Problem 8.10

8.10. a)

Taking the positive x axis towards the pitcher, we find that conservation of momentum on the $\hat{x}$ direction yields

p_ix + I_x  =  p_fx

$(.2 kg)(- 15 m/s)(\cos 45{}^{\circ}) + I_{x} ~=~ (.2 kg)(40 m/s)(\cos 30{}^{\circ})$

I_x  =  9.05 N s

And in the $\hat{y}$ direction,

p_iy + I_y  =  p_fy

$(.2 kg)(- 15 m/s)(\sin 45{}^{\circ}) + I_{y} ~=~ (.2 kg)(40 m/s)(\sin 30{}^{\circ})$

I_y  =  6.12 N s

$\vec{I} ~=~ (9.05 \hat{i} + 6.12 \hat{j}) N s$

b)

The impulse is given by the area under the curve. Averaging this area when we don't have a constant force gives us

$\vec{I} = \frac{1}{2} (0 + \vec{F_{m}})(4 ms) + \vec{F_{m}} (20 ms) + \frac{1}{2} (0 + \vec{F_{m}})(4 ms)$

$\vec{F_{m}} (24 \times 10^{-3} s) ~=~ \vec{I}$

So, the maximum force is given by

$\vec{F_{m}} ~=~ (377\hat{i} + 106\hat{j}) N$