Question Number 20038 by Tinkutara last updated on 20/Aug/17
$$\mathrm{In}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{shown},\:{m}\:\mathrm{slides}\:\mathrm{on} \\ $$$$\mathrm{inclined}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{wedge}\:{M}.\:\mathrm{If}\:\mathrm{velocity} \\ $$$$\mathrm{of}\:\mathrm{wedge}\:\mathrm{at}\:\mathrm{any}\:\mathrm{instant}\:\mathrm{be}\:{v},\:\mathrm{find} \\ $$$$\mathrm{velocity}\:\mathrm{of}\:{m}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{ground}. \\ $$
Commented by Tinkutara last updated on 20/Aug/17
Answered by ajfour last updated on 21/Aug/17
$$\:{v}=−\frac{{dy}}{{dt}}\:\:\:,\:\:\:{u}_{{rel}} =\frac{{dx}}{{dt}} \\ $$$${x}+{y}={l}\:\:\:\:\:\:\left({is}\:{constant}\right) \\ $$$${so}\:\:\:\:\:\:\frac{{dx}}{{dt}}=−\frac{{dy}}{{dt}}\:\:\:\Rightarrow\:\:\:\:{u}_{{rel}} ={v} \\ $$$$\:\:{u}=\sqrt{{u}_{{x}} ^{\mathrm{2}} +{u}_{{y}} ^{\mathrm{2}} } \\ $$$$\:\:\:{u}_{{x}} ={v}−{u}_{{rel}} \mathrm{cos}\:\mathrm{60}°\:={v}−\frac{{v}}{\mathrm{2}}=\frac{{v}}{\mathrm{2}} \\ $$$$\:\:\:{u}_{{y}} ={v}_{{rel}} \mathrm{sin}\:\mathrm{60}°=\:\frac{{v}\sqrt{\mathrm{3}}}{\mathrm{2}}\: \\ $$$$\:{Hence}\:{block}'{s}\:{speed}\:{with}\:{respect} \\ $$$${to}\:{ground}\:\:{u}=\sqrt{\left(\frac{{v}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{v}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:={v}\:. \\ $$$$ \\ $$
Commented by Tinkutara last updated on 21/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$