A shell of mass m is fired from a cannon at angle q to horizontal with a velocity V.The shell at the heighest point breaks into two fragments having masses in the ratio 2:3. The lighter fragment has a velocity zero immediately after explosion and falls vertically downward. the velocity of the other fragment just after explosion is 

Options

A. `5V cos q`
B. `3 V cos q`
C. `3/5 v cos q`
D. `5/3 v cos q`

Answer

To solve the problem, we will use the principle of conservation of momentum. Let's go through the steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the shell = \( m \)
- Angle of projection = \( \theta \)
- Initial velocity of the shell = \( V \)
- At the highest point, the vertical component of the velocity is zero, and the horizontal component is \( V \cos \theta \).

2. **Determine the Masses of the Fragments:**
- The shell breaks into two fragments with masses in the ratio of 2:3.
- Let the mass of the lighter fragment be \( m_1 = \frac{2}{5} m \) and the mass of the heavier fragment be \( m_2 = \frac{3}{5} m \).

3. **Analyze the Situation at the Highest Point:**
- At the highest point, the horizontal velocity of the shell is \( V \cos \theta \).
- The lighter fragment (mass \( m_1 \)) falls vertically downward, meaning its horizontal velocity after the explosion is 0.

4. **Apply Conservation of Momentum:**
- The total horizontal momentum before the explosion must equal the total horizontal momentum after the explosion.
- Before the explosion:
\[ \text{Initial momentum} = m \cdot (V \cos \theta) \]
- After the explosion:
\[ \text{Final momentum} = m_1 \cdot 0 + m_2 \cdot v_2 \]
- Where \( v_2 \) is the horizontal velocity of the heavier fragment after the explosion.

5. **Set Up the Equation:**
- From conservation of momentum:
\[ m \cdot (V \cos \theta) = 0 + \left(\frac{3}{5} m\right) v_2 \]

6. **Solve for \( v_2 \):**
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ V \cos \theta = \frac{3}{5} v_2 \]
- Rearranging gives:
\[ v_2 = \frac{5}{3} V \cos \theta \]

### Final Answer:
The velocity of the other fragment just after the explosion is:
\[ v_2 = \frac{5}{3} V \cos \theta \]