Question

The relation $$\overrightarrow{F}=m\overrightarrow{a}$$ cannot be deduced from Newton’s second law, if
A. Force depends on time.
B. Momentum depends on time.
C. Acceleration depends on time.
D. Mass depends on time.

Answer 1

Hint: We can derive the equation of Newton’s second law and check which of the following statements is right. Newton’s second law can be stated as the rate of change of momentum is directly proportional to the force applied on the body. Momentum is given as the product of mass and velocity.
Formula used:
 $$\begin{array}{rl}& F={\displaystyle \frac{dp}{dt}}\\ & p=mv\end{array}$$

Complete answer:
Newton’s second law can be expressed as
$$F={\displaystyle \frac{dp}{dt}}$$
Where p is the momentum and it is given as
$$p=mv$$
Where m is mass of the body and v is velocity. Substituting value of p in the Newton’s second law we get
$$\begin{array}{rl}& F={\displaystyle \frac{d(mv)}{dt}}\\ & F=m{\displaystyle \frac{dv}{dt}}+v{\displaystyle \frac{dm}{dt}}\end{array}$$
The rate of change of velocity with respect to time is acceleration i.e.
$$a={\displaystyle \frac{dv}{dt}}$$
Hence, the relation $$F=ma$$ is valid only when the rate of change of mass with respect of time is zero i.e. $${\displaystyle \frac{dm}{dt}}=0$$.
Hence, the relation can’t be deduced when the mass depends on time.

So, the correct answer is “Option D”.

Additional Information:
Newton’s first law of motion states that a body will remain in rest or motion until and unless a force is applied on it.
Newton’s second law can also be given as the product of mass and acceleration. From second law we can also say that the acceleration of a body depends on the mass and the force acting on it.
Newton’s third law of motion states that every action has an equal and opposite reaction.

Note:
The direction of change in momentum is given by the direction of force applied on it. The second law also implies the conservation of momentum. As force is given by differentiation of momentum with respect to time, therefore when net force is zero then the differential is zero which implies that momentum is constant.