Question

Newton second law of motion derivation
​

Answer 1

Answer:

Let 'm' = mass

'u' = initial speed

'v' = final speed

't' = time interval and

'F' = force.

~ is directly proportional.

Derivation:

Initial momentum of body = mu

Final momentum of body = mv

change in momentum ∆p = mu - mv

By newton's second law of motion,

Force, F ~ rate of change in momentum.

F ~ change in momentum / time.

F ~ mv - mu / t

F = km(v-u) / t

Here, k is constant k = 1.

F = m(v-u) / t

since, acceleration a= (v-u) / t.

F = m * a

Force = mass * acceleration.

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.