#### Answer

The speed of the first ball after the collision is 1.73 m/s at an angle of $54.7^{\circ}$ south of east.

#### Work Step by Step

Let $M$ be the mass of each ball.
We can find the east-west component of the initial momentum of the system of the two balls. Let east be the positive direction.
$p_{0x} = (M)(2.00~m/s)+(M)(-1.00~m/s)$
$p_{0x} = (M)(1.00~m/s)$
By conservation of momentum, the east-west component of the momentum after the collision must be equal to $p_{0x}$. We can find the east-west component $v_x$ of the first ball's velocity after the collision.
$p_{fx} = p_{0x}$
$(M)~v_x + (M)(0)= (M)(1.00~m/s)$
$v_x = 1.00~m/s$
The east-west component of the first ball's velocity after the collision is 1.00 m/s to the east.
The north-south component of the initial momentum is zero. By conservation of momentum, the north-south component of the momentum after the collision must be zero. We can find the north-south component $v_y$ of the first ball's velocity after the collision. Let north be the positive direction.
$p_{fy} = p_{0y}$
$p_{fy} = 0$
$(M)~v_y + (M)(1.41~m/s)= 0$
$v_y = -1.41~m/s$
The north-south component of the first ball's velocity after the collision is 1.41 m/s to the south.
We can find the speed of the first ball.
$v = \sqrt{(v_x)^2+(v_y)^2}$
$v = \sqrt{(1.00~m/s)^2+(-1.41~m/s)^2}$
$v = 1.73~m/s$
We can find the angle $\theta$ south of east.
$tan(\theta) = \frac{v_y}{v_x}$
$\theta = arctan(\frac{1.41~m/s}{1.00~m/s})$
$\theta = 54.7^{\circ}$
The speed of the first ball after the collision is 1.73 m/s at an angle of $54.7^{\circ}$ south of east.