Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.2 - Principle of Linear Impulse and Momentum for a System of Particles - Problems - Page 250: 19

$4m/s$

According to impulse momentum principle $mv_1+\Sigma \int_{t_1}^{t_2} F_x dt=mv_2$ $\implies mv_1+4(\int _{0}^{6}F_1 dt)-\int_{0}^{6} F_2 dt=mv_2$ We plug in the known values to obtain: $30(-5)+4[(2\times 10)+(\frac{1}{2}(10+30)(4-2))-(40\times 4+(\frac{1}{2}(40+10)(6-4))]=30v$ After simplifying the above equation, we obtain: $v=4m/s$