Answer
The magnitude of the change in momentum is $~~10.0~kg~m/s$
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:
$p_{f,y} = (0.30~kg)(0)$
$p_{f,y} = 0$
We can find the magnitude of the change in the vertical component of the momentum:
$\Delta p_y = \vert 0 - 2.58~kg~m/s \vert$
$\Delta p_y = 2.58~kg~m/s$
Let the horizontal direction away from the pitcher be the positive direction.
We can find the horizontal component of the initial momentum:
$p_{i,x} = (0.30~kg)(15~m/s)~cos~35^{\circ}$
$p_{i,x} = 3.69~kg~m/s$
We can find the horizontal component of the momentum as the ball leaves the bat:
$p_{f,x} = (0.30~kg)(-20~m/s)$
$p_{f,x} = -6.0~kg~m/s$
We can find the magnitude of the change in the horizontal component of the momentum:
$\Delta p_x = \vert -6.0~kg~m/s - 3.69~kg~m/s \vert$
$\Delta p_x = 9.69~kg~m/s$
We can find the magnitude of the change in momentum:
$\Delta p = \sqrt{(9.69~kg~m/s)^2+(2.58~kg~m/s)^2}$
$\Delta p = 10.0~kg~m/s$
The magnitude of the change in momentum is $~~10.0~kg~m/s$
