The impulse (the variation of momentum of the ball) is related to the force applied by
[tex]\Delta p = F \Delta t[/tex]
where [tex]\Delta p[/tex] is the variation of momentum, F is the intensity of the force and [tex]\Delta t[/tex] is the time of application of the force.
Using F=1000 N and [tex]\Delta t = 1.8 ms=1.8 \cdot 10^{-3}s[/tex], we can find the variation of momentum:
[tex]\Delta p = (1000 N)(1.8 \cdot 10^{-3} s)=1.8 kg m/s[/tex]
This [tex]\Delta p[/tex] can be rewritten as
[tex]\Delta p = p_f - p_i = mv_f - mv_i[/tex]
where [tex]p_f [/tex] and [tex]p_i[/tex] are the final and initial momentum. But the ball is initially at rest, so the initial momentum is zero, and
[tex]\Delta p = mv_f[/tex]
from which we find the final velocity of the ball:
[tex]v_f = \frac{\Delta p }{m}= \frac{1.8 kg m/s}{0.045 kg}= 40 m/s[/tex]
