Correct Answer - A
Conservation the momentum of two balls before and after colision in horizontal direction
`1xx6=(1+1)xxv`
`v=3ms^(-1)`
Now at the instant of maximum deflection ball will be moving horizontal with same speed as of trolley (i.e., velocity of ball w.r.t trolley will be `0`)
let be `v_(0)`
`2xx3=(4+2)xxv_(0)`
now conserving energy
`1/2xx2xx3^(2)-2xx10xx1.5(1-costheta)`
`=1/2xx2xxv^(2)+1/2xx4xxv^(2)`
`implies costheta=0.8impliestheta=37^(@)`