Question-223741

Question Number 223741 by mr W last updated on 04/Aug/25

Commented by mr W last updated on 03/Aug/25

$${as}\:{Q}\mathrm{223720},\:{but}\:{the}\:{question}\:{is}: \\ $$$$\underline{{after}\:{what}\:{time}}\:{will}\:{the}\:{mass}\:{M}\: \\ $$$${hit}\:{the}\:{wall}? \\ $$

Answered by fantastic last updated on 03/Aug/25

$${Im}\:{writing}\:{what}\:{is}\:{written}\:{in}\:{the}\:{book} \\ $$$$\:\: \\ $$$${After}\:{releasing}\:{mass}\:{m} \\ $$$$\:{the}\:{mass}\:{M}\:{will}\:{come}\:{down} \\ $$$${and}\:{will}\:{hit}\:{the}\:{ground} \\ $$$$\:{with}\:{the}\:{speed}\:{of}\:{v}\left({let}\right) \\ $$$${now}\:{right}\:{before}\:{hitting}\:{the}\:{wall} \\ $$$$\:{the}\:{direction}\: \\ $$$${of}\:{the}\:{speed}\:\:{of}\:{M}\:{will}\:{be}\:{horizontal}. \\ $$$${that}\:{time}\:{rope}\:{from}\:{M}\:{to}\:{the}\:{pully} \\ $$$$\:{makes}\:{angle}\:\theta\left({let}\right) \\ $$$$\therefore\:{that}\:{time}\:{speed}\:{of}\:{m}={v}\mathrm{cos}\:\theta \\ $$$${the}\:\:{mass}\:{m}\:{will}\:{go}\:{upwards}\:{h}\left({let}\right) \\ $$$${here}\:{E}_{{p}} \:{of}\:{M}\:{will}\:{decrease}\: \\ $$$${but}\:{E}_{{k}} \:{will}\:{increase}. \\ $$$${The}\:{E}_{{p}} \:{and}\:{E}_{{k}} \:{of}\:{m}\:{will}\:{both}\:{increase} \\ $$$${lets}\:\:{assume}\:{that}\:\: \\ $$$${the}\:{length}\:{of}\:{rope}\:{from}\:{pully}\:{to}\:{mass}\:{m}\:{is}\:{x} \\ $$$${So}\:\mathrm{2}+{x}=\mathrm{1}+\sqrt{\mathrm{5}}+{x}−{h} \\ $$$${or}\:{h}=\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){m} \\ $$$${decrease}\:{of}\:{E}_{{p}} \:{of}\:{mass}\:{M}=\mathrm{2}{g}×\mathrm{1}=\mathrm{2}{g} \\ $$$${increase}\:{of}\:{E}_{{k}} \:{of}\:{mass}\:{M}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×{v}^{\mathrm{2}} ={v}^{\mathrm{2}} \\ $$$${increase}\:{of}\:{E}_{{k}} \:{of}\:{mass}\:{m} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{0}.\mathrm{5}×{v}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta=\frac{{v}^{\mathrm{2}} }{\mathrm{5}}\left[\because\mathrm{cos}\:\theta=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right] \\ $$$${increase}\:{of}\:{E}_{{p}} \:{of}\:\:{mass}\:{m}=\mathrm{0}.\mathrm{5}×{g}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right) \\ $$$${from}\:{conservation}\:{of}\:{energy} \\ $$$$\mathrm{2}{g}={v}^{\mathrm{2}} +\mathrm{0}.\mathrm{5}×{g}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)+\frac{{v}^{\mathrm{2}} }{\mathrm{5}} \\ $$$${or}\:\mathrm{2}×\mathrm{9}.\mathrm{8}=\frac{\mathrm{6}}{\mathrm{5}}{v}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)×\mathrm{9}.\mathrm{8} \\ $$$${or}\:\:\frac{\mathrm{6}}{\mathrm{5}}{v}^{\mathrm{2}} =\mathrm{9}.\mathrm{8}\left[\mathrm{2}−\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}}\right] \\ $$$${v}^{\mathrm{2}} =\frac{\mathrm{9}.\mathrm{8}×\mathrm{5}}{\mathrm{6}}\left[\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right] \\ $$$${So}\:{v}\approx\mathrm{11}.\mathrm{3}\:{m}/{s} \\ $$

Commented by mr W last updated on 03/Aug/25

$${this}\:{is}\:{correct}.\:{that}\:{means} \\ $$$${v}=\sqrt{\frac{\mathrm{9}.\mathrm{8}×\mathrm{5}×\left(\mathrm{5}−\sqrt{\mathrm{5}}\right)}{\mathrm{12}}}\approx\mathrm{3}.\mathrm{36}\:{m}/{s} \\ $$$${it}'{s}\:{what}\:{i}\:{also}\:{got}. \\ $$

