A body of mass $2\text{kg}$ is moving on a smooth horizontal surface with initial velocity $10{\text{ms}}^{-1}$ along east. A uniform force $10\text{N}$ towards north is applied on the body. The final velocity of the body after $2\text{s}$ will be

Let the East direction be along the $+x$ -axis ($\hat{i}$) and North along the $+y$ -axis ($\hat{j}$).Initial velocity: ${\vec{v}}_x=10\hat{i}{\text{ms}}^{-1}$ Force applied towards North: $\vec{F}=10\hat{j}\text{N}$ Calculate the acceleration in the North direction:
$${\vec{a}}_y=\frac{\vec{F}}{m}=\frac{10\hat{j}}{2}=5\hat{j}{\text{ms}}^{-2}$$
Calculate the final velocity component in the North direction after $t=2\text{s}$:
$${\vec{v}}_y={\vec{u}}_y+{\vec{a}}_{y}t=0+(5)(2)\hat{j}=10\hat{j}{\text{ms}}^{-1}$$
Since there is no force acting along the East direction, ${\vec{v}}_x$ remains $10\hat{i}{\text{ms}}^{-1}$.The final velocity vector is:
$$\vec{v}=10\hat{i}+10\hat{j}$$
Magnitude of final velocity:
$$|\vec{v}|=\sqrt{10^2+10^2}=\sqrt{200}=10\sqrt{2}{\text{ms}}^{-1}$$
Direction angle $\theta$:
$$\tan \theta =\frac{v_y}{v_x}=\frac{10}{10}=1⟹\theta =45^{\circ }\text{ North of East}$$