Question Number 136930 by I want to learn more last updated on 27/Mar/21
Commented by I want to learn more last updated on 27/Mar/21
Commented by Olaf last updated on 28/Mar/21
$$\left({i}\right)\:{W}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{Kx}^{\mathrm{2}} \\ $$$${K}\:=\:\frac{\mathrm{2}{W}}{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}×\mathrm{10}}{\mathrm{0},\mathrm{05}^{\mathrm{2}} }\:=\:\mathrm{8000}\:\mathrm{N}{m}^{−\mathrm{1}} \\ $$$$\left({ii}\right)\:{F}\:=\:{Kx}\:=\:\mathrm{8000}×\mathrm{0},\mathrm{05}\:=\:\mathrm{400}\:\mathrm{N} \\ $$$$\left({iii}\right)\:{F}\:=\:{ma}\:\Rightarrow\:{a}\:=\:\frac{{F}}{{m}}\:=\:\frac{\mathrm{400}}{\mathrm{2}}\:=\:\mathrm{200}\:{ms}^{−\mathrm{2}} \\ $$$$\left({iv}\right)\:{V}_{{P}} \:=\:{at}\:=\:\mathrm{200}×\mathrm{0},\mathrm{25}\:=\:\mathrm{50}\:{ms}^{−\mathrm{1}} \\ $$$${M}_{{P}} {V}_{{P}} \:+\:{M}_{{Q}} {V}_{{Q}} \:=\:\left({M}_{{P}} +{M}_{{Q}} \right){V} \\ $$$$\mathrm{2}×\mathrm{50}+\mathrm{4}×\mathrm{0}\:=\:\left(\mathrm{2}+\mathrm{4}\right){V} \\ $$$${V}\:=\:\mathrm{16},\mathrm{7}\:{ms}^{−\mathrm{1}} \\ $$
Commented by mr W last updated on 28/Mar/21
$${all}\:{is}\:{correct}\:{except}\:\left({iv}\right),\:{since}\:{the} \\ $$$${block}\:{P}\:{doesn}'{t}\:{move}\:{with}\:{constant} \\ $$$${acceleration}\:{after}\:{release}. \\ $$$${period}\:{T}=\mathrm{2}\pi\sqrt{\frac{{m}}{{k}}}=\mathrm{2}\pi\sqrt{\frac{\mathrm{2}}{\mathrm{8000}}}\approx\mathrm{0}.\mathrm{1}\:{s} \\ $$$${for}\:{the}\:{block}\:{P}\:{to}\:{return}\:{back}\:{to}\:{its} \\ $$$${initial}\:{position}\:{it}\:{needs}\:\frac{{T}}{\mathrm{4}}\approx\mathrm{0}.\mathrm{025}\:{s}. \\ $$$${at}\:{this}\:{position}\:{the}\:{spring}\:{is}\: \\ $$$${completely}\:{relaxed}\:{and}\:{losses}\: \\ $$$${connection}\:{with}\:{the}\:{block}\:{P}\:{which}\:{then} \\ $$$${moves}\:{with}\:{a}\:{constant}\:{velocity}\:{v}_{{P}} : \\ $$$${v}_{{P}} =\sqrt{\frac{\mathrm{2}{W}}{{m}}}=\sqrt{\frac{\mathrm{2}×\mathrm{10}}{\mathrm{2}}}=\sqrt{\mathrm{10}}\:{m}/{s} \\ $$$${also}\:{at}\:{time}\:{t}=\mathrm{0}.\mathrm{25}\:{s},\:{the}\:{block}\:{P}\:{has} \\ $$$${this}\:{velocity}\:{till}\:{it}\:{collides}\:{with}\:{block} \\ $$$${Q}.\:{the}\:{velocity}\:{after}\:{collision}\:{is} \\ $$$${v}_{{P}+{Q}} =\frac{{M}_{{P}} {v}_{{P}} }{{M}_{{P}} +{M}_{{Q}} }=\frac{\mathrm{2}×\sqrt{\mathrm{10}}}{\mathrm{2}+\mathrm{4}}=\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}\approx\mathrm{1}.\mathrm{05}\:{m}/{s} \\ $$
Commented by I want to learn more last updated on 28/Mar/21
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sirs}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Commented by Olaf last updated on 28/Mar/21
$$\mathrm{I}\:\mathrm{uninstall}\:\mathrm{this}\:\mathrm{app}. \\ $$$$\mathrm{Good}\:\mathrm{continuation}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 14/Sep/21
$$\mathrm{nice} \\ $$