## College Physics (4th Edition)

(a) The magnitude of the change in momentum is $3.75~kg~m/s$ and the direction is an angle of $36.9^{\circ}$ above the horizontal. (b) The average force of the bat on the ball is $75~N$
(a) We can find the horizontal component of the change in momentum: $\Delta p_x = m~\Delta v_x = (0.15~kg)(-20~m/s) = -3.0~kg~m/s$ We can find the vertical component of the change in momentum: $\Delta p_y = m~\Delta v_y = (0.15~kg)(15~m/s) = 2.25~kg~m/s$ We can find the magnitude of the change in momentum: $\Delta p = \sqrt{p_x^2+p_y^2} = \sqrt{(-3.0~kg~m/s)^2+(2.25~kg~m/s)^2} = 3.75~kg~m/s$ We can find the angle $\theta$ above the horizontal of the change in momentum: $tan~\theta = \frac{2.25}{3.0}$ $\theta = tan^{-1}(\frac{2.25}{3.0})$ $\theta = 36.9^{\circ}$ The magnitude of the change in momentum is $3.75~kg~m/s$ and the direction is an angle of $36.9^{\circ}$ above the horizontal. (b) The impulse on the ball is equal to the ball's change in momentum: $F~t = \Delta p$ $F = \frac{\Delta p}{t}$ $F = \frac{3.75~kg~m/s}{0.050~s}$ $F = 75~N$ The average force of the bat on the ball is $75~N$