Answer
The magnitude of the change in momentum is $~~5.0~kg~m/s$
Work Step by Step
We can find the vertical component of the initial momentum:
$p_{i,y} = (0.30~kg)(15~m/s)~sin~35^{\circ}$
$p_{i,y} = 2.58~kg~m/s$ (down)
We can find the vertical component of the momentum as the ball leaves the bat:
$p_{f,y} = (0.30~kg)(20~m/s)$
$p_{f,y} = 6.0~kg~m/s$ (down)
We can find the change in the vertical component of the momentum:
$\Delta p_y = 6.0~kg~m/s - 2.58~kg~m/s$
$\Delta p_y = 3.42~kg~m/s$ (down)
We can find the horizontal component of the initial momentum:
$p_{i,x} = (0.30~kg)(15~m/s)~cos~35^{\circ}$
$p_{i,x} = 3.69~kg~m/s$ (toward the batter)
We can find the horizontal component of the momentum as the ball leaves the bat:
$p_{f,x} = (0.30~kg)(0)$
$p_{f,x} = 0$
We can find the change in the horizontal component of the momentum:
$\Delta p_x = 0 - 3.69~kg~m/s$
$\Delta p_x = -3.69~kg~m/s$
We can find the magnitude of the change in momentum:
$\Delta p = \sqrt{(-3.69~kg~m/s)^2+(3.42~kg~m/s)^2}$
$\Delta p = 5.0~kg~m/s$
The magnitude of the change in momentum is $~~5.0~kg~m/s$
