Question Number 28709 by Tinkutara last updated on 29/Jan/18
Answered by ajfour last updated on 29/Jan/18
$${lets}\:{say}\:{v}_{\mathrm{0}} '\:={u}\:>\:{v}_{\mathrm{0}} \:\left({just}\:{to}\:{imagine}\right) \\ $$$${As}\:{block}\:{climbs}\:{up}\:{along} \\ $$$${vertical}\:{surface}\:{of}\:{sledge}, \\ $$$${after}\:{the}\:{bend}\:{both}\:{sledge}\:{and} \\ $$$${block}\:{possess}\:{the}\:{same}\:{horizontal} \\ $$$${component}\:{of}\:{velocity}\:{thereafter}, \\ $$$$\:{say}\:{v}_{{x}} .\: \\ $$$$\Rightarrow\:\:{m}\left({u}−{v}_{\mathrm{0}} \right)=\mathrm{2}{mv}_{{x}} \\ $$$${or}\:\:\:\:\:{v}_{{x}} =\frac{{u}−{v}_{\mathrm{0}} }{\mathrm{2}}\:\:\:\:\:\:\:\:....\left({i}\right) \\ $$$${let}\:{v}_{{y}} \:{be}\:{block}'{s}\:{vertical}\:{component} \\ $$$${of}\:{velocity}\:{at}\:{the}\:{instant}\:{of}\:{escape}, \\ $$$${then} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{m}\left({u}^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} \right)=\left(\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{x}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{y}} ^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{mgh}+\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{x}} ^{\mathrm{2}} \\ $$$$\Rightarrow\:{u}^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} =\mathrm{2}{v}_{{x}} ^{\mathrm{2}} +\mathrm{2}{gh}+{v}_{{y}} ^{\mathrm{2}} \:\:\:\:\: \\ $$$${u}^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} =\frac{\left({u}−{v}_{\mathrm{0}} \right)^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{gh}+{v}_{{y}} ^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\boldsymbol{{v}}_{\boldsymbol{{y}}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{{u}}^{\mathrm{2}} +\boldsymbol{{v}}_{\mathrm{0}} ^{\mathrm{2}} \right)+\boldsymbol{{uv}}_{\mathrm{0}} −\mathrm{2}\boldsymbol{{gh}} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{v}}_{\boldsymbol{{y}}} ^{\mathrm{2}} \:\:=\frac{\left(\boldsymbol{{u}}+\boldsymbol{{v}}_{\mathrm{0}} \right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\boldsymbol{{gh}}\:\:\:\:\:\:...\left({ii}\right) \\ $$$${And}\:{for}\:{critical}\:{value}\:{of}\:{u}={v}_{\mathrm{0}} ' \\ $$$${just}\:{necessary}\:{to}\:{escape}\: \\ $$$${we}\:{set}\:{v}_{{y}} =\mathrm{0} \\ $$$${Then},\:\:\:\left(\boldsymbol{{u}}+\boldsymbol{{v}}_{\mathrm{0}} \right)=\mathrm{2}\sqrt{\boldsymbol{{gh}}} \\ $$$$\Rightarrow\:\:\:\:\:{u}_{{critical}} \:=\mathrm{2}\sqrt{{gh}}−{v}_{\mathrm{0}} \\ $$$$\:\:\bar {\boldsymbol{{v}}}_{{escape}} \:=\boldsymbol{{v}}_{\boldsymbol{{x}}} \hat {\boldsymbol{{i}}}+\boldsymbol{{v}}_{\boldsymbol{{y}}} \hat {\boldsymbol{{j}}}\:\:\:;\:{from}\:\left({i}\right),\:\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\left(\frac{{u}−{v}_{\mathrm{0}} }{\mathrm{2}}\right)\hat {{i}}+\left(\sqrt{\frac{\left({u}+{v}_{\mathrm{0}} \right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{gh}}\:\right)\hat {{j}} \\ $$$${speed}\:{of}\:{block}\:{when}\:{it}\:{escapes}: \\ $$$$\:\mid{v}_{{escape}} \mid=\left(\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}{uv}_{\mathrm{0}} }{\mathrm{2}}−\mathrm{2}{gh}\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$\:\boldsymbol{{Y}}{es}\:{it}\:{will}\:{land}\:{back}\:{again}\:{on} \\ $$$${sledge},\:{since}\:{horizontal}\:{velocity} \\ $$$${of}\:{neither}\:{block}\:{nor}\:{sledge}\:{changes} \\ $$$${while}\:{the}\:{block}\:{is}\:{in}\:{air}\:{and} \\ $$$${initially}\:\left({at}\:{the}\:{instant}\:{of}\:{escape}\right) \\ $$$${they}\:{were}\:{the}\:{same}. \\ $$
Commented by Tinkutara last updated on 29/Jan/18
There are two different sets of answers given in book in different places. Please check which is correct.
Commented by Tinkutara last updated on 29/Jan/18
Commented by ajfour last updated on 29/Jan/18
please post.
Commented by Tinkutara last updated on 29/Jan/18
$${This}\:{is}\:{another}\:{answer}: \\ $$$${Speed}\:{of}\:{block}\:{when}\:{it}\:{escapes}\:{is} \\ $$$$=\sqrt{\frac{\mathrm{5}{v}_{\mathrm{1}} }{\mathrm{4}}+\frac{\mathrm{5}{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}{v}_{\mathrm{0}} {v}_{\mathrm{1}} }{\mathrm{2}}−\mathrm{4}{gh}} \\ $$$${where}\:{v}_{\mathrm{1}} ={v}_{\mathrm{0}} ^{'} \\ $$$${For}\:{block}\:{to}\:{escape}\:{v}_{\mathrm{0}} ^{'} =\mathrm{2}\sqrt{{gh}}−{v}_{\mathrm{0}} \\ $$
Commented by Tinkutara last updated on 29/Jan/18
Which of these answers are correct?
Commented by ajfour last updated on 29/Jan/18
to me my answer seems correct..