Question

Three bodies A, B, and C of masses 2 kg, 3 kg, and 5 kg respectively are projected simultaneously with the same speed from the roof of a tower. The body A is thrown vertically upwards, body B is thrown vertically downwards and body C is projected horizontally. The acceleration of the centre of mass of the system of three bodies is (Acceleration due to gravity \( = 10 \, m \, s^{-2} \))

• A. \( 6 \, m \, s^{-2} \)
• B. \( 10 \, m \, s^{-2} \)
• C. \( 8 \, m \, s^{-2} \)
• D. \( 12 \, m \, s^{-2} \)

Hint:

The launch directions, weights, and initial velocities of the bodies are extra data meant to complicate the problem. As long as gravity is the only external force acting on the components of a system, the center of mass will always accelerate at exactly \( g \).

Answer 1

The Correct Option is B

Concept: The acceleration of the center of mass \( (\vec{a}_{\text{cm}}) \) of any multi-body system depends entirely on the net external forces acting on the system: \[ \vec{a}_{\text{cm}} = \frac{m_1\vec{a}_1 + m_2\vec{a}_2 + \dots + m_n\vec{a}_n}{m_1 + m_2 + \dots + m_n} \]

Step 1: Analyzing internal vs external force variables.
Once all three bodies are launched into the air, they become free-falling objects. Neglecting atmospheric air resistance, the only external force acting on each body is its weight due to gravity. Therefore, every single body experiences an identical gravitational acceleration pointing straight down: \[ \vec{a}_1 = \vec{g}, \quad \vec{a}_2 = \vec{g}, \quad \vec{a}_3 = \vec{g} \]

Step 2: Substituting values into the center of mass equation.
\[ \vec{a}_{\text{cm}} = \frac{m_1\vec{g} + m_2\vec{g} + m_3\vec{g}}{m_1 + m_2 + m_3} = \frac{(m_1 + m_2 + m_3)\vec{g}}{m_1 + m_2 + m_3} = \vec{g} \] Thus, the acceleration of the center of mass is exactly equal to \( g = 10 \, m \, s^{-2} \).