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Numerical Based Question |
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Answer» Let the veloercity of the particle after collision be v As the horizontal components before and after collision remain unchanged then ucosQ = v cos(90-Q)=vsinQ. ....(1) Again the coefficint of restitution e=(vsin(90-Q))/(usinQ)=3/4 .......(2) Comparing (1) and (2) we get (v/u)"2=3/4 =>v= (√3/2)u Option (1) is correct |
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