Answer:
4.44 i + 11.11 j , 11.96 m/s
Explanation:
m1 = 0.2 kg, v1 = 10 m/s along X axis, v1 = 10 i
m2 = 0.25 kg, v2 = 20 m/s along north, v2 = 20 j
After collision they stick together and let they move with velocity v.
Use the conservation of momentum
m1 x v1 + m2 x v2 = (m1 + m2) x v
0.2 x (10 i) + 0.25 x (20 j) = (0.2 + 0.25) x v
2i + 5j = 0.45 v
v = 4.44 i + 11.11 j
magnitude of velocity = [tex]\sqrt{(4.44)^{2}+(11.11)^{2}}[/tex] = 11.96 m/s
Thus, the velocity of 0.250 kg hockey puck is 4.44 i + 11.11 j  metre per second and the magnitude of velocity is 11.96 m/s.
