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A string with one end fixed on a rigid wall, passing over a fixed frictionless pulley at a distance of 2 m from the wall, has a point mass M of 2 kg attached to it at a distance of 1 m from the wall. A mass in of 0.5 kg is attached to the free end. The system is initially held at rest so that the string is horizontal between wall and pulley and vertical beyond the pulley as shown in Fig. The system is released from the rest from the position as shown. The ratio of velocity of M and m when M strikes the wall is

Answer» <html><body><p>`(sqrt(5))/2`<br/>`2/sqrt(5)`<br/>`3/sqrt(5)`<br/>`(sqrt(5))/3`</p>Solution :When `M` strikes the wall, vertically downward component of its displacement from initital position is `1 m` and its distance from pulley `B` is <br/> `C'B=sqrt(1^(2)+2^(2))=sqrt(5)m` <br/> When its <a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> distance from the pulley was `CB=1m`. It means vertically upward displacement of mass `m` is `(sqrt(5)-1)m`. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C01_E01_314_S01.png" width="80%"/> <br/> Let `M` strike the wall velocity `v`. Since the string between the two blocks always <a href="https://interviewquestions.tuteehub.com/tag/remains-621920" style="font-weight:bold;" target="_blank" title="Click to know more about REMAINS">REMAINS</a> taut, therefore at any instant speed of `m` is <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> to that component of velocity of `M` which which is along the string `C'B`. Hence , velocity of `m` when `M` strikes the wall is `vcostheta`, where<br/> `costheta=2/sqrt(5)` <br/> `:. (v_(M))/(v_(m))=v/(vcostheta)=(sqrt(5))/2` <br/>According to law of energy, <a href="https://interviewquestions.tuteehub.com/tag/loss-1079380" style="font-weight:bold;" target="_blank" title="Click to know more about LOSS">LOSS</a> of potential energy of `M=`increase in `<a href="https://interviewquestions.tuteehub.com/tag/pe-590719" style="font-weight:bold;" target="_blank" title="Click to know more about PE">PE</a>` of `m+KE` of `M+KE` of `m` <br/> `Mgxx1=mg(sqrt(5)-1)+1/2Mv^(2)+1/2m(vcostheta)^(2)` <br/> `v=5sqrt((5-sqrt(5))/6)m//s`</body></html>


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