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, we need to apply the principle of conservation of momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the initial conditions
We have a projectile with an initial mass \( M \) moving with a velocity \( v \). The total mass can be considered as \( 4m \) (where \( m \) is a unit mass), since the two parts are in the ratio \( 1:3 \).

### Step 2: Define the masses of the parts
Let the smaller mass be \( m_1 = m \) and the larger mass be \( m_2 = 3m \).

### Step 3: Apply the conservation of momentum
The momentum before the burst must equal the momentum after the burst.

- Initial momentum \( P_{initial} = (4m) \cdot v = 4mv \)
- After the burst, the smaller mass \( m_1 \) becomes stationary, so its momentum is \( 0 \).
- The larger mass \( m_2 \) will have some velocity \( v' \).

Thus, the final momentum \( P_{final} = m_1 \cdot 0 + m_2 \cdot v' = 3m \cdot v' \).

### Step 4: Set up the equation
Using the conservation of momentum:
\[ P_{initial} = P_{final} \]
\[ 4mv = 3m \cdot v' \]

### Step 5: Solve for \( v' \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 4v = 3v' \]
Now, solving for \( v' \):
\[ v' = \frac{4v}{3} \]

### Conclusion
The velocity of the larger part after the burst is \( \frac{4v}{3} \).