(N/A) Let an unbalanced force $F$ act on a body of mass $m$ moving with an initial velocity $u \, m s^{-1}$. The velocity changes to $v \, m s^{-1}$ after time $t$ seconds.
Initial momentum of the body $(p_1) = m \times u = mu$.
Final momentum of the body after $t$ seconds $(p_2) = m \times v = mv$.
Change in momentum $= p_2 - p_1 = mv - mu = m(v - u)$.
Rate of change of momentum $= \frac{m(v - u)}{t}$.
According to Newton's second law of motion, the rate of change of momentum is directly proportional to the applied force $F$:
$F \propto \frac{m(v - u)}{t}$.
Since acceleration $a = \frac{v - u}{t}$, we can write:
$F \propto ma$.
$F = kma$, where $k$ is a constant of proportionality.
By defining the unit of force such that $1 \, N$ of force produces an acceleration of $1 \, m s^{-2}$ in a body of $1 \, kg$ mass, we get $k = 1$.
Therefore, the mathematical formula for force is $F = ma$.