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Hint: Define centre of mass. Obtain the mathematical expression for acceleration of centre of mass. The acceleration on each ball thrown upward will be the acceleration due to gravity. Using these values on the equation for acceleration of centre of mass, we will get the required answer.

Complete step by step answer:

Centre of mass can be defined as a position of a system of masses where the weighted relative position of masses sums to zero.

Acceleration of the centre of mass of a system of mass can be expressed mathematically as,

$\begin{align}

& {{a}_{cm}}=\dfrac{{{a}_{1}}{{m}_{1}}+{{a}_{2}}{{m}_{2}}+{{a}_{3}}{{m}_{3}}+......+{{a}_{n}}{{m}_{n}}}{{{m}_{1}}+{{m}_{2}}+.....+{{m}_{n}}} \\

& {{a}_{cm}}M={{a}_{1}}{{m}_{1}}+{{a}_{2}}{{m}_{2}}+{{a}_{3}}{{m}_{3}}+......+{{a}_{n}}{{m}_{n}} \\

\end{align}$

Where, ${{a}_{cm}}$ is the acceleration of the centre of mass of the system.

Here, ${{a}_{1}}{{m}_{1}}$ is the net force on particle 1 and so on and ${{a}_{cm}}M$ is the force on the centre of mass of the system if all the mass is located at the COM of the system.

Now, in the above question, we have two balls thrown upward.

Let the mass of the balls are ${{m}_{1}}$ and ${{m}_{2}}$ .

Now, when we throw an object upward, gravity acts on it and the acceleration acts on it will be negative of acceleration due to gravity i.e. deceleration.

So, the acceleration on each ball will be g.

Applying these to the equation for acceleration of centre of mass,

$\begin{align}

& {{a}_{cm}}=\dfrac{{{m}_{1}}g+{{m}_{2}}g}{{{m}_{1}}+{{m}_{2}}} \\

& {{a}_{cm}}=\dfrac{\left( {{m}_{1}}+{{m}_{2}} \right)g}{{{m}_{1}}+{{m}_{2}}} \\

& {{a}_{cm}}=g \\

\end{align}$

So, the acceleration of the centre of mass of the two balls thrown upward will be g.

The correct option is (A).

Note: If we consider any number of balls or particles thrown upward, then also the acceleration for centre of mass will be g. This is because the acceleration acting on each ball will be equal to acceleration due to gravity and the acceleration on the centre of mass will be also equal to g.

Complete step by step answer:

Centre of mass can be defined as a position of a system of masses where the weighted relative position of masses sums to zero.

Acceleration of the centre of mass of a system of mass can be expressed mathematically as,

$\begin{align}

& {{a}_{cm}}=\dfrac{{{a}_{1}}{{m}_{1}}+{{a}_{2}}{{m}_{2}}+{{a}_{3}}{{m}_{3}}+......+{{a}_{n}}{{m}_{n}}}{{{m}_{1}}+{{m}_{2}}+.....+{{m}_{n}}} \\

& {{a}_{cm}}M={{a}_{1}}{{m}_{1}}+{{a}_{2}}{{m}_{2}}+{{a}_{3}}{{m}_{3}}+......+{{a}_{n}}{{m}_{n}} \\

\end{align}$

Where, ${{a}_{cm}}$ is the acceleration of the centre of mass of the system.

Here, ${{a}_{1}}{{m}_{1}}$ is the net force on particle 1 and so on and ${{a}_{cm}}M$ is the force on the centre of mass of the system if all the mass is located at the COM of the system.

Now, in the above question, we have two balls thrown upward.

Let the mass of the balls are ${{m}_{1}}$ and ${{m}_{2}}$ .

Now, when we throw an object upward, gravity acts on it and the acceleration acts on it will be negative of acceleration due to gravity i.e. deceleration.

So, the acceleration on each ball will be g.

Applying these to the equation for acceleration of centre of mass,

$\begin{align}

& {{a}_{cm}}=\dfrac{{{m}_{1}}g+{{m}_{2}}g}{{{m}_{1}}+{{m}_{2}}} \\

& {{a}_{cm}}=\dfrac{\left( {{m}_{1}}+{{m}_{2}} \right)g}{{{m}_{1}}+{{m}_{2}}} \\

& {{a}_{cm}}=g \\

\end{align}$

So, the acceleration of the centre of mass of the two balls thrown upward will be g.

The correct option is (A).

Note: If we consider any number of balls or particles thrown upward, then also the acceleration for centre of mass will be g. This is because the acceleration acting on each ball will be equal to acceleration due to gravity and the acceleration on the centre of mass will be also equal to g.

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