## Engineering Mechanics: Statics & Dynamics (14th Edition)

$I=15KN\cdot s$ in both cases
The required impulse can be determined as follows: We apply the principle of impulse and momentum in the x-direction $mv_{x_1}+\Sigma \int F_xdt=mv_{x_2}$ We plug in the known values to obtain: $20000(0.75)-\int F_x dt=0$ This simplifies to: $I=15KN\cdot s$