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A ball moving with a velocity of 12ms^(-1)collideswith another identical stationary ball . After the collision they move as shown in figure . Find the speed of the balls after the collision . Also , decide whether the condition is elastic or not .

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Supposethe mass of each ball <a href="https://interviewquestions.tuteehub.com/tag/bem-2465851" style="font-weight:bold;" target="_blank" title="Click to know more about BEM">BEM</a> . Accordingto the law of conservation of momentum bytakingX - components , <br/> `mv_(1) =mv_(1) cos 30^(@) +mv_(2)cos 30^(@)` <br/> ` :. v_(1) = (sqrt(3))/2 v_(1)+(sqrt(3))/2 v_(2)""....(1)`<br/> TakingY - components<br/> ` p = mv_(1) sin 30^(@) - mv_(2) sin 30^(@)`<br/> ` 0 = (v_(1))/2 - (v_(2))/2 ` <br/> ` :. 0 = v_(1)-v_(2)` <br/> ` :. ` From equation (1) , <br/> Thevelocity before collision`v_(1) =12 m//sand v_(2) = 0 ` <br/> ` :. ` The kinetic energy of system before collision<br/> `K_(1)=1/2 mv_(1)^(2) +1/2 mv_(2)^(2)` <br/> `K_(1) =1/2 xxmxx (12)^(2)`<br/> ` K_(1) =72 m J ""....(1)` <br/> Velocity after collision `v_(1) =4sqrt(3) m//s and v_(2) = 4sqrt(3) m//s ` <br/> ` :. ` The kinetic after of system after collision . <br/> `K_(2) =1/2 m xx 48 +1/2 m xx 48` <br/> ` :. K_(2) =1/2 m xx 48 +1/2 m xx48`<br/> ` :. K_(2) = 24 m + 24 m ` <br/> ` :. K_(2) = 48 m J "".....(2)`<br/> ` :. ` From equation (1) and (2) , <br/> `K_(2) gt K_(2)` so the kinetic energy is not <a href="https://interviewquestions.tuteehub.com/tag/conserved-2539138" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVED">CONSERVED</a> , the collision is not elastic .</body></html>


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