Question Number 225866 by BHOOPENDRA last updated on 15/Nov/25
$${Question}\:\mathrm{222520} \\ $$
Answered by BHOOPENDRA last updated on 15/Nov/25
$$\frac{−{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}+{b}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${y}\left({x}\right)=−{b}+{b}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$${The}\:{ball}\:{is}\:{released}\:{from}\:{rest}\:{at}\: \\ $$$${point}\:{A}\left({d},{h}\right)\:{d}\neq\mathrm{0} \\ $$$${collision}\:{occurs}\:{at}\:{point}\:{p}=\left({d},{y}\left({d}\right)\right) \\ $$$${y}\left({d}\right)=−{b}+{b}\sqrt{\mathrm{1}+\frac{{d}^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$${The}\:{velocity}\:{just}\:{before}\:{impact} \\ $$$${is}\:{purely}\:{vertical} \\ $$$${V}_{\mathrm{1}} =\left(\mathrm{0},−{v}\right)\:{v}=\sqrt{\mathrm{2}{g}\left({h}−{y}\left({d}\right)\right)} \\ $$$${Velocity}\:{reflection}\:{at}\:{smooth} \\ $$$${elastic}\:{surface} \\ $$$${m}={y}'\left({d}\right)=\frac{{bd}}{{a}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{{d}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}} \\ $$$${unit}\:{normal}\:{vector}\:{is} \\ $$$${n}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}\left(−{m},\mathrm{1}\right). \\ $$$${for}\:{a}\:{perfectly}\:{elastic}\:{collision} \\ $$$${V}_{\mathrm{1}} ^{+} ={V}_{\mathrm{1}} ^{−} −\mathrm{2}\left({V}_{\mathrm{1}} ^{−} .{n}\right){n}. \\ $$$${because}\:{V}_{\mathrm{1}} ^{−} =\left(\mathrm{0},−{v}\right)\:{and}\:{n}\:{has}\:{both} \\ $$$${x}\:{and}\:{y}\:{components}\:,{the}\:{reflected} \\ $$$${velocity}\:{becomes} \\ $$$${V}_{\mathrm{1}} ^{+} =\left({v}_{{x}} ,{v}_{{y}} \right)\:\:\:{v}_{{x}} \neq\mathrm{0} \\ $$$${thus}\:{the}\:{ball}\:{accuires}\:{horizontal} \\ $$$${velocity}\:{immediately} \\ $$$${after}\:{the}\:{first}\:{bounce} \\ $$$${Motion}\:{after}\:{first}\:{bounce} \\ $$$${between}\:{bounces}\:{the}\:{motion}\:{is}\:{projectile} \\ $$$${x}\left({t}\right)={x}_{\mathrm{0}\:} +{v}_{{x}} {t}\:\:\:\:{y}\left({t}\right)={y}_{\mathrm{0}} +{v}_{{y}} {t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${at}\:{each}\:{point}\:{P}_{{k}} \:{the}\:{velocity} \\ $$$${reflects}\:{agin} \\ $$$${V}_{{k}\:\:} +\mathrm{1}^{+} ={V}_{{k}} ^{−} −\mathrm{2}\left({V}_{{k}} ^{−} .{n}\left({p}_{{k}} \right){n}\left({p}_{{k}} \right).\right. \\ $$$${This}\:{defines}\:{a}\:{discrete}\:{dynamical} \\ $$$${map} \\ $$$${F}:{P}_{{k}} \rightarrow{P}_{{k}+\mathrm{1}} \\ $$$${It}\:{is}\:{an}\:{area}\:{preserving}\:,{energy} \\ $$$${preserving}\:{reflection}\:{map}\:{on}\:{the} \\ $$$${curve} \\ $$$${Now}\:{total}\:{mechanical}\:{energy}\:{contraint} \\ $$$${E}=\frac{\mathrm{1}}{\mathrm{2}}{m}\left({v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} \right)+{mgy} \\ $$$${initially}\: \\ $$$${E}={mgh} \\ $$$${thus},{at}\:{any}\:\:{later}\:{time} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{m}\left({v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} \right)+{mgy}={mgh} \\ $$$${therefore} \\ $$$${y}={h}\:\Rightarrow{v}_{{x}\:\:} ={v}_{{y}} =\mathrm{0} \\ $$$${The}\:{ball}\:{can}\:{reach}\:{height}\:{h}\:{if}\:{and} \\ $$$${only}\:{if}\:{it}\:{has}\:{zero}\:{velocity}\:{at}\:{that} \\ $$$${moment} \\ $$$${suppose}\:{ball}\:{returns}\:{to}\:\left({d},{h}\right){at}\: \\ $$$${some}\:{time}\:{T}>\mathrm{0} \\ $$$${x}\left({T}\right)={d},{y}\left({T}\right)={h},\:\overset{.