A bomb is kept stationary at a point. It suddenly explodes into two fragments of masses 1 g and 3 g. the total K.E. of the fragments is `6.4xx10^(4)J`. What is the K.E. of the smaller fragment

Options

A. `2.5xx10^(4)J`
B. `3.5xx10^4J`
C. `4.8xx10^(4)J`
D. `5.2xx10^(4)J`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
A bomb explodes into two fragments with masses 1 g (0.001 kg) and 3 g (0.003 kg). The total kinetic energy (K.E.) of the fragments after the explosion is given as \(6.4 \times 10^4 \, \text{J}\). We need to find the kinetic energy of the smaller fragment (1 g).

### Step 2: Conservation of Momentum
Since the bomb was initially at rest, the total initial momentum is zero. According to the law of conservation of momentum, the total momentum after the explosion must also be zero:
\[ m_1 v_1 + m_2 v_2 = 0 \]
Where:
- \(m_1 = 0.001 \, \text{kg}\) (mass of the smaller fragment)
- \(m_2 = 0.003 \, \text{kg}\) (mass of the larger fragment)
- \(v_1\) = velocity of the smaller fragment
- \(v_2\) = velocity of the larger fragment

From this, we can express \(v_1\) in terms of \(v_2\):
\[ 0.001 v_1 + 0.003 v_2 = 0 \implies v_1 = -3 v_2 \]

### Step 3: Express Kinetic Energies
The kinetic energy of each fragment can be expressed as:
\[ \text{K.E. of smaller fragment} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} (0.001) v_1^2 \]
\[ \text{K.E. of larger fragment} = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (0.003) v_2^2 \]

### Step 4: Substitute \(v_1\) in K.E. Equation
Substituting \(v_1 = -3 v_2\) into the kinetic energy of the smaller fragment:
\[ \text{K.E. of smaller fragment} = \frac{1}{2} (0.001) (-3 v_2)^2 = \frac{1}{2} (0.001) (9 v_2^2) = 0.0045 v_2^2 \]

### Step 5: Total Kinetic Energy
The total kinetic energy is given as:
\[ \text{Total K.E.} = \text{K.E. of smaller fragment} + \text{K.E. of larger fragment} \]
\[ 6.4 \times 10^4 = 0.0045 v_2^2 + \frac{1}{2} (0.003) v_2^2 \]
\[ = 0.0045 v_2^2 + 0.0015 v_2^2 = 0.006 v_2^2 \]

### Step 6: Solve for \(v_2^2\)
Now, we can solve for \(v_2^2\):
\[ 0.006 v_2^2 = 6.4 \times 10^4 \]
\[ v_2^2 = \frac{6.4 \times 10^4}{0.006} = 1.06667 \times 10^7 \]

### Step 7: Calculate K.E. of Smaller Fragment
Now we can find the kinetic energy of the smaller fragment:
\[ \text{K.E. of smaller fragment} = 0.0045 v_2^2 = 0.0045 \times 1.06667 \times 10^7 \]
\[ = 4.8 \times 10^4 \, \text{J} \]

### Final Answer
The kinetic energy of the smaller fragment (1 g) is:
\[ \text{K.E.} = 4.8 \times 10^4 \, \text{J} \]

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