A hockey puck of mass 0.16 kg, sliding on a nearly frictionless surface of ice with a velocity of 2,0 m/s[E]...?

Momentum is conserved in elastic collision

Let velocity in East and North direction be positive, while West and South direction be negative

Conservation of momentum in East-West direction

m₁ux₁ + m₂ux₂ = m₁vx₁ + m₂vx ₂

(0.16 kg)(2.0 m/s) + (0.17 kg)(0) = (0.16 kg)(1.5sin31 m/s) + (0.17 kg)(vx ₂) 

0.32 = 0.24sin31 + 0.17 vx ₂

vx ₂ = (0.32 - 0.24sin31)/0.17

vx ₂ = 1.16 m/s

Conservation of momentum in North-South direction

m₁uy₁ + m₂uy₂ = m₁vy₁ + m₂vy ₂

(0.16 kg)(0) + (0.17 kg)(0) = (0.16 kg)(1.5cos31 m/s) + (0.17 kg)(vy ₂) 

0 = 0.24cos31 + 0.17 vy ₂

vy ₂ = (- 0.24cos31)/0.17

vy ₂ = -1.2 m/s

Resultant velocity = √[(1.16)² + (-1.2)²] = 1.67 m/s

angle θ = arctan(-1.2/1.16) = -46°

This represents E 46 S 

The second puck moves, after collision, at 1.67 m/s at [E 46 S]