} {{x}}\left({T}\right)=\mathrm{0} \\ $$$$\overset{.} {{y}}\left({T}\right)=\mathrm{0} \\ $$$${This}\:{is}\:{complete}\:{specification}\:{of} \\ $$$${mechanical}\:{state}. \\ $$$${Thus}\:{the}\:{state}\:{at}\:{time}\:{T}\:{is}\:{identical} \\ $$$${to}\:{the}\:{state}\:{at}\:{time}\:\mathrm{0} \\ $$$${hence}\:{the}\:{trajectory}\:{must}\:{be}\:{strictly} \\ $$$${periodic} \\ $$$${vertical}\:{descent}\:{only}\:\left({v}_{{x}} =\mathrm{0}\right) \\ $$$${and}\:{must}\:{reach}\:{exactly}\:{the}\:{same} \\ $$$${point}\:{and}\:{reverse}\:{the}\:{motion} \\ $$$${however}\:{after}\:{the}\:{first}\:{impact} \\ $$$${the}\:{velocity}\:{has}\:{a}\:{non}\:{zero}\:{horizontal} \\ $$$${component}. \\ $$$${v}_{{x}} ^{+} =−\mathrm{2}\left({V}_{\mathrm{1}} ^{−} .{n}\right)\frac{−{m}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}\:\neq\mathrm{0}\:{for}\:{d}\neq\mathrm{0} \\ $$$${since}\:{n}\:{has}\:{an}\:{x}\:{component}\:{whenever} \\ $$$${d}\neq\mathrm{0}\:{therefore} \\ $$$${The}\:{system}\:{can}\:{never}\:{return}\:{to} \\ $$$${a}\:{state}\:{with}\:{v}_{{x}} =\mathrm{0}. \\ $$$${it}\:{can}\:{never}\:{return}\:{to}\:\left({d},{h}\right){with}\:{zero} \\ $$$${velocity} \\ $$$${no}\:{periodic}\:{orbit}\:{exists}\:{for}\:{d}\neq\mathrm{0} \\ $$$${so}\:{for}\:{d}\neq\mathrm{0}\:{the}\:{ball}\:{never}\:{returns} \\ $$$${to}\:\left({d},{h}\right) \\ $$$${The}\:{only}\:{special}\:{case} \\ $$$${d}=\mathrm{0}\:\:\:{m}={y}^{'} \left(\mathrm{0}\right)=\mathrm{0} \\ $$$${so}\:{the}\:{surface}\:{is}\:{horizontal}\:{at} \\ $$$${vertex}\:{and}\:{the}\:{vertical}\:{motion} \\ $$$${symmetric} \\ $$$${v}_{{x}} =\mathrm{0}\forall{t}. \\ $$$$ \\ $$
Commented by BHOOPENDRA last updated on 15/Nov/25
$${After}\:{first}\:{bounce}\:{the}\:{ball} \\ $$$${has}\:{non}\:{zero}\:{horizontal}\:{speed} \\ $$$${there}\:{is}\:{no}\:{force}\:{to}\:{remove}\:{sideways} \\ $$$${component}\left({no}\:{friction}\right) \\ $$$${v}_{{x}} \:{never}\:{become}\:{identically}\:{zero} \\ $$$${again}\:{at}\:{right}\:{height}\:{and}\:{place} \\ $$$${therefore}\:{the}\:{ball}\:{never}\:{returns}\:{at} \\ $$$${rest}\:{at}\left({d},{h}\right) \\ $$$${The}\:{only}\:{periodic}\:{case}\:{is}\:{d}=\mathrm{0} \\ $$
Commented by mr W last updated on 16/Nov/25
$${thanks}\:{sir}! \\ $$$${at}\:{least}\:{there}\:{exist}\:{a}\:{point}\:{for}\:{any} \\ $$$${given}\:{d}.\:{when}\:{the}\:{ball}\:{is}\:{dropped} \\ $$$${from}\:{here},\:{it}\:{will}\:{return}\:{back}\:{to} \\ $$$${this}\:{point}. \\ $$$${see}\:{Q}\mathrm{225941} \\ $$