Commented by fantastic last updated on 03/Aug/25

$${you}\:{want}\:{to}\:{see}\:{what}\:{they}\:{did}? \\ $$

Commented by fantastic last updated on 03/Aug/25

Commented by fantastic last updated on 03/Aug/25

$$:\left(\right. \\ $$

Commented by mr W last updated on 04/Aug/25

$${i}\:{see}.\: \\ $$$${they}\:{mistook}\:{v}^{\mathrm{2}} \:{as}\:{answer}\:{for}\:{v}. \\ $$$${i}\:{have}\:{changed}\:{the}\:{question}\:{to}\:{ask} \\ $$$${the}\:{time}\:{which}\:{the}\:{mass}\:{M}\:{takes} \\ $$$${to}\:{hit}\:{the}\:{wall}.\:{if}\:{you}\:{like},\:{give}\:{a}\:{try}! \\ $$

Commented by fantastic last updated on 03/Aug/25

$${sorry}.\:{I}\:{have}\:{exams}\:{after}\:\mathrm{5}\:{or}\:\mathrm{6} \\ $$$${days}.\:{I}\:{will}\:{be}\:{a}\:{lot}\:{more}\:{inactive} \\ $$$${in}\:{this}\:{app}.{I}\:{have}\:{to}\:{focus} \\ $$$${in}\:{my}\:{studies}\:{these}\:{days}. \\ $$$${however}\:{i}\:{will}\:{give}\:{it}\:{a}\:{try}. \\ $$$$ \\ $$

Commented by fantastic last updated on 03/Aug/25

$${I}\:{request}\:{you}\:{whenever}\:{i} \\ $$$${give}\:{any}\:{physics}\:{question} \\ $$$${please}\:{explain}\:{it}\:{in}\:{breif}. \\ $$$${Otherwise}\:{my}\:{microscopic} \\ $$$${brain}\:{will}\:{not}\:{understand} \\ $$$$ \\ $$

Commented by fantastic last updated on 03/Aug/25

$${Sir}\:{can}\:{you}\:{please}\:{expplain}\:{me}\:{how} \\ $$$${velocity}\:{of}\:{m}={v}\mathrm{cos}\:\theta \\ $$$${I}\:{understand}\:{other}\:{things}\:{well} \\ $$

Commented by mr W last updated on 03/Aug/25

$${in}\:{basic}\:{physics}\:{we}\:{assume}\:{that} \\ $$$${such}\:{a}\:{string}\:\left({rope}\right)\:{is}\:{not}\:{stretchable}. \\ $$$${that}\:{means}\:{in}\:{axial}\:{direction}\:{of} \\ $$$${the}\:{string},\:{any}\:{two}\:{points}\:{must}\: \\ $$$${have}\:{the}\:{same}\:{displacement},\:{thus} \\ $$$${also}\:{the}\:{same}\:{velocity}\:{and}\:{the}\:{same} \\ $$$${acceleration}.\:{in}\:{following}\:{picture}: \\ $$$${d}_{\mathrm{1}} \:\mathrm{cos}\:\alpha_{\mathrm{1}} ={d}_{\mathrm{2}} \:\mathrm{cos}\:\alpha_{\mathrm{2}} \\ $$$${v}_{\mathrm{1}} \:\mathrm{cos}\:\alpha_{\mathrm{1}} ={v}_{\mathrm{2}} \:\mathrm{cos}\:\alpha_{\mathrm{2}} \\ $$$${a}_{\mathrm{1}} \:\mathrm{cos}\:\alpha_{\mathrm{1}} ={a}_{\mathrm{2}} \:\mathrm{cos}\:\alpha_{\mathrm{2}} \\ $$

Commented by mr W last updated on 03/Aug/25

Commented by mr W last updated on 03/Aug/25

Commented by mr W last updated on 03/Aug/25

$${in}\:{current}\:{case}\:{that}\:{means}: \\ $$$$\mathrm{v}_{\mathrm{1}} \mathrm{cos}\:\theta\:=\mathrm{v}_{\mathrm{2}} =\mathrm{v}_{\mathrm{3}} \\ $$

Commented by fantastic last updated on 03/Aug/25

$${thanks}\:{sir}. \\ $$

Commented by fantastic last updated on 03/Aug/25

$${Wait}.{Sir}\:{do}\:{you}\:{understand} \\ $$$${the}\:{language}\:{after}\:\sqrt{\mathrm{5}}\:,\mathrm{2},\mathrm{1}?? \\ $$

