(N/A) Statement: The time rate of change of momentum of a body is directly proportional to the resultant force acting on it,and this change occurs in the direction of the resultant force.
Let a resultant force $\vec{F}$ act on a body of mass $m$ for a time interval $\Delta t$. During this,let its velocity change from $\vec{v}$ to $\vec{v} + \Delta \vec{v}$.
Initial momentum: $\vec{p}_i = m\vec{v}$
Final momentum: $\vec{p}_f = m(\vec{v} + \Delta \vec{v})$
Change in momentum:
$\Delta \vec{p} = \vec{p}_f - \vec{p}_i = m(\vec{v} + \Delta \vec{v}) - m\vec{v} = m\Delta \vec{v}$
From the second law of motion:
$\vec{F} \propto \frac{\Delta \vec{p}}{\Delta t} \implies \vec{F} = k \frac{\Delta \vec{p}}{\Delta t}$
Taking the limit $\Delta t \to 0$:
$\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}$ (assuming constant mass).
Important points:
$(i)$ Newton's second law is $\vec{F} = \frac{d\vec{p}}{dt}$. If mass is constant,$\vec{F} = m\vec{a}$.
$(ii)$ If the resultant external force is zero,then $\vec{a} = 0$,meaning velocity is constant,which is consistent with Newton's first law.
$(iii)$ The law provides the magnitude of force. It is a vector quantity with components: $F_x = ma_x, F_y = ma_y, F_z = ma_z$. If force acts at an angle,only the component in the direction of velocity changes it.
$(iv)$ The equation $\vec{F} = m\vec{a}$ applies to point objects,rigid bodies,or systems of particles where $\vec{F}$ is the resultant force and $\vec{a}$ is the acceleration of the center of mass.
$(v)$ The law relates force and acceleration at a specific instant; it does not depend on the history of the object's motion.