Given:
m1 = mass 1 = 1kg
v1= initial velocity 1 = 12 m/s
m2= mass 2 = 3 kg
P after = momentum after collision = 15 kgm/s
(i)
Momentum of Ball A before collision
Momentum = mass x velocity
Pa = m1 v1
Replacing with the values given:
Pa = (1 kg) (12 m/s) = 12 kg m/s
(ii)
Momentum before = momentum after
Pa + Pb = P after
12 + Pb = 15
Since The ball B is travelling North, the distances travelled form a right triangle:
Apply pythagorean theorem:
c^2 = a^2 + b^2
Where c is the hypotenuse= P after = 15
a & b are the other 2 legs of the triangle = Pa and Pb
Replacing:
15^2 = 12^2 + Pb^2
Solve for Pb
15^2 - 12^2 = pb^2
√15^2 -12^2 = Pb
pb= 9 kgms^2