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Question Number 28709 by Tinkutara last updated on 29/Jan/18

Answered by ajfour last updated on 29/Jan/18

$$letssay{v}_0^{\prime }=u>v_0\left(justtoimagine\right)$$$$Asblockclimbsupalong$$$$verticalsurfaceofsledge,$$$$afterthebendbothsledgeand$$$$blockpossessthesamehorizontal$$$$componentofvelocitythereafter,$$$$say{v}_x.$$$$\Rightarrow m(u-v_0)=2{mv}_x$$$$or{v}_x=\frac{u-v_0}{2}....\left(i\right)$$$$let{v}_{y}be{block}^{\prime }sverticalcomponent$$$$ofvelocityattheinstantofescape,$$$$then$$$$\frac{1}{2}m(u^2+v_0^2)=(\frac{1}{2}{mv}_x^2+\frac{1}{2}{mv}_y^2)$$$$+mgh+\frac{1}{2}{mv}_x^2$$$$\Rightarrow u^2+v_0^2=2v_x^2+2gh+v_y^2$$$$u^2+v_0^2=\frac{{(u-v_0)}^2}{2}+2gh+v_y^2$$$$\Rightarrow {\mathit{v}}_{\mathit{y}}^2=\frac{1}{2}({\mathit{u}}^2+{\mathit{v}}_0^2)+{\mathit{u}\mathit{v}}_0-2\mathit{g}\mathit{h}$$$${\mathit{v}}_{\mathit{y}}^2=\frac{{(\mathit{u}+{\mathit{v}}_0)}^2}{2}-2\mathit{g}\mathit{h}...\left(ii\right)$$$$Andforcriticalvalueofu=v_0^{\prime }$$$$justnecessarytoescape$$$$weset{v}_y=0$$$$Then,(\mathit{u}+{\mathit{v}}_0)=2\sqrt{\mathit{g}\mathit{h}}$$$$\Rightarrow u_{critical}=2\sqrt{gh}-v_0$$$${\overline{\mathit{v}}}_{escape}={\mathit{v}}_{\mathit{x}}\hat{\mathit{i}}+{\mathit{v}}_{\mathit{y}}\hat{\mathit{j}};from\left(i\right),\left(ii\right)$$$$=\left(\frac{u-v_0}{2}\right)\hat{i}+\left(\sqrt{\frac{{(u+v_0)}^2}{2}-2gh}\right)\hat{j}$$$$speedofblockwhenitescapes:$$$$\mid v_{escape}\mid ={(\frac{3u^2}{4}+\frac{3v_0^2}{4}+\frac{3{uv}_0}{2}-2gh)}^{1/2}$$$$\mathit{Y}esitwilllandbackagainon$$$$sledge,sincehorizontalvelocity$$$$ofneitherblocknorsledgechanges$$$$whiletheblockisinairand$$$$initially\left(attheinstantofescape\right)$$$$theywerethesame.$$

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

$$Thisisanotheranswer:$$$$Speedofblockwhenitescapesis$$$$=\sqrt{\frac{5v_1}{4}+\frac{5v_0^2}{4}+\frac{3v_0{v}_1}{2}-4gh}$$$$where{v}_1=v_0^{{}^{\prime }}$$$$Forblocktoescape{v}_0^{{}^{\prime }}=2\sqrt{gh}-v_0$$

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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