Answer
The magnitude of the change in momentum is $~~5.86~kg~m/s$
Work Step by Step
We can find the vertical component of the initial momentum:
$p_{i,y} = -(0.300~kg)(12.0~m/s)~sin~35.0^{\circ}$
$p_{i,y} = -2.065~kg~m/s$
We can find the vertical component of the momentum as the ball leaves the bat:
$p_{f,y} = (0.300~kg)(10.0~m/s)$
$p_{f,y} = 3.0~kg~m/s$
We can find the change in the vertical component of the momentum:
$\Delta p_y = 3.0~kg~m/s - (-2.065~kg~m/s)$
$\Delta p_y = 5.065~kg~m/s$
We can find the horizontal component of the initial momentum:
$p_{i,x} = -(0.300~kg)(12.0~m/s)~cos~35^{\circ}$
$p_{i,x} = -2.949~kg~m/s$
We can find the horizontal component of the momentum as the ball leaves the bat:
$p_{f,x} = (0.300~kg)(0)$
$p_{f,x} = 0$
We can find the change in the horizontal component of the momentum:
$\Delta p_x = 0 - (-2.949~kg~m/s)$
$\Delta p_x = 2.949~kg~m/s$
We can find the magnitude of the change in momentum:
$\Delta p = \sqrt{(5.065~kg~m/s)^2+(2.949~kg~m/s)^2}$
$\Delta p = 5.86~kg~m/s$
The magnitude of the change in momentum is $~~5.86~kg~m/s$
