Question
Prove mathematically F = ma
✓
Answer
Derivation of F = ma from Newton's Second Law of Motion:
Newton introduced the concept of momentum and say "The momentum of a moving body is defined as the product of its mass and velocity."
Thus, $p=m v$, where $p=$ momentum of body
$m =$ mass of body
$v=$ velocity of body
Suppose the velocity of body of mass $m$ changes from $u$ to $v$ in time $t$.
Initial momentum, $p _1= mu$
and final momentum, $p _2= mv$
the change in momentum, $\left(p_2-p_1\right)$ takes place in time $t$. Then according to Newton's second law of motion, the magnitude of force $F$ is:
$\frac{p_2-p_1}{t} \propto F, \text { or } F=\frac{km(v-u)}{t}$
where $k=$ constant of proportionality
Now, $a \frac{(v-u)}{t}$, where $a=$ acceleration of body
$\begin{array}{l}
\therefore F=k m a \\
\Rightarrow F=m a
\end{array}
$
$
(\because k=1, \text { constant })
$
This relation holds good when mass of the body remains constant.
Newton introduced the concept of momentum and say "The momentum of a moving body is defined as the product of its mass and velocity."
Thus, $p=m v$, where $p=$ momentum of body
$m =$ mass of body
$v=$ velocity of body
Suppose the velocity of body of mass $m$ changes from $u$ to $v$ in time $t$.
Initial momentum, $p _1= mu$
and final momentum, $p _2= mv$
the change in momentum, $\left(p_2-p_1\right)$ takes place in time $t$. Then according to Newton's second law of motion, the magnitude of force $F$ is:
$\frac{p_2-p_1}{t} \propto F, \text { or } F=\frac{km(v-u)}{t}$
where $k=$ constant of proportionality
Now, $a \frac{(v-u)}{t}$, where $a=$ acceleration of body
$\begin{array}{l}
\therefore F=k m a \\
\Rightarrow F=m a
\end{array}
$
$
(\because k=1, \text { constant })
$
This relation holds good when mass of the body remains constant.
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