sosi

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.