Question-23294

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 u, 𝛉 . Striker has mass M, radius R. queen has mass m, radius r. friction coefficient between caromboard surface and queen, or striker is 𝛍. Coefficient of restitution for the hit is e. (Let the curved surfaces of queen and striker be smooth.)

$${The}\:{player}\:{failed}\:{to}\:{pocket}\:{the} \\ $$$${queen}\:{and}\:{it}\:{could}\:{reach}\:{only} \\ $$$${till}\:{P}.\:{Find}\:\boldsymbol{{u}},\:\boldsymbol{\theta}\:. \\ $$$${Striker}\:{has}\:{mass}\:\boldsymbol{{M}},\:{radius}\:\boldsymbol{{R}}. \\ $$$${queen}\:{has}\:{mass}\:\boldsymbol{{m}},\:{radius}\:\boldsymbol{{r}}. \\ $$$${friction}\:{coefficient}\:{between} \\ $$$${caromboard}\:{surface}\:{and}\:{queen}, \\ $$$${or}\:{striker}\:{is}\:\boldsymbol{\mu}.\:{Coefficient}\:{of} \\ $$$${restitution}\:{for}\:{the}\:{hit}\:{is}\:\boldsymbol{{e}}. \\ $$$$\left({Let}\:{the}\:\:{curved}\:{surfaces}\:{of}\:{queen}\right. \\ $$$$\left.{and}\:{striker}\:{be}\:{smooth}.\right) \\ $$

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))))]))

$$\mathrm{this}\:\mathrm{is}\:\mathrm{my}\:\mathrm{try}… \\ $$$$ \\ $$$$\mathrm{tan}\:\theta=\frac{\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{R}+\mathrm{r}}{\:\sqrt{\mathrm{2}}}}{\mathrm{d}−\frac{\mathrm{R}+\mathrm{r}}{\:\sqrt{\mathrm{2}}}}=\frac{\mathrm{a}−\left(\mathrm{R}+\mathrm{r}\right)\sqrt{\mathrm{2}}}{\mathrm{2d}−\left(\mathrm{R}+\mathrm{r}\right)\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{a}−\left(\mathrm{R}+\mathrm{r}\right)\sqrt{\mathrm{2}}}{\mathrm{2d}−\left(\mathrm{R}+\mathrm{r}\right)\sqrt{\mathrm{2}}}\right) \\ $$$$ \\ $$$$\mathrm{v}=\mathrm{velocity}\:\mathrm{of}\:\mathrm{m}\:\mathrm{after}\:\mathrm{collision} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{mv}^{\mathrm{2}} =\mu\mathrm{mg}\frac{\mathrm{a}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{v}=\sqrt{\mu\mathrm{ga}\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$$\mathrm{u}_{\mathrm{1}} =\mathrm{velocity}\:\mathrm{of}\:\mathrm{M}\:\mathrm{before}\:\mathrm{collision} \\ $$$$\mathrm{u}_{\mathrm{2}} =\mathrm{velocity}\:\mathrm{of}\:\mathrm{M}\:\mathrm{after}\:\mathrm{collision}\:\mathrm{along}\:\mathrm{v} \\ $$$$\mathrm{let}\:\mathrm{b}=\sqrt{\left(\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{R}+\mathrm{r}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\left(\mathrm{d}−\frac{\mathrm{R}+\mathrm{r}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{M}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{u}_{\mathrm{1}} ^{\mathrm{2}} \right)=\mu\mathrm{Mgb} \\ $$$$\Rightarrow\mathrm{u}_{\mathrm{1}} =\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{2}\mu\mathrm{gb}} \\ $$$$ \\ $$$$\mathrm{v}−\mathrm{u}_{\mathrm{2}} =\mathrm{eu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Rightarrow\mathrm{u}_{\mathrm{2}} =\mathrm{v}−\mathrm{eu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$$$\mathrm{mv}+\mathrm{Mu}_{\mathrm{2}} =\mathrm{Mu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{mv}+\mathrm{Mv}−\mathrm{Meu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right)=\mathrm{Mu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\left(\mathrm{M}+\mathrm{m}\right)\mathrm{v}=\left(\mathrm{1}+\mathrm{e}\right)\mathrm{Mu}_{\mathrm{1}} \mathrm{cos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\left(\mathrm{M}+\mathrm{m}\right)\sqrt{\mu\mathrm{ga}\sqrt{\mathrm{2}}}=\left(\mathrm{1}+\mathrm{e}\right)\mathrm{Mcos}\:\left(\theta−\frac{\pi}{\mathrm{4}}\right)\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{2}\mu\mathrm{gb}} \\ $$$$\Rightarrow\mathrm{u}=\sqrt{\mu\mathrm{g}\left[\mathrm{2b}+\frac{\mathrm{a}\left(\mathrm{M}+\mathrm{m}\right)^{\mathrm{2}} \sqrt{\mathrm{2}}}{\left(\mathrm{1}+\mathrm{e}\right)^{\mathrm{2}} \mathrm{M}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\left(\theta−\frac{\pi}{\mathrm{4}}\right)}\right]} \\ $$

Commented by mrW1 last updated on 28/Oct/17

thank you for this help. I had difficulty how to handle the coefficient of restitution and in which direction the striker moves finally.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{this}\:\mathrm{help}.\:\mathrm{I}\:\mathrm{had}\:\mathrm{difficulty} \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{handle}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of} \\ $$$$\mathrm{restitution}\:\mathrm{and}\:\mathrm{in}\:\mathrm{which}\:\mathrm{direction} \\ $$$$\mathrm{the}\:\mathrm{striker}\:\mathrm{moves}\:\mathrm{finally}. \\ $$

Commented by ajfour last updated on 29/Oct/17