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

Step 1: Recall how centre of mass moves.
The acceleration of the centre of mass of a group of bodies depends only on the outside forces on the system. We use: \[ \vec{a}_{cm} = \frac{m_1\vec{a}_1 + m_2\vec{a}_2 + m_3\vec{a}_3}{m_1 + m_2 + m_3} \]
Step 2: Identify the outside force on each body.
Once each body leaves the hand, the only outside force on it is gravity. We ignore air. So each body has the same downward acceleration $g$, no matter which way it was thrown.
Step 3: Note the directions do not matter.
Body A goes up, B goes down, C goes sideways. But the only acceleration each feels is $g$ downward. Their throw directions change their paths, not their acceleration.
Step 4: Put equal accelerations into the formula.
Since $\vec{a}_1 = \vec{a}_2 = \vec{a}_3 = \vec{g}$: \[ \vec{a}_{cm} = \frac{(m_1 + m_2 + m_3)\vec{g}}{m_1 + m_2 + m_3} \]
Step 5: Cancel the masses.
The total mass cancels top and bottom: \[ \vec{a}_{cm} = \vec{g} \]
Step 6: State the answer.
So the centre of mass accelerates straight down at: \[ \boxed{a_{cm} = 10\ ms^{-2}} \]