Answer
After the collision, the second puck's speed is $0.27~m/s$ and it moves at an angle of $53.3^{\circ}$ to the left.
Work Step by Step
Let's assume the first puck is moving in the +x-direction.
Let $M$ be the mass of each puck.
By conservation of momentum, the final momentum of the system is equal to the initial momentum.
We can find the horizontal component $v_x$ of the final velocity of the second puck:
$M~v_x +M~(0.36~m/s)~cos~37^{\circ} = M~(0.45~m/s)$
$v_x = (0.45~m/s)-(0.36~m/s)~cos~37^{\circ}$
$v_x = 0.162~m/s$
We can find the vertical component $v_y$ of the final velocity of the second puck:
$M~v_y -M~(0.36~m/s)~sin~37^{\circ} = M~(0)$
$v_y = (0.36~m/s)~sin~37^{\circ}$
$v_y = 0.217~m/s$
We can find the magnitude of the second puck's velocity after the collision:
$\sqrt{v_x^2+v_y^2} = \sqrt{(0.162~m/s)^2+(0.217~m/s)^2} = 0.27~m/s$
We can find the direction to the left:
$tan~\theta = \frac{0.217}{0.162}$
$\theta = tan^{-1}(\frac{0.217}{0.162})$
$\theta = 53.3^{\circ}$
After the collision, the second puck's speed is $0.27~m/s$ and it moves at an angle of $53.3^{\circ}$ to the left.
