Question

give derivation of Newton's second law of motion

Answer 1

Newton's second law of motion :- Newton said " if body changes momentum with respect to time, force must be experienced by body and that force is the rate of change of momentum per unit time."

Let mass of body m is moving with velocity $\vec{u}$ and after t times , velocity of it became $\vec{v}$.
then, initial momentum = $m\vec{u}$
final momentum = $m\vec{v}$
change in momentum = $m\vec{v}-m\vec{u}$

according to Newton's law, force = $\frac{\Delta{P}}{\Delta{t}}$

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

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

we know, from 1st kinematics equation.
$\vec{a}=\frac{\vec{v}-\vec{u}}{t}$

so, $F=m\vec{a}$

hence, $\boxed{F=ma}$ is the formula generated with help of 2nd law of motion

Answer 2

Newton's 2nd law of motion states that ;

" The rate of change of momentum is directly proportional to the unbalance force in the direction of force "

$\sf \: Force \propto \dfrac{Change \: in \: momentum}{Time \: taken}$

Consider a body of Mass m having an initial velocity u. The initial momentum of this body will be mu. Suppose a force F acts on this body for time t & causes the final velocity to become v. The final momentum of this body will be mv. Now,the change in momentum of this body is mv - mu & the time taken for this change is t. So, According to Newton's First Law of Motion :

$\large \: \sf \: F \propto \dfrac{ mv \: - \: mu}{t}$

$\implies\large \: \sf \: F \propto \dfrac{ m(v - u)}{t}$

Recall the first equation of motion  

v = u + at

$\implies\tt{a=\dfrac{v-u}{t}}$

Substitute this value in above one

Hence,

$\tt{ F \propto ma }$

But we need to remove the proportionality symbol ,

In order to remove it we need to add an proportionality constant.

So,

$\tt{ F =k* ma }$

k = 1

So,

$\tt{F=m*a}$

Derived.