Question

Derivation for Newton's second law of motion

Answer 1

According to the Newton’s 2nd Law of motion, the rate of change of linear momentum of a body is directly proportional to the applied external force and in the direction of force.

It means that the linear momentum will change faster when a bigger force is applied.

Consider a body of mass ‘m’ moving with velocity v.

The linear momentum of a body is given by:

p = mv

Now According to Newton’s 2nd Law of Motion:

Force is directly proportional to rate of change of momnetum, that is

F α dp/dt

F  = k dp/dt

F = k d(mv)/dt

F = k md(v)/dt

F = k ma

Experimentally k =1

F = k ma

Which is the required equation of force.

hope it helps

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.