Question

A body of mass ${{m}_{1}}$​ exerts a force on another body of mass ${{m}_{2}}$​. If the magnitude of acceleration of ${{m}_{2}}$​ is ${{a}_{2}}$, then the magnitude of the acceleration of ${{m}_{1}}$​ is (considering only two bodies in space):
A: zero
B: $\dfrac{{{m}_{2}}{{a}_{2}}}{{{m}_{1}}}$
C: $\dfrac{{{m}_{1}}{{a}_{2}}}{{{m}_{2}}}$
D: ${{a}_{2}}$

Answer 1

Hint:We know that every action has an equal and opposite reaction. Hence, when the first body exerts a force on the second, it is capable of exerting an equal and opposite force on the first body. This is to be kept in mind while solving the problem.

Formulas used:
Newton’s third law of motion is used here. That is,
$|\overrightarrow{{{F}_{BA}}}|=|\overrightarrow{-{{F}_{AB}}}|$, that is the force exerted by A on B is equal and opposite to the force exerted on B by A.

Complete step by step answer:
In the question, we are given that a body of mass ${{m}_{1}}$​ exerts a force on another body of mass ${{m}_{2}}$​ and the magnitude of acceleration of ${{m}_{2}}$​ is${{a}_{2}}$.
We have to find the magnitude of the acceleration of ${{m}_{1}}$​ , that is ${{a}_{1}}$
We know that $F=ma$
By Newton’s third law of motion we know that
$|\overrightarrow{{{F}_{12}}}|=|\overrightarrow{-{{F}_{21}}}|$.
Where the force exerted by the first body on second ${{F}_{12}}={{m}_{1}}{{a}_{1}}$.
Similarly force exerted by the second body on first ${{F}_{21}}={{m}_{2}}{{a}_{2}}$.
This means, ${{m}_{1}}{{a}_{1}}={{m}_{2}}{{a}_{2}}$.
Therefore, the value of ${{a}_{1}}$ is $\dfrac{{{m}_{2}}{{a}_{2}}}{{{m}_{1}}}$.

Hence, option B is the correct answer among the given options.

Note:The solving procedure varies with the conditions in which the two bodies lie. Sometimes, when additional forces like frictional force and gravitational force come into play, the total forces acting on the system varies and the result also varies. The students should check whether additional conditions aren’t specified.