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Question-28709

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Question Number 28709 by Tinkutara last updated on 29/Jan/18
Answered by ajfour last updated on 29/Jan/18
lets say v_0 ′ =u > v_0 (just to imagine) 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 . ⇒ m(u−v_0 )=2mv_x or v_x =((u−v_0 )/2) ....(i) let v_y be block′s vertical component of velocity at the instant of escape, then (1/2)m(u^2 +v_0 ^2 )=((1/2)mv_x ^2 +(1/2)mv_y ^2 ) +mgh+(1/2)mv_x ^2 ⇒ u^2 +v_0 ^2 =2v_x ^2 +2gh+v_y ^2 u^2 +v_0 ^2 =(((u−v_0 )^2 )/2)+2gh+v_y ^2 ⇒ v_y ^2 =(1/2)(u^2 +v_0 ^2 )+uv_0 −2gh v_y ^2 =(((u+v_0 )^2 )/2)−2gh ...(ii) And for critical value of u=v_0 ′ just necessary to escape we set v_y =0 Then, (u+v_0 )=2(√(gh)) ⇒ u_(critical) =2(√(gh))−v_0 v_(escape) ^� =v_x i^� +v_y j^� ; from (i), (ii) = (((u−v_0 )/2))i^� +((√((((u+v_0 )^2 )/2)−2gh)) )j^� speed of block when it escapes: ∣v_(escape) ∣=(((3u^2 )/4)+((3v_0 ^2 )/4)+((3uv_0 )/2)−2gh)^(1/2) Yes it will land back again on sledge, since horizontal velocity of neither block nor sledge changes while the block is in air and initially (at the instant of escape) they were the same.
$${lets}\:{say}\:{v}_{\mathrm{0}} '\:={u}\:>\:{v}_{\mathrm{0}} \:\left({just}\:{to}\:{imagine}\right) \\ $$$${As}\:{block}\:{climbs}\:{up}\:{along} \\ $$$${vertical}\:{surface}\:{of}\:{sledge}, \\ $$$${after}\:{the}\:{bend}\:{both}\:{sledge}\:{and} \\ $$$${block}\:{possess}\:{the}\:{same}\:{horizontal} \\ $$$${component}\:{of}\:{velocity}\:{thereafter}, \\ $$$$\:{say}\:{v}_{{x}} .\: \\ $$$$\Rightarrow\:\:{m}\left({u}−{v}_{\mathrm{0}} \right)=\mathrm{2}{mv}_{{x}} \\ $$$${or}\:\:\:\:\:{v}_{{x}} =\frac{{u}−{v}_{\mathrm{0}} }{\mathrm{2}}\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$${let}\:{v}_{{y}} \:{be}\:{block}'{s}\:{vertical}\:{component} \\ $$$${of}\:{velocity}\:{at}\:{the}\:{instant}\:{of}\:{escape}, \\ $$$${then} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{m}\left({u}^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} \right)=\left(\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{x}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{y}} ^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{mgh}+\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{x}} ^{\mathrm{2}} \\ $$$$\Rightarrow\:{u}^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} =\mathrm{2}{v}_{{x}} ^{\mathrm{2}} +\mathrm{2}{gh}+{v}_{{y}} ^{\mathrm{2}} \:\:\:\:\: \\ $$$${u}^{\mathrm{2}} +{v}_{\mathrm{0}} ^{\mathrm{2}} =\frac{\left({u}−{v}_{\mathrm{0}} \right)^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{gh}+{v}_{{y}} ^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\boldsymbol{{v}}_{\boldsymbol{{y}}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{{u}}^{\mathrm{2}} +\boldsymbol{{v}}_{\mathrm{0}} ^{\mathrm{2}} \right)+\boldsymbol{{uv}}_{\mathrm{0}} −\mathrm{2}\boldsymbol{{gh}} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{v}}_{\boldsymbol{{y}}} ^{\mathrm{2}} \:\:=\frac{\left(\boldsymbol{{u}}+\boldsymbol{{v}}_{\mathrm{0}} \right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\boldsymbol{{gh}}\:\:\:\:\:\:…\left({ii}\right) \\ $$$${And}\:{for}\:{critical}\:{value}\:{of}\:{u}={v}_{\mathrm{0}} ' \\ $$$${just}\:{necessary}\:{to}\:{escape}\: \\ $$$${we}\:{set}\:{v}_{{y}} =\mathrm{0} \\ $$$${Then},\:\:\:\left(\boldsymbol{{u}}+\boldsymbol{{v}}_{\mathrm{0}} \right)=\mathrm{2}\sqrt{\boldsymbol{{gh}}} \\ $$$$\Rightarrow\:\:\:\:\:{u}_{{critical}} \:=\mathrm{2}\sqrt{{gh}}−{v}_{\mathrm{0}} \\ $$$$\:\:\bar {\boldsymbol{{v}}}_{{escape}} \:=\boldsymbol{{v}}_{\boldsymbol{{x}}} \hat {\boldsymbol{{i}}}+\boldsymbol{{v}}_{\boldsymbol{{y}}} \hat {\boldsymbol{{j}}}\:\:\:;\:{from}\:\left({i}\right),\:\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\left(\frac{{u}−{v}_{\mathrm{0}} }{\mathrm{2}}\right)\hat {{i}}+\left(\sqrt{\frac{\left({u}+{v}_{\mathrm{0}} \right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{gh}}\:\right)\hat {{j}} \\ $$$${speed}\:{of}\:{block}\:{when}\:{it}\:{escapes}: \\ $$$$\:\mid{v}_{{escape}} \mid=\left(\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}{uv}_{\mathrm{0}} }{\mathrm{2}}−\mathrm{2}{gh}\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$\:\boldsymbol{{Y}}{es}\:{it}\:{will}\:{land}\:{back}\:{again}\:{on} \\ $$$${sledge},\:{since}\:{horizontal}\:{velocity} \\ $$$${of}\:{neither}\:{block}\:{nor}\:{sledge}\:{changes} \\ $$$${while}\:{the}\:{block}\:{is}\:{in}\:{air}\:{and} \\ $$$${initially}\:\left({at}\:{the}\:{instant}\:{of}\:{escape}\right) \\ $$$${they}\:{were}\:{the}\:{same}. \\ $$
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 =(√(((5v_1 )/4)+((5v_0 ^2 )/4)+((3v_0 v_1 )/2)−4gh)) where v_1 =v_0 ^′ For block to escape v_0 ^′ =2(√(gh))−v_0
$${This}\:{is}\:{another}\:{answer}: \\ $$$${Speed}\:{of}\:{block}\:{when}\:{it}\:{escapes}\:{is} \\ $$$$=\sqrt{\frac{\mathrm{5}{v}_{\mathrm{1}} }{\mathrm{4}}+\frac{\mathrm{5}{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}{v}_{\mathrm{0}} {v}_{\mathrm{1}} }{\mathrm{2}}−\mathrm{4}{gh}} \\ $$$${where}\:{v}_{\mathrm{1}} ={v}_{\mathrm{0}} ^{'} \\ $$$${For}\:{block}\:{to}\:{escape}\:{v}_{\mathrm{0}} ^{'} =\mathrm{2}\sqrt{{gh}}−{v}_{\mathrm{0}} \\ $$
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..

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