}\:{the}\:\:{curved}\:{surfaces}\:{of}\:{queen}\right. \\ $$$$\left.{and}\:{striker}\:{be}\:{smooth}.\right) \\ $$
Commented by Physics lover last updated on 28/Oct/17
$${thats}\:{really}\:{one}\:{of}\:{the}\:{toughest} \\ $$$${questions}\:{i}\:{hav}\:{ever}\:{seen}. \\ $$
Answered by mrW1 last updated on 29/Oct/17
![this is my try... tan θ=(((a/2)−((R+r)/(√2)))/(d−((R+r)/(√2))))=((a−(R+r)(√2))/(2d−(R+r)(√2))) ⇒θ=tan^(−1) (((a−(R+r)(√2))/(2d−(R+r)(√2)))) v=velocity of m after collision (1/2)mv^2 =μmg(a/(√2)) ⇒v=(√(μga(√2))) u_1 =velocity of M before collision u_2 =velocity of M after collision along v let b=(√(((a/2)−((R+r)/(√2)))^2 +(d−((R+r)/(√2)))^2 )) (1/2)M(u^2 −u_1 ^2 )=μMgb ⇒u_1 =(√(u^2 −2μgb)) v−u_2 =eu_1 cos (θ−(π/4)) ⇒u_2 =v−eu_1 cos (θ−(π/4)) mv+Mu_2 =Mu_1 cos (θ−(π/4)) mv+Mv−Meu_1 cos (θ−(π/4))=Mu_1 cos (θ−(π/4)) (M+m)v=(1+e)Mu_1 cos (θ−(π/4)) (M+m)(√(μga(√2)))=(1+e)Mcos (θ−(π/4))(√(u^2 −2μgb)) ⇒u=(√(μg[2b+((a(M+m)^2 (√2))/((1+e)^2 M^2 cos^2 (θ−(π/4))))]))](Q23338.png)
$$\mathrm{this}\:\mathrm{is}\:\mathrm{my}\:\mathrm{try}... \\ $$$$ \\ $$$$\mathrm{tan}\:\theta=\frac{\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{R}+\mathrm{r}}{\sqrt{\mathrm{2}}}}{\mathrm{d}−\frac{\mathrm{R}+\mathrm{r}}{\sqrt{\mathrm{2}}}}=\frac{\mathrm{a}−\left(\mathrm{R}+\mathrm{r}\right)\sqrt{\mathrm{2}}}{\mathrm{2d}−\left(\mathrm{R}+\mathrm{r}\right)\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{a}−\left(\mathrm{R}+\mathrm{r}\right)\sqrt{\mathrm{2}}}{\mathrm{2d}−\left(\mathrm{R}+\mathrm{r}\right)\sqrt{\mathrm{2}}}\right) \\ $$$$ \\ $$$$\mathrm{v}=\mathrm{velocity}\:\mathrm{of}\:\mathrm{m}\:\mathrm{after}\:\mathrm{collision} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{mv}^{\mathrm{2}} =\mu\mathrm{mg}\frac{\mathrm{a}}{\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{v}=\sqrt{\mu\mathrm{ga}\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$$\mathrm{u}_{\mathrm{1}} =\mathrm{velocity}\:\mathrm{of}\:\mathrm{M}\:\mathrm{before}\:\mathrm{collision} \\ $$$$\mathrm{u}_{\mathrm{2}} =\mathrm{velocity}\:\mathrm{of}\:\mathrm{M}\:\mathrm{after}\:\mathrm{collision}\:\mathrm{along}\:\mathrm{v} \\ $$$$\mathrm{let}\:\mathrm{b}=\sqrt{\left(\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{R}+\mathrm{r}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\left(\mathrm{d}−\frac{\mathrm{R}+\mathrm{r}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{M}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{u}_{\mathrm{1}} ^{\mathrm{2}} \right)=\mu\mathrm{Mgb} \\ $$$$\Rightarrow\mathrm{u}_{\mathrm{1}} =\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{2}\mu\mathrm{gb}} \\ $$$$ \\ $$$$\mathrm{v}−\mathrm{u}_{\mathrm{2}} =\mathrm{eu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Rightarrow\mathrm{u}_{\mathrm{2}} =\mathrm{v}−\mathrm{eu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$$$\mathrm{mv}+\mathrm{Mu}_{\mathrm{2}} =\mathrm{Mu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{mv}+\mathrm{Mv}−\mathrm{Meu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right)=\mathrm{Mu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\left(\mathrm{M}+\mathrm{m}\right)\mathrm{v}=\left(\mathrm{1}+\mathrm{e}\right)\mathrm{Mu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\left(\mathrm{M}+\mathrm{m}\right)\sqrt{\mu\mathrm{ga}\sqrt{\mathrm{2}}}=\left(\mathrm{1}+\mathrm{e}\right)\mathrm{Mcos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right)\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{2}\mu\mathrm{gb}} \\ $$$$\Rightarrow\mathrm{u}=\sqrt{\mu\mathrm{g}\left[\mathrm{2b}+\frac{\mathrm{a}\left(\mathrm{M}+\mathrm{m}\right)^{\mathrm{2}} \sqrt{\mathrm{2}}}{\left(\mathrm{1}+\mathrm{e}\right)^{\mathrm{2}} \mathrm{M}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\left(\theta−\frac{\pi}{\mathrm{4}}\right)}\right]} \\ $$
Commented by mrW1 last updated on 28/Oct/17
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{this}\:\mathrm{help}.\:\mathrm{I}\:\mathrm{had}\:\mathrm{difficulty} \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{handle}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of} \\ $$$$\mathrm{restitution}\:\mathrm{and}\:\mathrm{in}\:\mathrm{which}\:\mathrm{direction} \\ $$$$\mathrm{the}\:\mathrm{striker}\:\mathrm{moves}\:\mathrm{finally}. \\ $$
Commented by ajfour last updated on 29/Oct/17
$$\mathcal{G}{reat}\:{Sir},\:{i}\:{m}\:{so}\:{happy};\:{thanks} \\ $$$${for}\:{the}\:{awesome}\:{effort},\:{time}, \\ $$$${and}\:{thinking}. \\ $$
Commented by Physics lover last updated on 29/Oct/17
$${Can}\:{you}\:{please}\:{explain}\:{how}\:{did}\: \\ $$$${you}\:{write}\:{tan}\:\theta\:{and}\:{b}\:. \\ $$
Commented by mrW1 last updated on 29/Oct/17
