#### Answer

The speed of the blob is 0.85 m/s and it is moving at an angle of $72.2^{\circ}$ below the positive x-axis.

#### Work Step by Step

We can find the horizontal component of the momentum of the system.
$p_x = (0.040~kg)(4.0~m/s)[cos(45^{\circ})]+(0.030~kg)(-3.0~m/s)+(0.020~kg)(0)$
$p_x = 0.023~N~s$
We can find the horizontal component of the velocity of the 90-gram blob of clay.
$m~v_x = p_x$
$v_x = \frac{p_x}{m}$
$v_x = \frac{0.023~N~s}{0.090~kg}$
$v_x = 0.26~m/s$
We can find the vertical component of the momentum of the system.
$p_y = (0.040~kg)(-4.0~m/s)[sin(45^{\circ})]+(0.030~kg)(0)+(0.020~kg)(2.0~m/s)$
$p_y = -0.073~N~s$
We can find the vertical component of the velocity of the 90-gram blob of clay.
$m~v_y = p_y$
$v_y = \frac{p_y}{m}$
$v_y = \frac{-0.073~N~s}{0.090~kg}$
$v_y = -0.81~m/s$
We can find the speed of the blob of clay.
$v = \sqrt{(v_x)^2+(v_y)^2}$
$v = \sqrt{(0.26~m/s)^2+(-0.81~m/s)^2}$
$v = 0.85~m/s$
We can find the angle $\theta$ below the positive x-axis.
$tan(\theta) = \frac{v_y}{v_x}$
$\theta = arctan(\frac{0.81~m/s}{0.26~m/s})$
$\theta = 72.2^{\circ}$
The speed of the blob is 0.85 m/s and it is moving at an angle of $72.2^{\circ}$ below the positive x-axis.