Question

A 40.0 g ball of clay traveling east at 9.00 m/s collides and sticks together with a 20.0 g ball of clay traveling north at 9.00 m/s . What is the speed of the resulting ball of clay? What is the direction of the resulting ball of clay?

Answer 1

Since they stick together, this is a "perfectly inelastic collision."
The formula for inelastic collisions is:
(m1)(v1)i + (m2)(v2)i = (m1 + m2)vf

You're going to have to run through the equation twice, once using the north/south speeds for velocity, and once using the east/west speeds.

Let's start with the north/south speeds.

(40.0)(0) + (20.0)(9) = (40.0 + 20.0)(vf)
0 + 60 = (110)(vf)
vf = 3 m/s to the north
That's just the north/south velocity. We're going to solve for the east/west velocity now:

(40.0)(9) + (20.0)(0) = (40.0 + 20.0)(vf)
vf = 6 m/s to the east

Now we're going to use the Pythagorean Theorem to find the resultant vector:
a^2 + b^2 = c^2
(6)^2 + (3)^2 = c^2
45 = c^2
c = 6.71 m/s

now you need to find the angle:
inverse sin of (opp/hyp) will give us the angle:
inverse sin of 3/6 = 30 degrees north of east

So the answer is:
Vf = 6.71 m/s at 30 degrees north of east