Answer
B's speed just before the collision was $4.0~m/s$
Work Step by Step
By conservation of momentum, the west component of momentum before the collision must be equal to the west component of momentum after the collision.
We can find B's speed $v_B$ just before the collision:
$(300~g)~v_B = (220~g+300~g)(3.13~m/s)~cos~42.5^{\circ}$
$v_B = \frac{(220~g+300~g)(3.13~m/s)~cos~42.5^{\circ}}{300~g}$
$v_B = 4.0~m/s$
B's speed just before the collision was $4.0~m/s$.
