One projectile moving with velocity v in space, gets burst into 2 parts of masses in the ratio `1:3`. The smaller part becomes stationary. What is the velocity of thhe other part?

Options

A. 4v
B. v
C. `4v/3`
D. `3v/4`

Explanation

To solve the problem step-by-step, we will use the principles of conservation of momentum.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- A projectile with an initial mass \(4m\) is moving with a velocity \(v\).
- It breaks into two parts with masses in the ratio \(1:3\).
- The smaller part (mass \(m\)) becomes stationary (velocity = 0).
- We need to find the velocity of the larger part (mass \(3m\)).

2. **Initial Momentum Calculation:**
- The initial momentum \(P_i\) of the system can be calculated using the formula:
\[ P_i = \text{mass} \times \text{velocity} = 4m \cdot v \]

3. **Final Momentum Calculation:**
- After the explosion, the smaller part has a mass \(m\) and is stationary, so its momentum is:
\[ P_{\text{small}} = m \cdot 0 = 0 \]
- The larger part has a mass \(3m\) and we denote its velocity as \(V'\). Thus, its momentum is:
\[ P_{\text{large}} = 3m \cdot V' \]

4. **Applying Conservation of Momentum:**
- According to the law of conservation of momentum, the total initial momentum must equal the total final momentum:
\[ P_i = P_{\text{small}} + P_{\text{large}} \]
- Substituting the values we have:
\[ 4m \cdot v = 0 + 3m \cdot V' \]

5. **Solving for \(V'\):**
- Rearranging the equation gives:
\[ 4m \cdot v = 3m \cdot V' \]
- Dividing both sides by \(3m\):
\[ V' = \frac{4m \cdot v}{3m} \]
- The \(m\) cancels out:
\[ V' = \frac{4}{3} v \]

6. **Conclusion:**
- The velocity of the larger part after the explosion is:
\[ V' = \frac{4}{3} v \]

### Final Answer:
The velocity of the other part (larger mass) is \( \frac{4}{3} v \).

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