#### Answer

The 50-gram ball of clay is moving at a speed of 1.7 m/s at an angle $45^{\circ}$ north of east.

#### Work Step by Step

We can find the east component of momentum.
$p_x = (0.020~kg)(3.0~m/s)$
$p_x = 0.060~N~s$
We can find the north component of momentum.
$p_y = (0.030~kg)(2.0~m/s)$
$p_x = 0.060~N~s$
We can find the magnitude of the momentum of the 50-gram ball of clay.
$p = \sqrt{(p_x)^2+(p_y)^2}$
$p = \sqrt{(0.060~N~s)^2+(0.060~N~s)^2}$
$p = 0.085~N~s$
We can use the momentum to find the speed $v$ of the 50-gram ball of clay.
$m~v = p$
$v = \frac{p}{m}$
$v = \frac{0.085~N~s}{0.050~kg}$
$v = 1.7~m/s$
Since the east component and the north component of the momentum are equal, the direction is $45^{\circ}$ north of east.