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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.
$${thats}\:{really}\:{one}\:{of}\:{the}\:{toughest} \\ $$$${questions}\:{i}\:{hav}\:{ever}\:{seen}. \\ $$
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
Great Sir, i m so happy; thanks  for the awesome effort, time,  and thinking.
$$\mathcal{G}{reat}\:{Sir},\:{i}\:{m}\:{so}\:{happy};\:{thanks} \\ $$$${for}\:{the}\:{awesome}\:{effort},\:{time}, \\ $$$${and}\:{thinking}. \\ $$
Commented by Physics lover last updated on 29/Oct/17
Can you please explain how did   you write tan θ and b .
$${Can}\:{you}\:{please}\:{explain}\:{how}\:{did}\: \\ $$$${you}\:{write}\:{tan}\:\theta\:{and}\:{b}\:. \\ $$
Commented by mrW1 last updated on 29/Oct/17

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