The momentum of the first ball is 4*15 kg m/s
This is equal to the momentum of the 14 kg ball after it starts moving:
4*15 = 14*v
where v is the velocity of the second ball
I suspect your teacher did not realize what he wrote: commpletely elastic, and transfers all of its momentum....
Rotational energy, and rotational momentum being conserved is an issue here, I think. Nothing on the size of either ball is given...so my guess is your teacher MEANT to say, the collision in translational motion was elastic, and all translational momentum was transfered. Now in reality, it would be a feat to do this with balls of different mass, as any pool player can attest, but they don't play on a frictionaless table, where balls theoritically move without rolling.
So considering translational motion only.
KE before= KE after
1/2 4 * 15^2=1/2 14 vf^2 give you the velocity of the second ball.
Now if you consider momentum
momentum before=momentum after
Notice it is a different answer than in the KE solution.
Thus, as pool players recognize, it is not possible. Maybe your teacher had something in mind here, but it escapes me.