A bomb at rest explodes into two parts of masses `m_(1)` and `m_(2)`. If the momentums of the two parts be `P_(1)` and `P_(2)`, then their kinetic energies will be in the ratio of:

Options

A. `m_(1)//m_(2)`
B. `m_(2)//m_(1)`
C. `P_(1)//P_(2)`
D. `P_(2)//P_(1)`

Answer

To solve the problem of finding the ratio of the kinetic energies of two parts of a bomb that explodes into two masses \( m_1 \) and \( m_2 \), we can follow these steps:

### Step 1: Understand the Initial Conditions
The bomb is at rest before the explosion. Therefore, the initial momentum of the system is zero.

### Step 2: Apply Conservation of Momentum
According to the conservation of momentum, the total momentum before the explosion must equal the total momentum after the explosion. Since the initial momentum is zero, we have:
\[ P_1 + P_2 = 0 \]
This implies:
\[ P_1 = -P_2 \]

### Step 3: Express Kinetic Energy in Terms of Momentum
The kinetic energy \( E \) of an object can be expressed in terms of its momentum \( P \) and mass \( m \) as follows:
\[ E = \frac{P^2}{2m} \]
Thus, for the two parts, the kinetic energies \( E_1 \) and \( E_2 \) can be written as:
\[ E_1 = \frac{P_1^2}{2m_1} \]
\[ E_2 = \frac{P_2^2}{2m_2} \]

### Step 4: Find the Ratio of Kinetic Energies
To find the ratio of the kinetic energies \( \frac{E_1}{E_2} \), we can substitute the expressions for \( E_1 \) and \( E_2 \):
\[ \frac{E_1}{E_2} = \frac{\frac{P_1^2}{2m_1}}{\frac{P_2^2}{2m_2}} = \frac{P_1^2 \cdot m_2}{P_2^2 \cdot m_1} \]

### Step 5: Substitute the Relationship Between Momenta
Since \( P_1 = -P_2 \), we have \( P_1^2 = P_2^2 \). Therefore, we can simplify the ratio:
\[ \frac{E_1}{E_2} = \frac{m_2}{m_1} \]

### Conclusion
Thus, the ratio of the kinetic energies of the two parts after the explosion is:
\[ \frac{E_1}{E_2} = \frac{m_2}{m_1} \]

### Final Answer
The correct answer is option B: \( \frac{m_2}{m_1} \).

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