1.

Numerical Based Question

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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