Answer:
Let F be external force applied on the body in the direction of motion of the body for time interval [tex]\sf \Delta t[/tex], the the velocity of a body of mass m changes from [tex]\sf v [/tex] to [tex]\sf v + \Delta v[/tex] i.e. change in momentum, [tex]\sf \Delta p = m \Delta v [/tex].
According to Newton's second law :
[tex]:\implies \sf F \propto \dfrac{\Delta p}{\Delta t} \\ \\ \\ [/tex]
[tex]:\implies \sf F = k \: \dfrac{\Delta p}{\Delta t} \\ \\ \\ [/tex]
Where k is a constant of proportionality.
If limit [tex]\sf \Delta t \rightarrow 0,[/tex] then the term [tex]\sf \dfrac{\Delta p}{\Delta t}[/tex] becomes the derivative [tex]\sf \dfrac{dp}{dt}[/tex].
Thus,
[tex]:\implies \sf F = k \: \dfrac{dp}{dt} \\ \\ \\ [/tex]
For a body of fixed mass (m), we have :
[tex]:\implies \sf F = k \dfrac{d(mv)}{dt} \\ \\ \\ [/tex]
[tex]:\implies \sf F = km \: \dfrac{dv}{dt} \\ \\ \\ [/tex]
[tex]:\implies \sf F = kma \\ \\ \\ [/tex]
If v is fixed and m is variable then :
[tex]:\implies \sf F = \dfrac{kd(mv)}{dt} \\ \\ \\ [/tex]
[tex]:\implies \sf F = \dfrac{kvdm}{dt} \\ \\ \\ [/tex]
because, k = 1 then :
[tex]:\implies \sf F =\dfrac{vdm}{dt}[/tex]
Now, a unit force may be defined as the force which produces unit acceleration in a body of unit mass :]
So,
F = 1
m = 1
a = 1
k = 1
So,
[tex]:\implies \underline{ \boxed{ \sf F = ma}}[/tex]
