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Question: According to Newton’s law of motion correct equation is: (where symbols have their usual meanings) ...

According to Newton’s law of motion correct equation is: (where symbols have their usual meanings)
A) F=dPdtF = \dfrac{{dP}}{{dt}}
B) F=maF = ma
C) F=vdmdtF = v\dfrac{{dm}}{{dt}}
D) All of these

Explanation

Solution

Newton’s second law of motion gives the relation between applied force and momentum of the body.

Complete step by step solution:
As we know that second law of Newton is defined as
The rate of change of linear momentum of a body is directly proportional to the applied force and the change takes place in the direction of the applied force.
F=dpdtF = \dfrac{{dp}}{{dt}} ................. (1)
If a body has mass mm moving with velocity vv then the linear momentum is
p=mv\Rightarrow p = mv
Differentiating both sides with respect to time tt
dpdt=d(mv)dt\dfrac{{dp}}{{dt}} = \dfrac{{d\left( {mv} \right)}}{{dt}}
dpdt=mdvdt+vdmdt\dfrac{{dp}}{{dt}} = m\dfrac{{dv}}{{dt}} + v\dfrac{{dm}}{{dt}}
From equation (1)
F=mdvdt+vdmdt\Rightarrow F = m\dfrac{{dv}}{{dt}} + v\dfrac{{dm}}{{dt}}................. (2)
This equation gives the applied force on body
Most of time in question the mass of body is constant so in that situation dmdt=0\dfrac{{dm}}{{dt}} = 0 then above equation (2) becomes
F=mdvdt\Rightarrow F = m\dfrac{{dv}}{{dt}}
We know dvdt=a\dfrac{{dv}}{{dt}} = a, acceleration of body
F=ma\therefore F = ma It is also known as Newton’s law
If the mass of body is varying with time and velocity is constant then dvdt=0\dfrac{{dv}}{{dt}} = 0 equation (2) becomes
F=vdmdtF = v\dfrac{{dm}}{{dt}}, By this we also calculate the force
But the original form of Newton’s second law is F=dpdtF = \dfrac{{dp}}{{dt}}

So in this question option (A) is correct.

Note: Students may be confused with other options why F=maF = ma is not correct. As shown in above this formula comes only when mass of the body is constant. Means it is a special case of Newton’s second law in which mass is constant. If mass is varying then this formula is not valid.
But F=dpdtF = \dfrac{{dp}}{{dt}} is always valid. So the answer must be option A.