A body of mass `2 kg` moving with a velocity `3 m//s` collides with a body of mass `1 kg` moving with a velocity of `4 m//s` in opposite direction. If the collision is head on and completely inelastic, then

Options

A. both particles move together with velocity `(2/3) m//s`
B. the momentum of system is `2 kg m//s` throughout
C. the momentum of system is `10 kg m//s`
D. the loss of `KE` of system is `(49//3) J`

Explanation

To solve the problem step by step, we will follow the principles of conservation of momentum and kinetic energy for a completely inelastic collision.

### Step 1: Identify the masses and velocities
- Mass of body 1, \( m_1 = 2 \, \text{kg} \)
- Velocity of body 1, \( v_1 = 3 \, \text{m/s} \)
- Mass of body 2, \( m_2 = 1 \, \text{kg} \)
- Velocity of body 2, \( v_2 = -4 \, \text{m/s} \) (negative because it is in the opposite direction)

### Step 2: Calculate the initial momentum
The total initial momentum \( p_{\text{initial}} \) of the system can be calculated using the formula:
\[ p_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot v_2 \]
Substituting the values:
\[ p_{\text{initial}} = (2 \, \text{kg} \cdot 3 \, \text{m/s}) + (1 \, \text{kg} \cdot -4 \, \text{m/s}) = 6 \, \text{kg m/s} - 4 \, \text{kg m/s} = 2 \, \text{kg m/s} \]

### Step 3: Apply the conservation of momentum
In a completely inelastic collision, the two bodies stick together after the collision. Let \( v' \) be the common velocity after the collision. The total momentum after the collision \( p_{\text{final}} \) is given by:
\[ p_{\text{final}} = (m_1 + m_2) \cdot v' \]
Setting the initial momentum equal to the final momentum:
\[ 2 \, \text{kg m/s} = (2 \, \text{kg} + 1 \, \text{kg}) \cdot v' \]
\[ 2 \, \text{kg m/s} = 3 \, \text{kg} \cdot v' \]
Solving for \( v' \):
\[ v' = \frac{2 \, \text{kg m/s}}{3 \, \text{kg}} = \frac{2}{3} \, \text{m/s} \]

### Step 4: Calculate the initial kinetic energy
The initial kinetic energy \( KE_{\text{initial}} \) of the system is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Calculating each term:
\[ KE_{\text{initial}} = \frac{1}{2} \cdot 2 \cdot (3)^2 + \frac{1}{2} \cdot 1 \cdot (-4)^2 \]
\[ = \frac{1}{2} \cdot 2 \cdot 9 + \frac{1}{2} \cdot 1 \cdot 16 \]
\[ = 9 + 8 = 17 \, \text{J} \]

### Step 5: Calculate the final kinetic energy
The final kinetic energy \( KE_{\text{final}} \) after the collision is:
\[ KE_{\text{final}} = \frac{1}{2} (m_1 + m_2) (v')^2 \]
Substituting the values:
\[ KE_{\text{final}} = \frac{1}{2} \cdot 3 \cdot \left(\frac{2}{3}\right)^2 \]
\[ = \frac{1}{2} \cdot 3 \cdot \frac{4}{9} = \frac{12}{18} = \frac{2}{3} \, \text{J} \]

### Step 6: Calculate the loss of kinetic energy
The loss of kinetic energy \( \Delta KE \) is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 17 \, \text{J} - \frac{2}{3} \, \text{J} \]
Converting \( 17 \) to a fraction:
\[ 17 = \frac{51}{3} \, \text{J} \]
Thus,
\[ \Delta KE = \frac{51}{3} - \frac{2}{3} = \frac{49}{3} \, \text{J} \]

### Final Answer
The loss of kinetic energy during the collision is \( \frac{49}{3} \, \text{J} \).

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