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 253: 30

$16.1m/s$

We know that $v_1=40Km/h=11.11m/s$ Now we apply the principle of impulse and momentum in the x-direction $mv_{x_1}+\Sigma \int ^{t_2}_{t_1} F_x dt=mv_{x_2}$ We plug in the known values to obtain: $7\times 10^{3}(11.11)-\frac{1}{2}(5\times 10^3)+\frac{1}{2}(15+5)(5-2)\times 10^3=7\times 10^3 v_2$ This simplifies to: $v_2=16.1m/s$