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A particle of mass 1 kg is attached to a string of length 5 m. The string is attached to a fixed point O. It is released from theposition as shown in Fig. Calculate a. the impulse developed in the string when it becorries taut, b. the velocity of the particle just after the string becomes taut, c. the impulse developed in this string PQ at this instant.

Answer» <html><body><p></p>Solution :From `M` to `<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>` the ball will fall a distance o `4m`, then impulse will be developed in the string ON. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C01_E01_047_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of the ball just before the string <a href="https://interviewquestions.tuteehub.com/tag/becomes-1994370" style="font-weight:bold;" target="_blank" title="Click to know more about BECOMES">BECOMES</a> tight <br/>` u_(0)=sqrt(2xx10xx4)=sqrt(80)m//s` <br/> velocity of the ball in the <a href="https://interviewquestions.tuteehub.com/tag/direction-1696" style="font-weight:bold;" target="_blank" title="Click to know more about DIRECTION">DIRECTION</a> perpendicular to the impulse will remain same (just after and just before the impulse developed.) <br/> `v=u_(0)sintheta=sqrt(80)xx3/5=15/(sqrt(5))m//s` <br/> `J=mu_0costheta=1sqrt(80)xx4/5=16/(sqrt(5)) kgm//s` <br/> `J'=Jsintheta=4/(5sqrt(5))kgm//s` <br/> and impulse given by earth <br/> `J''=Jcostheta=64/(5sqrt(5))kgm//s`</body></html>


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