(B) The initial velocity of the centre of mass $(V_{cm})$ is given by:$V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{1 \times 4 + 2 \times 0}{1

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The centre of mass first moves up and then comes down.

(B) The initial velocity of the centre of mass $(V_{cm})$ is given by:$V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{1 \times 4 + 2 \times 0}{1 + 2} = \frac{4}{3}\, m/s$ (upwards).Since both balls are under the influence of gravity,the acceleration of each ball is $g$ downwards.The acceleration of the centre of mass $(a_{cm})$ is given by:$a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = \frac{1 \times (-g) + 2 \times (-g)}{1 + 2} = -g$ (downwards).Since the centre of mass has an initial upward velocity and a constant downward acceleration,it will first move up,reach a maximum height,and then come down.

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

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$A$ ball of mass $1\, kg$ is thrown up with an initial speed of $4\, m/s$. $A$ second ball of mass $2\, kg$ is released from rest from some height as shown in the figure. Select the correct alternative.

A
The centre of mass of the two balls comes down with acceleration $g/3$.
B
The centre of mass first moves up and then comes down.
C
The acceleration of the centre of mass is first $g$ downwards and then $g$ upwards.
D
The centre of mass of the two balls remains stationary.

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System of Particles and Rotational Motion

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Motion of Centre of Mass

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Two bodies of masses $2 \ kg$ and $4 \ kg$ are moving with velocities $20 \ m/s$ and $10 \ m/s$ towards each other due to mutual gravitational attraction. What is the velocity of their centre of mass (in $m/s$)?

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$A$ ball of mass $1\, kg$ is thrown up with an initial speed of $4\, m/s$. $A$ second ball of mass $2\, kg$ is released from rest from some height as shown in the figure. Select the correct alternative.

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