(C) The acceleration of the centre of mass $a_{cm}$ is given by the formula: $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$.Since both balls are in th

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is equal to $g$ (acceleration due to gravity)

(C) The acceleration of the centre of mass $a_{cm}$ is given by the formula: $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$.Since both balls are in the air,they are under the influence of gravity alone,so $a_1 = -g$ and $a_2 = -g$.Substituting these values: $a_{cm} = \frac{m_1(-g) + m_2(-g)}{m_1 + m_2}$.$a_{cm} = -g \frac{(m_1 + m_2)}{(m_1 + m_2)} = -g$.Therefore,the magnitude of the acceleration of the centre of mass is equal to $g$,which is the acceleration due to gravity.

Class 11 Physics System of Particles and Rotational Motion Motion of Centre of Mass

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Two balls are thrown simultaneously in the air. The acceleration of the centre of mass of the two balls when in the air,

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A
depends on the masses of the two balls
B
depends on the speeds of the two balls
C
is equal to $g$ (acceleration due to gravity)
D
depends on the direction of motion of the two balls.

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System of Particles and Rotational Motion

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Motion of Centre of Mass

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