## College Physics (4th Edition)

B's speed just before the collision was $4.0~m/s$
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 $4.0~m/s$.