a particle has 2 velocities of eqaul magnitude inclined to ecah othr at an angle @ if one of them is halved the angle betwn d othr and the original resultant velocity is bisected by d new resulatnt then @ is
???
1cudnt get the problem clearly ..
finally after one of the velocities is halved we get
tan(α/4)=(vsin(α/2)/2)/(vcosα+v/2cosα)
thus
tan(α/4)=tan(α/2)/3
substitute
tan(α/2)=2tan(α/4)/(1-tan2(α/4)
ou will get
tanα/4=1/√3
thus α/4=30 and α=120
1ya d thing is at that tym meko woh formula ni pata tha
nw i noe so thanks
bt it wud b definately helpful if u post ur fig i mean d diagrm...lolz or u'll post ur fig ;)
1
the dotted lines represent the x and y components after the transformation
