Answer
$v_B{_{_f}} =4.3_{m/s}$
Work Step by Step
$\Delta K_f = \Delta K_i$
$\frac{1}{2}m(v_B{_{_f}}^2- v_A{_{_f}}^2 ) = \frac{1}{2}m(v_B{_{_i}}^2- v_A{_{_i}}^2 ) $
$m(v_B{_{_f}}^2- v_A{_{_f}}^2 ) = m(v_B{_{_i}}^2- v_A{_{_i}}^2 ) $
$(v_B{_{_f}}^2- v_A{_{_f}}^2 ) = (v_B{_{_i}}^2- v_A{_{_i}}^2 ) $
$v_B{_{_f}}^2- 16_{m/s}^2 = 2.6_{m/s}^2- 2_{m/s}^2 $
$v_B{_{_f}}^2 =( 2.6_{m/s}^2- 2_{m/s}^2) + 16_{m/s}^2 $
$v_B{_{_f}} =\sqrt {18.76_{m^2/s^2}} $
$v_B{_{_f}} =4.3_{m/s}$
