Answer
The velocity is: $~~(2.67~m/s)~\hat{i} + (-3.00~m/s)~\hat{j}$
Work Step by Step
$v_1 = (-4.00~m/s)~\hat{i} + (-5.00~m/s)~\hat{j}$
We can find the momentum of the 2.00-kg particle:
$p_1 = (2.00~kg)[(-4.00~m/s)~\hat{i} + (-5.00~m/s)~\hat{j}]$
$p_1 = (-8.00~kg~m/s)~\hat{i} + (-10.00~kg~m/s)~\hat{j}$
$v_2 = (6.00~m/s)~\hat{i} + (-2.00~m/s)~\hat{j}$
We can find the momentum of the 4.00-kg particle:
$p_2 = (4.00~kg)[(6.00~m/s)~\hat{i} + (-2.00~m/s)~\hat{j}]$
$p_2 = (24.00~kg~m/s)~\hat{i} + (-8.00~kg~m/s)~\hat{j}$
We can use conservation of momentum to express the momentum after the collision in unit-vector notation:
$p_f = p_1+p_2$
$p_f = (16.00~kg~m/s)~\hat{i} + (-18.00~kg~m/s)~\hat{j}$
We can express the velocity in unit-vector notation:
$m~v_f = p_f$
$v_f = \frac{p_f}{m}$
$v_f = \frac{(16.00~kg~m/s)~\hat{i} + (-18.00~kg~m/s)~\hat{j}}{6.00~kg}$
$v_f = (2.67~m/s)~\hat{i} + (-3.00~m/s)~\hat{j}$
The velocity is: $~~(2.67~m/s)~\hat{i} + (-3.00~m/s)~\hat{j}$
