(a) 96 m/s
The impulse theorem states that the impulse exerted on the ball is equal to the change in momentum of the ball:
[tex]I=\Delta p=m \Delta v = m(v-u)[/tex]
where
[tex]I=32.4 Ns[/tex] is the impulse exerted on the ball
[tex]m=0.40 kg[/tex] is the mass of the ball
[tex]v[/tex] is the final speed of the ball
[tex]u=16 m/s[/tex] is the initial speed of the ball
Solving the equation for v, we find
[tex]v=\frac{I}{m}+u=\frac{32.4 Ns}{0.40 kg}+16 m/s=96 m/s[/tex]
(b) 952.9 N
The impulse is also equal to the product between the average force and the contact time:
[tex]I=F\Delta t[/tex]
where here we have
[tex]I=32.4 Ns[/tex] is the impulse
[tex]F[/tex] is the average force
[tex]\Delta t=34 ms=0.034 s[/tex] is the contact time
Solving the equation for F, we find the force:
[tex]F=\frac{I}{\Delta t}=\frac{32.4 Ns}{0.034 s}=952.9 N[/tex]
