Answer:
(a). The speed of the first ball after the collision is 1.95 m/s.
(b). The direction of the first ball after the collision is 44.16° due south of east.
Explanation:
Given that,
Velocity of one ball u₁= 2.2i m/s
Velocity of second ball u₂=- 0.80i m/s
Final velocity of the second ball v₂= 1.36j m/s
The mass of the identical balls are

(a). We need to calculate the speed of the first ball after the collision
Using law of conservation of momentum

Along X- axis


Put the value into the formula


Along Y-axis



Put the value into the formula

Then the final speed of the first ball


(b) We need to calculate the direction of the first ball after the collision
Using formula of direction




Negative sign shows the direction of first ball .
Hence, (a). The speed of the first ball after the collision is 1.95 m/s.
(b). The direction of the first ball after the collision is 44.16° due south of east.