This is the case of oblique elastic collision.
Let the first ball have velocity initial velocity before the collision = u and final velocity after the collision = v at 30.9° above the horizontal
and the second ball have velocity w after the collision at an angle θ below the horizontal.
By the law of conservation of linear momentum applied in the vertical direction,
vsin(30.9°) = wsinθ … (1) [mass of both the balls being equal cancel out]
Applying the law of conservation of momntum in the horizontal direction,
u + 0 = vcos(30.9°) + wcosθ
=> u – vcos(30.9°) = wcosθ … (2)
Squarring and adding eqns. (1) and (2),
u^2 + v^2 – 2uvcos(30.9°) = w^2 … (3)
For elastic collision, kinetic energy is also conserved
=> (1/2) mu^2 = (1/2)mv^2 + (1/2)mw^2
=> u^2 = v^2 + w^2
=> u^2 – v^2 = w^2 … (4)
Subtracting eqn. (4) from (3),
2v^2 – 2uvcos(30.9°) = 0
=> v = ucos(30.9°)
=> Kinetic energy of the first ball after the collision
= (1/2) mv^2
= (1/2) * (0.200) * ucos(30.9°)
= (0.100) * √(294) cos(30.9°) … [plugging the value of u obtained in the first step]
= 1.47 J.
