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 \( m_2 \) with momenta \( P_1 \) and \( P_2 \), we can follow these steps:

### Step 1: Understand the conservation of momentum
Since the bomb is initially at rest, the total initial momentum is zero. When it explodes, the momentum of the two parts must balance out:
\[ P_1 + P_2 = 0 \]
This implies:
\[ P_1 = -P_2 \]

### Step 2: Express kinetic energy in terms of momentum
The kinetic energy \( K \) of an object can be expressed as:
\[ K = \frac{1}{2} mv^2 \]
We can also express \( v \) in terms of momentum \( P \):
\[ P = mv \implies v = \frac{P}{m} \]
Substituting this into the kinetic energy formula gives:
\[ K = \frac{1}{2} m \left(\frac{P}{m}\right)^2 = \frac{P^2}{2m} \]

### Step 3: Write the kinetic energies of the two parts
For the two masses, the kinetic energies can be expressed as:
\[ K_1 = \frac{P_1^2}{2m_1} \quad \text{and} \quad K_2 = \frac{P_2^2}{2m_2} \]

### Step 4: Find the ratio of the kinetic energies
The ratio of the kinetic energies \( \frac{K_1}{K_2} \) can be written as:
\[ \frac{K_1}{K_2} = \frac{\frac{P_1^2}{2m_1}}{\frac{P_2^2}{2m_2}} = \frac{P_1^2 \cdot m_2}{P_2^2 \cdot m_1} \]

### Step 5: Substitute \( P_1 \) and \( P_2 \)
Since \( P_1 = -P_2 \), we have:
\[ P_1^2 = P_2^2 \]
Thus, the ratio simplifies to:
\[ \frac{K_1}{K_2} = \frac{m_2}{m_1} \]

### Conclusion
The ratio of the kinetic energies of the two parts of the bomb after the explosion is:
\[ \frac{K_1}{K_2} = \frac{m_2}{m_1} \]