Answer
$p = (-2.44~kg~m/s)~\hat{i}+(2.80~kg~m/s)~\hat{j}$
Work Step by Step
$F_x = (0.280~N/s)~t$
$J_x = \int_{0}^{2.00}F_x~dt$
$J_x = \int_{0}^{2.00}(0.280~N/s)~t~dt$
$J_x = (0.140~N/s)~t^2\vert_{0}^{2.00}$
$J_x = 0.560~kg~m/s$
$p_{x2} = p_{x1}+J_x$
$p_{x2} = -3.00~kg~m/s+0.560~kg~m/s$
$p_{x2} = -2.44~kg~m/s$
$F_y = (-0.450~N/s^2)~t^2$
$J_y = \int_{0}^{2.00}F_y~dt$
$J_y = \int_{0}^{2.00}(-0.450~N/s)~t^2~dt$
$J_y = (-0.150~N/s)~t^3\vert_{0}^{2.00}$
$J_y = -1.20~kg~m/s$
$p_{y2} = p_{y1}+J_y$
$p_{y2} = (4.00~kg~m/s)+(-1.20~kg~m/s)$
$p_{y2} = 2.80~kg~m/s$
$p = (-2.44~kg~m/s)~\hat{i}+(2.80~kg~m/s)~\hat{j}$
