## Physics: Principles with Applications (7th Edition)

The impulse given to the ball is $2.4~kg\cdot m/s$ in the horizontal direction away from the wall.
Note that only the horizontal component of velocity changes direction, so we only need to consider the horizontal change in momentum. $F\cdot t = \Delta p = m \Delta v$ $F\cdot t = (0.060~kg)(2)\times (28~m/s)~cos(45^{\circ})$ $F\cdot t = 2.4~kg\cdot m/s$ The impulse given to the ball is $2.4~kg\cdot m/s$ in the horizontal direction away from the wall.