# Question-65589

Question Number 65589 by mr W last updated on 31/Jul/19
Commented by mr W last updated on 31/Jul/19
$${Answer}\:{to}\:{Q}\mathrm{65277} \\$$
Commented by Tanmay chaudhury last updated on 31/Jul/19
$${sir}\:{pls}\:{explain}\:{if}\:{possible}.. \\$$
Commented by mr W last updated on 12/Feb/21
$${since}\:{the}\:{collision}\:{is}\:{elastic},\:{the} \\$$$${velocity}\:{of}\:{the}\:{ball}\:{after}\:{the}\:{collision} \\$$$${is}\:{the}\:{same}\:{as}\:{before}\:{the}\:{collision}. \\$$$${that}\:{means}\:{the}\:{ball}\:{returns}\:{to}\:{its} \\$$$${start}\:{position}\:{following}\:{the}\:{same} \\$$$${trace}\:{as}\:{it}\:{came}.\:{that}\:{means}\:{also} \\$$$${that}\:{the}\:{direction}\:{of}\:{the}\:{velocity}\:{at}\:{the} \\$$$${point}\:{of}\:{collision}\:{must}\:{be}\:{perpendicular} \\$$$${to}\:{the}\:{surface}. \\$$$$\\$$$${let}\:{t}={time}\:{from}\:{A}\:{to}\:{P} \\$$$${t}=\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)}{{u}\:\mathrm{cos}\:\theta} \\$$$${at}\:{point}\:{P}: \\$$$${v}_{{x}} ={u}\:\mathrm{cos}\:\theta \\$$$${v}_{{y}} =−{u}\:\mathrm{sin}\:\theta+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)}{{u}\:\mathrm{cos}\:\theta} \\$$$${v}_{{y}} ={v}_{{x}} \mathrm{tan}\:\alpha \\$$$$\Rightarrow−{u}\:\mathrm{sin}\:\theta+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)}{{u}\:\mathrm{cos}\:\theta}={u}\:\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha \\$$$$\Rightarrow\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)}{{u}^{\mathrm{2}} \:\mathrm{cos}\:\theta}=\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha \\$$$${with}\:\lambda=\frac{{gR}}{{u}^{\mathrm{2}} } \\$$$$\Rightarrow\frac{\lambda\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)}{\mathrm{cos}\:\theta}=\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha \\$$$$\Rightarrow\lambda=\frac{\mathrm{cos}\:\theta\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha\right)}{\mathrm{1}+\mathrm{cos}\:\alpha} \\$$$$\Rightarrow\frac{{u}^{\mathrm{2}} }{{gR}}=\frac{\mathrm{1}+\mathrm{cos}\:\alpha}{\left(\mathrm{tan}\:\theta+\mathrm{tan}\:\alpha\right)\mathrm{cos}^{\mathrm{2}} \:\theta}\:\:\:\:…\left({i}\right) \\$$$$\\$$$$−{R}\:\mathrm{sin}\:\alpha={u}\:\mathrm{sin}\:\theta\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\$$$$\Rightarrow{t}=\frac{{u}\:\mathrm{sin}\:\theta+\sqrt{{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gR}\:\mathrm{sin}\:\alpha}}{{g}} \\$$$$\Rightarrow\frac{{u}\:\mathrm{sin}\:\theta+\sqrt{{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}{gR}\:\mathrm{sin}\:\alpha}}{{g}}=\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)}{{u}\:\mathrm{cos}\:\theta} \\$$$$\Rightarrow\mathrm{sin}\:\theta+\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}\frac{{gR}}{{u}^{\mathrm{2}} }\:\mathrm{sin}\:\alpha}=\frac{{gR}}{{u}^{\mathrm{2}} }×\frac{\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)}{\mathrm{cos}\:\theta} \\$$$$\Rightarrow\mathrm{sin}\:\theta+\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}\lambda\:\mathrm{sin}\:\alpha}=\frac{\lambda\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)}{\mathrm{cos}\:\theta} \\$$$$\Rightarrow\mathrm{sin}\:\theta+\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}\lambda\:\mathrm{sin}\:\alpha}=\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha \\$$$$\Rightarrow\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}\lambda\:\mathrm{sin}\:\alpha}=\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha \\$$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{2}\lambda\:\mathrm{sin}\:\alpha=\left(\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha\right)^{\mathrm{2}} \\$$$$\Rightarrow\lambda=\frac{\left(\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha\right)^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}\:\mathrm{sin}\:\alpha} \\$$$$\Rightarrow\frac{\left(\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha\right)^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{cos}\:\theta\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha\right)}{\mathrm{1}+\mathrm{cos}\:\alpha} \\$$$$\Rightarrow\frac{\mathrm{cos}\:\theta\:\mathrm{tan}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{2}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{cos}\:\alpha} \\$$$$\Rightarrow\left(\mathrm{tan}\:\alpha−\mathrm{sin}\:\alpha\right)\:\mathrm{cos}\:\theta−\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)\mathrm{sin}\:\theta=\mathrm{0} \\$$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{tan}\:\alpha−\mathrm{sin}\:\alpha}{\mathrm{1}+\mathrm{cos}\:\alpha}\:\:\:\:…\left({ii}\right) \\$$$${put}\:{this}\:{into}\:\left({i}\right): \\$$$$\frac{{u}^{\mathrm{2}} }{{gR}}=\frac{\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)}{\left(\mathrm{tan}\:\alpha+\mathrm{tan}\:\theta\right)} \\$$$$\Rightarrow\frac{{u}^{\mathrm{2}} }{{gR}}=\frac{\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\left(\mathrm{tan}\:\alpha−\mathrm{sin}\:\alpha\right)^{\mathrm{2}} }{\mathrm{2}\:\mathrm{tan}\:\alpha} \\$$$${minimum}\:{of}\:{u}\:{is}\:{at}: \\$$$$\alpha=\mathrm{64}.\mathrm{2057}° \\$$$$\theta=\mathrm{39}.\mathrm{159}° \\$$$${u}_{{min}} =\mathrm{0}.\mathrm{9098}\sqrt{{gR}} \\$$
Commented by Tanmay chaudhury last updated on 31/Jul/19
$${thank}\:{you}\:{sir}…{this}\:{problem}\:{remain}\:{unsolved} \\$$$${till}\:{date}…{you}\:{did}\:{it}\:{excellent}… \\$$
Commented by mr W last updated on 01/Aug/19
Commented by mr W last updated on 01/Aug/19
Commented by mr W last updated on 01/Aug/19
Commented by mr W last updated on 01/Aug/19