A-particle-P-is-sliding-down-a-frictionless-hemispherical-bowl-It-passes-the-point-A-at-t-0-At-this-instant-of-time-the-horizontal-component-of-its-velocity-is-v-A-bead-Q-of-the-same-mass-as-P-i

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Question Number 19223 by Tinkutara last updated on 07/Aug/17

$$\mathrm{A}\mathrm{particle}P\mathrm{is}\mathrm{sliding}\mathrm{down}\mathrm{a}\mathrm{frictionless}$$$$\mathrm{hemispherical}\mathrm{bowl}.\mathrm{It}\mathrm{passes}\mathrm{the}\mathrm{point}$$$$A\mathrm{at}t=0.\mathrm{At}\mathrm{this}\mathrm{instant}\mathrm{of}\mathrm{time},\mathrm{the}$$$$\mathrm{horizontal}\mathrm{component}\mathrm{of}\mathrm{its}\mathrm{velocity}\mathrm{is}$$$$v.\mathrm{A}\mathrm{bead}Q\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{mass}\mathrm{as}P\mathrm{is}$$$$\mathrm{ejected}\mathrm{from}A\mathrm{at}t=0\mathrm{along}\mathrm{the}$$$$\mathrm{horizontal}\mathrm{direction},\mathrm{with}\mathrm{the}\mathrm{speed}v.$$$$\mathrm{Friction}\mathrm{between}\mathrm{the}\mathrm{bead}\mathrm{and}\mathrm{the}$$$$\mathrm{string}\mathrm{may}\mathrm{be}\mathrm{neglected}.\mathrm{Let}{t}_P\mathrm{and}{t}_Q$$$$\mathrm{be}\mathrm{the}\mathrm{respective}\mathrm{times}\mathrm{taken}\mathrm{by}P\mathrm{and}$$$$Q\mathrm{to}\mathrm{reach}\mathrm{the}\mathrm{point}B.\mathrm{Then}$$$$\left(a\right)t_P<t_Q$$$$\left(b\right)t_P=t_Q$$$$\left(c\right)t_P>t_Q$$$$\left(d\right)\frac{t_P}{t_Q}=\frac{\mathrm{length}\mathrm{of}\mathrm{at}\mathrm{arc}ACB}{\mathrm{length}\mathrm{of}\mathrm{chord}AB}$$

Commented by Tinkutara last updated on 07/Aug/17

Commented by Tinkutara last updated on 08/Aug/17

$$\mathrm{Help}.$$

Commented by ajfour last updated on 08/Aug/17

$$\int \frac{\mathrm{dx}}{\sqrt{\mathrm{a}+\cos \mathrm{x}}}=?$$$${\int }_0^{{\mathrm{t}}_{\mathrm{P}}}{\mathrm{dt}}_{\mathrm{P}}={\int }_{-\alpha }^{\alpha }\frac{\mathrm{Rd}\theta }{\sqrt{{\mathrm{v}}_0^2+2\mathrm{gR}(\cos \theta -\cos \alpha )}}$$

Commented by Tinkutara last updated on 08/Aug/17

$$\mathrm{This}\mathrm{integral}\mathrm{is}\mathrm{undefined}\mathrm{except}\mathrm{for}\mathrm{a}$$$$=1\mathrm{as}\mathrm{I}\mathrm{put}\mathrm{this}\mathrm{expression}\mathrm{in}\mathrm{Geogebra}$$$$\mathrm{by}\mathrm{taking}\mathrm{many}\mathrm{values}.$$

Commented by Tinkutara last updated on 08/Aug/17

$$\mathrm{So}{t}_P{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{be}\mathrm{evaluated}?$$

Commented by ajfour last updated on 08/Aug/17

$$\mathrm{I}\mathrm{cannot}.$$

Commented by ajfour last updated on 08/Aug/17

$$\mathrm{i}\mathrm{came}\mathrm{across}\mathrm{it}\mathrm{in}\mathrm{trying}\mathrm{to}$$$$\mathrm{evaluate}{\mathrm{t}}_{\mathrm{P}}.$$

Answered by ajfour last updated on 08/Aug/17

$$\left(\mathrm{a}\right){\mathrm{t}}_{\mathrm{P}}<{\mathrm{t}}_{\mathrm{Q}}$$

Commented by Tinkutara last updated on 08/Aug/17

$$\mathrm{Explain}.$$

Commented by ajfour last updated on 08/Aug/17

Commented by ajfour last updated on 08/Aug/17

$$\mathrm{v}>{\mathrm{v}}_0\mathrm{and}\cos \theta >\cos \alpha$$$${\mathrm{v}}_{\mathrm{x}}=\mathrm{vcos}\theta \mathrm{u}={\mathrm{v}}_0\cos \alpha$$$$\Rightarrow {\mathrm{v}}_{\mathrm{x}}>{\mathrm{v}}_0\cos \alpha (=\mathrm{u}).$$

Commented by Tinkutara last updated on 08/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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