Let $m = 2000$ kg, initial speed $v_i = 12$ m/s (north), and final speed $v_f = 16$ m/s (east).

[Q1] The change in kinetic energy $\Delta K$ is a scalar quantity.

$$\Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \frac{1}{2}m(v_f^2 - v_i^2)$$ $$\Delta K = \frac{1}{2}(2000 \text{ kg})((16 \text{ m/s})^2 - (12 \text{ m/s})^2) = 1000(256 - 144) \text{ J}$$

[Q2] The change in momentum $\Delta \vec{p}$ is a vector quantity, $\Delta \vec{p} = \vec{p}_f - \vec{p}_i$. Let east be the +x direction and north be the +y direction. The initial momentum is $\vec{p}_i = m\vec{v}_i = (2000)(12\hat{j}) = 24000\hat{j}$ kg·m/s. The final momentum is $\vec{p}_f = m\vec{v}_f = (2000)(16\hat{i}) = 32000\hat{i}$ kg·m/s. The change in momentum is $\Delta \vec{p} = \vec{p}_f - \vec{p}_i = 32000\hat{i} - 24000\hat{j}$ kg·m/s. The magnitude is found using the Pythagorean theorem:

$$|\Delta \vec{p}| = \sqrt{(32000)^2 + (-24000)^2} \text{ kg·m/s}$$

The direction $\theta$, measured south from the east direction, is:

$$\theta = \arctan\left(\frac{|\Delta p_y|}{|\Delta p_x|}\right) = \arctan\left(\frac{24000}{32000}\right) = \arctan(0.75)$$