A particle of mass `m` is thrown horizontally from the top of a tower and anoher particle of mass `2m` is thrown vertically upward. The acceleration of centre of mass is

Options

A. g
B. `(9g)/(3)`
C. `(2g)/(3)`
D. `(g)/(2)`

Answer

To find the acceleration of the center of mass of the system consisting of two particles, we can follow these steps:

### Step 1: Identify the masses and their accelerations
- Let the mass of the first particle (thrown horizontally) be \( m_1 = m \).
- Let the mass of the second particle (thrown vertically upward) be \( m_2 = 2m \).
- The acceleration of the first particle (horizontal throw) is \( \vec{a_1} = 0 \) (since it is thrown horizontally, it has no vertical acceleration initially).
- The acceleration of the second particle (thrown vertically upward) is affected by gravity, so \( \vec{a_2} = -g \hat{j} \) (where \( g \) is the acceleration due to gravity).

### Step 2: Use the formula for the acceleration of the center of mass
The acceleration of the center of mass \( \vec{a_{cm}} \) is given by the formula:
\[ \vec{a_{cm}} = \frac{m_1 \vec{a_1} + m_2 \vec{a_2}}{m_1 + m_2} \]

### Step 3: Substitute the values into the formula
Substituting the values we identified:
\[ \vec{a_{cm}} = \frac{m \cdot 0 + 2m \cdot (-g \hat{j})}{m + 2m} \]
This simplifies to:
\[ \vec{a_{cm}} = \frac{0 - 2mg \hat{j}}{3m} \]

### Step 4: Simplify the expression
Now, simplifying the expression:
\[ \vec{a_{cm}} = \frac{-2mg \hat{j}}{3m} = -\frac{2g}{3} \hat{j} \]

### Step 5: Conclusion
Thus, the acceleration of the center of mass is:
\[ \vec{a_{cm}} = -\frac{2g}{3} \hat{j} \]
This means the center of mass accelerates downward with an acceleration of \( \frac{2g}{3} \).