## College Physics (4th Edition)

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.
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.