Commented by mr W last updated on 03/Aug/25

$${no},\:{i}\:{don}'{t}\:{understand}\:{Bengali}.\:{why}? \\ $$

Commented by fantastic last updated on 04/Aug/25

$${I}\:{was}\:{curious} \\ $$

Answered by mr W last updated on 04/Aug/25

Commented by mr W last updated on 06/Aug/25

$${u}=\frac{{d}\phi}{{dt}} \\ $$$$\mathrm{tan}\:\varphi=\frac{\mathrm{sin}\:\phi}{\mathrm{2}−\mathrm{cos}\:\phi} \\ $$$${x}=\frac{\mathrm{sin}\:\phi}{\mathrm{sin}\:\varphi}−\mathrm{1} \\ $$$${v}={u}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\phi−\varphi\right)={u}\:\mathrm{sin}\:\left(\phi+\varphi\right) \\ $$$$\:\:\:={u}\:\left(\mathrm{sin}\:\phi\:\mathrm{cos}\:\varphi+\mathrm{cos}\:\phi\:\mathrm{sin}\:\varphi\right) \\ $$$$\:\:\:=\frac{{u}\left[\mathrm{sin}\:\phi\:\left(\mathrm{2}−\mathrm{cos}\:\phi\right)+\mathrm{cos}\:\phi\:\mathrm{sin}\:\phi\right]}{\:\sqrt{\mathrm{sin}^{\mathrm{2}} \:\phi+\left(\mathrm{2}−\mathrm{cos}\:\phi\right)^{\mathrm{2}} }} \\ $$$$\:\:\:=\frac{\mathrm{2}{u}\:\mathrm{sin}\:\phi}{\:\sqrt{\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\phi}} \\ $$$$\frac{{Mu}^{\mathrm{2}} }{\mathrm{2}}+\frac{{mv}^{\mathrm{2}} }{\mathrm{2}}={Mg}\:\mathrm{sin}\:\phi−{mg}\:{x} \\ $$$$\frac{\mathrm{2}{u}^{\mathrm{2}} \left(\mathrm{6}−\mathrm{4}\:\mathrm{cos}\:\phi−\mathrm{cos}^{\mathrm{2}} \:\phi\right)}{\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\phi}={g}\left(\mathrm{1}+\mathrm{4}\:\mathrm{sin}\:\phi−\sqrt{\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\phi}\right) \\ $$$${u}^{\mathrm{2}} =\frac{{g}\left(\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\phi\right)\left(\mathrm{1}+\mathrm{4}\:\mathrm{sin}\:\phi−\sqrt{\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\phi}\right)}{\mathrm{2}\left(\mathrm{6}−\mathrm{4}\:\mathrm{cos}\:\phi−\mathrm{cos}^{\mathrm{2}} \:\phi\right)} \\ $$$$\frac{{d}\phi}{{dt}}={u}=\sqrt{\frac{{g}\left(\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\phi\right)\left(\mathrm{1}+\mathrm{4}\:\mathrm{sin}\:\phi−\sqrt{\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\phi}\right)}{\mathrm{2}\left(\mathrm{6}−\mathrm{4}\:\mathrm{cos}\:\phi−\mathrm{cos}^{\mathrm{2}} \:\phi\right)}} \\ $$$${t}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\frac{\mathrm{2}\left(\mathrm{6}−\mathrm{4}\:\mathrm{cos}\:\phi−\mathrm{cos}^{\mathrm{2}} \:\phi\right)}{{g}\left(\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\phi\right)\left(\mathrm{1}+\mathrm{4}\:\mathrm{sin}\:\phi−\sqrt{\mathrm{5}−\mathrm{4}\:\mathrm{cos}\:\phi}\right)}}\:{d}\phi \\ $$$$\:\:\:=\mathrm{0}.\mathrm{67092204}\:{s} \\ $$$$\:\:\:>\:\mathrm{0}.\mathrm{592}\:{s}\:\left({see}\:{below}\right)\: \\ $$$${this}\:{is}\:{reasonable},\:{since}\:{the}\:{mass}\:{m} \\ $$$${brakes}\:{the}\:{mass}\:{M}. \\ $$

Commented by fantastic last updated on 05/Aug/25

$${wow}\:{sir} \\ $$

Commented by mr W last updated on 06/Aug/25

Commented by mr W last updated on 06/Aug/25

$${u}={R}\frac{{d}\phi}{{dt}} \\ $$$$\frac{{mu}^{\mathrm{2}} }{\mathrm{2}}={mgR}\mathrm{sin}\:\phi \\ $$$${u}={R}\frac{{d}\phi}{{dt}}=\sqrt{\mathrm{2}{gR}\:\mathrm{sin}\:\phi} \\ $$$$\frac{{d}\phi}{{dt}}=\sqrt{\frac{\mathrm{2}{g}}{{R}}\:\mathrm{sin}\:\phi} \\ $$$${t}_{\mathrm{1}} =\sqrt{\frac{{R}}{{g}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\phi}{\:\sqrt{\mathrm{2}\:\mathrm{sin}\:\phi}}\approx\mathrm{1}.\mathrm{854075}\sqrt{\frac{{R}}{{g}}} \\ $$$${with}\:{R}=\mathrm{1}: \\ $$$${t}_{\mathrm{1}} \approx\mathrm{0}.\mathrm{592}\:{s} \\ $$

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