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A ball of mass m is pushed with a horizontal velocity v_(0) from one end of a sledge of mass M and length l. if the ball stops after is first collision with the sledge, find the speeds of the ball ad sledge after the second collision of the ball with the sledge.

Answer» <html><body><p></p>Solution :a. Using C.O.L.M, `Mv_(2)=mv_(0)` ………..i <br/> Using Newton's restitution equation <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C01_S01_073_S01.png" width="80%"/> <br/> `v_(2)-0=ev_(0)` <br/> Here `m/Mv_(0)=ev_(0)` <br/> which gives `e=m/M` ……….ii <br/> Now the 2nd collisiion left wall the sledge and ball C.O.L.M. <br/> `0+Mv_(2)=mv_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)+MV_(2)^(')` <br/> `M(m/Mv_(0))=mv_(1)+Mv_(2)^(')` <br/> `implies mv_(1)+Mv_(2)^(')=mv_(0)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C01_S01_073_S02.png" width="80%"/> <br/> Again using Newton's restitution <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> <br/> `v_(2)^(')-v_(1)=e(0-m/Mv_(0))` <br/> `v_(2)^(')-v_(1)=m/M(-m/Mv_(0))` <br/> `v_(1)-v_(2)^(')=(m/M)v_(0)`..........<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a> <br/> From eqn iii and iv we get <br/> `v_(1)=(2v_(0))/((M+m))` and `v_(2)^(')=(mv_(0))/M((M-m)/(M+m))`</body></html>


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