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Question Number 19362 by Tinkutara last updated on 10/Aug/17

$$\mathrm{The}\mathrm{block}Q\mathrm{moves}\mathrm{to}\mathrm{the}\mathrm{right}\mathrm{with}\mathrm{a}$$$$\mathrm{constant}\mathrm{velocity}{v}_0\mathrm{as}\mathrm{shown}\mathrm{in}\mathrm{figure}.$$$$\mathrm{The}\mathrm{relative}\mathrm{velocity}\mathrm{of}\mathrm{body}P\mathrm{with}$$$$\mathrm{respect}\mathrm{to}Q\mathrm{is}(\mathrm{assume}\mathrm{all}\mathrm{pulleys}\mathrm{and}$$$$\mathrm{strings}\mathrm{are}\mathrm{ideal})$$

Commented by Tinkutara last updated on 10/Aug/17

Commented by ajfour last updated on 10/Aug/17

$${2\mathrm{x}}_{\mathrm{P}}={3\mathrm{x}}_{\mathrm{Q}}$$$$\Rightarrow {2\mathrm{v}}_{\mathrm{P}}={3\mathrm{v}}_{\mathrm{Q}}={3\mathrm{v}}_0$$$${\mathrm{v}}_{\mathrm{P}}=\frac{3}{2}{\mathrm{v}}_0\Rightarrow {\mathrm{v}}_{\mathrm{P}}-{\mathrm{v}}_{\mathrm{Q}}=\frac{3}{2}{\mathrm{v}}_0-{\mathrm{v}}_0$$$${\mathrm{v}}_{\mathrm{rel}}=\frac{{\mathrm{v}}_0}{2}.$$

Commented by Tinkutara last updated on 11/Aug/17

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

Commented by ajfour last updated on 10/Aug/17

$$\mathrm{towards}\mathrm{Q}.$$

Commented by ajfour last updated on 11/Aug/17

$$\mathrm{let}+\mathrm{ve}\mathrm{direction}\mathrm{is}\mathrm{towards}\mathrm{right}.$$$${\mathrm{v}}_{\mathrm{Q}}={\mathrm{v}}_0,{\mathrm{v}}_{\mathrm{P}}=\frac{3}{2}{\mathrm{v}}_0$$$${\mathrm{v}}_{\mathrm{rel}}\mathrm{of}\mathrm{P}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{Q}\mathrm{is}={\mathrm{v}}_{\mathrm{P}}-{\mathrm{v}}_{\mathrm{Q}}$$$$=\frac{3}{2}{\mathrm{v}}_0-{\mathrm{v}}_0=\frac{{\mathrm{v}}_0}{2}(+\mathrm{ve})$$$$\mathrm{hence}\mathrm{towards}\mathrm{right}\mathrm{means}\mathrm{towards}\mathrm{Q}.$$

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