# Why do we add k in derivation of 2nd law of motion when it's value is 1?

Let us understand the derivation of Newton's 2nd law of motion.

"the magnitude of the applied force upon a body is directly proportional to the rate of the change of the momentum with time."

Or

If $F$ force is applied over a body whose initial velocity is $u$ and it becomes $v$ with respect to the time $t$.

Then, initial momentum $P_1=mass\times volume=mu$

And final momentum $P_2=mv$

Therefore, change in momentum $\delta P=P_2-P_1$

$=mv-mu$

And rate of the change of the momentum with respect to the time $t=\frac{\delta P}{\delta t}$

$=\frac{mv-mu}{t}$

$=m\frac{v-u}{t}$

$=ma$     [Because rate of the change of velocity with respect to the time $a=\frac{v-u}{t}$]

According to Newton's 2nd law of motion:

$F\propto\frac{\delta P}{\delta t}$

Or $F\propto ma$

Let $F=k\frac ma$

$k$ is a constant. We choose the value of $m$ , $u$, $v$ and time $t$ such that we find the value of $k=1$

So $F=ma$

Here we find that the value of $k$ has taken $1$ to make the derived equation of 2nd law of motion simple.

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Updated on: 10-Oct-2022

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