We can use conservation of momentum to solve this problem. The total momentum before the collision is given by:
p_before = m1 * v1 + m2 * v2
where m1 = 20 g = 0.02 kg, v1 = 5.5 m/s to the east, m2 = 55 g = 0.055 kg, and v2 = 2.0 m/s to the north. We can resolve the velocity of the second ball into its east and north components:
v2_east = 0
v2_north = 2.0 m/s
So we have:
p_before = 0.02 kg * 5.5 m/s + 0.055 kg * 2.0 m/s = 0.2215 kg·m/s to the east + 0.11 kg·m/s to the north
After the collision, the two balls stick together and move off in some direction at some speed v. The total momentum after the collision is given by:
p_after = (m1 + m2) * v
where m1 + m2 = 0.075 kg.
Since momentum is conserved, we can equate p_before and p_after:
p_before = p_after
0.2215 kg·m/s to the east + 0.11 kg·m/s to the north = 0.075 kg * v
To solve for v, we can take the magnitude of both sides of the equation:
sqrt[(0.2215 kg·m/s)^2 + (0.11 kg·m/s)^2] = 0.075 kg * v
v = sqrt[(0.2215 kg·m/s)^2 + (0.11 kg·m/s)^2] / 0.075 kg
v = 3.27 m/s
So the speed of the resulting ball of clay is 3.27 m/s. To find its direction, we can use trigonometry. Let θ be the angle between the direction of motion of the resulting ball and the eastward direction. Then we have:
tan(θ) = (0.11 kg·m/s) / (0.2215 kg·m/s)
θ = tan^(-1)[(0.11 kg·m/s) / (0.2215 kg·m/s)]
θ = 25.2 degrees north of east
So the direction of the resulting ball of clay is 25.2 degrees north of east.