A body of mass `4 kg` moving with velocity `12 m//s` collides with another body of mass `6 kg` at rest. If two bodies stick together after collision , then the loss of kinetic energy of system is

Options

A. zero
B. `288 J`
C. `172.8 J`
D. `144 J`

Explanation

To solve the problem step by step, we will follow the principles of conservation of momentum and the calculation of kinetic energy.

### Step 1: Identify the given data
- Mass of the first body, \( m_1 = 4 \, \text{kg} \)
- Velocity of the first body, \( v_1 = 12 \, \text{m/s} \)
- Mass of the second body, \( m_2 = 6 \, \text{kg} \)
- Velocity of the second body, \( v_2 = 0 \, \text{m/s} \) (at rest)

### Step 2: Calculate the initial momentum of the system
The initial momentum \( p_{initial} \) of the system can be calculated using the formula:
\[ p_{initial} = m_1 v_1 + m_2 v_2 \]
Substituting the values:
\[ p_{initial} = (4 \, \text{kg} \times 12 \, \text{m/s}) + (6 \, \text{kg} \times 0 \, \text{m/s}) = 48 \, \text{kg m/s} \]

### Step 3: Apply conservation of momentum to find the final velocity
After the collision, the two bodies stick together, so their combined mass \( m_f \) is:
\[ m_f = m_1 + m_2 = 4 \, \text{kg} + 6 \, \text{kg} = 10 \, \text{kg} \]
Let \( v_f \) be the final velocity of the combined mass. According to the conservation of momentum:
\[ p_{initial} = p_{final} \]
Thus,
\[ 48 \, \text{kg m/s} = 10 \, \text{kg} \times v_f \]
Solving for \( v_f \):
\[ v_f = \frac{48 \, \text{kg m/s}}{10 \, \text{kg}} = 4.8 \, \text{m/s} \]

### Step 4: Calculate the initial kinetic energy of the system
The initial kinetic energy \( KE_{initial} \) is given by:
\[ KE_{initial} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the values:
\[ KE_{initial} = \frac{1}{2} \times 4 \, \text{kg} \times (12 \, \text{m/s})^2 + \frac{1}{2} \times 6 \, \text{kg} \times (0 \, \text{m/s})^2 \]
Calculating:
\[ KE_{initial} = \frac{1}{2} \times 4 \times 144 + 0 = 288 \, \text{J} \]

### Step 5: Calculate the final kinetic energy of the system
The final kinetic energy \( KE_{final} \) is given by:
\[ KE_{final} = \frac{1}{2} m_f v_f^2 \]
Substituting the values:
\[ KE_{final} = \frac{1}{2} \times 10 \, \text{kg} \times (4.8 \, \text{m/s})^2 \]
Calculating:
\[ KE_{final} = \frac{1}{2} \times 10 \times 23.04 = 115.2 \, \text{J} \]

### Step 6: Calculate the loss of kinetic energy
The loss in kinetic energy \( \Delta KE \) is given by:
\[ \Delta KE = KE_{initial} - KE_{final} \]
Substituting the values:
\[ \Delta KE = 288 \, \text{J} - 115.2 \, \text{J} = 172.8 \, \text{J} \]

### Final Answer
The loss of kinetic energy of the system is \( 172.8 \, \text{J} \).

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