A stationary body explodes into two fragments of masses `m_(1)` and `m_(2)`. If momentum of one fragment is `p`, the energy of explosion is

Options

A. `(p^(2))/(2(m_(1)+m_(2)))`
B. `(p^(2))/(2sqrt(m_(1)m_(2)))`
C. `(p^(2)(m_(1)+m_(2)))/(2m_(1)m_(2))`
D. `(p^(2))/(2(m_(1)-m_(2)))`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the conservation of momentum
Since the body is initially stationary, its initial momentum is zero. According to the law of conservation of momentum, the total momentum after the explosion must also be zero. Therefore, if one fragment has momentum \( p \), the second fragment must have momentum \( -p \) to ensure the total momentum remains zero.

### Step 2: Define the momenta of the fragments
Let:
- \( m_1 \) be the mass of the first fragment.
- \( m_2 \) be the mass of the second fragment.
- The momentum of the first fragment is \( p \).
- The momentum of the second fragment is \( -p \).

From the momentum definition, we have:
- For the first fragment: \( p = m_1 v_1 \) (where \( v_1 \) is the velocity of the first fragment).
- For the second fragment: \( -p = m_2 v_2 \) (where \( v_2 \) is the velocity of the second fragment).

### Step 3: Relate velocities to momenta
From the equations above, we can express the velocities in terms of the momenta:
- \( v_1 = \frac{p}{m_1} \)
- \( v_2 = -\frac{p}{m_2} \)

### Step 4: Calculate the kinetic energies of the fragments
The kinetic energy (KE) of each fragment can be calculated using the formula \( KE = \frac{1}{2} mv^2 \).

For the first fragment:
\[ KE_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 \left(\frac{p}{m_1}\right)^2 = \frac{p^2}{2m_1} \]

For the second fragment:
\[ KE_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_2 \left(-\frac{p}{m_2}\right)^2 = \frac{1}{2} m_2 \left(\frac{p^2}{m_2^2}\right) = \frac{p^2}{2m_2} \]

### Step 5: Calculate the total energy of the explosion
The total energy of the explosion is the sum of the kinetic energies of both fragments:
\[ E = KE_1 + KE_2 = \frac{p^2}{2m_1} + \frac{p^2}{2m_2} \]
Factoring out \( \frac{p^2}{2} \):
\[ E = \frac{p^2}{2} \left(\frac{1}{m_1} + \frac{1}{m_2}\right) \]

### Final Result
Thus, the energy of the explosion is:
\[ E = \frac{p^2}{2} \left(\frac{1}{m_1} + \frac{1}{m_2}\right) \]