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

Options

A. both particles will move together with velocity `(2//3) m//sec` after collision
B. the momentum of the system is `2 kg m//sec` throughout
C. the momentum of the system is `10 kg m//sec` throughout
D. The loss in KE of the system is `(49//3)` Joule

Explanation

To solve the problem of two bodies colliding in a perfectly inelastic manner, we will follow these steps:

### Step 1: Identify the masses and velocities of the bodies
- Mass of body 1 (m1) = 2 kg, Velocity of body 1 (u1) = 3 m/s (moving in the positive direction)
- Mass of body 2 (m2) = 1 kg, Velocity of body 2 (u2) = -4 m/s (moving in the negative direction)

### Step 2: Calculate the initial momentum of the system
The initial momentum (P_i) can be calculated using the formula:
\[ P_i = m_1 \cdot u_1 + m_2 \cdot u_2 \]

Substituting the values:
\[ P_i = (2 \, \text{kg} \cdot 3 \, \text{m/s}) + (1 \, \text{kg} \cdot -4 \, \text{m/s}) \]
\[ P_i = 6 \, \text{kg m/s} - 4 \, \text{kg m/s} \]
\[ P_i = 2 \, \text{kg m/s} \]

### Step 3: Apply conservation of momentum
In a perfectly inelastic collision, the two bodies stick together after the collision. The final momentum (P_f) is given by:
\[ P_f = (m_1 + m_2) \cdot v_f \]
where \( v_f \) is the final velocity of the combined mass.

Since momentum is conserved:
\[ P_i = P_f \]
\[ 2 \, \text{kg m/s} = (2 \, \text{kg} + 1 \, \text{kg}) \cdot v_f \]
\[ 2 \, \text{kg m/s} = 3 \, \text{kg} \cdot v_f \]

### Step 4: Solve for the final velocity
Rearranging the equation to solve for \( v_f \):
\[ v_f = \frac{2 \, \text{kg m/s}}{3 \, \text{kg}} \]
\[ v_f = \frac{2}{3} \, \text{m/s} \]

### Step 5: Calculate the final momentum of the system
The final momentum (P_f) can be calculated as:
\[ P_f = (m_1 + m_2) \cdot v_f \]
\[ P_f = 3 \, \text{kg} \cdot \frac{2}{3} \, \text{m/s} \]
\[ P_f = 2 \, \text{kg m/s} \]

### Step 6: Calculate the initial and final kinetic energies
Initial kinetic energy (KE_i):
\[ KE_i = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]
\[ KE_i = \frac{1}{2} (2 \, \text{kg}) (3 \, \text{m/s})^2 + \frac{1}{2} (1 \, \text{kg}) (-4 \, \text{m/s})^2 \]
\[ KE_i = \frac{1}{2} (2) (9) + \frac{1}{2} (1) (16) \]
\[ KE_i = 9 + 8 = 17 \, \text{J} \]

Final kinetic energy (KE_f):
\[ KE_f = \frac{1}{2} (m_1 + m_2) v_f^2 \]
\[ KE_f = \frac{1}{2} (3 \, \text{kg}) \left(\frac{2}{3} \, \text{m/s}\right)^2 \]
\[ KE_f = \frac{1}{2} (3) \left(\frac{4}{9}\right) \]
\[ KE_f = \frac{3 \cdot 4}{18} = \frac{12}{18} = \frac{2}{3} \, \text{J} \]

### Step 7: Calculate the loss of kinetic energy
Loss of kinetic energy:
\[ \text{Loss} = KE_i - KE_f \]
\[ \text{Loss} = 17 \, \text{J} - \frac{2}{3} \, \text{J} \]
Converting 17 J to a fraction:
\[ 17 = \frac{51}{3} \]
\[ \text{Loss} = \frac{51}{3} - \frac{2}{3} = \frac{49}{3} \, \text{J} \]

### Summary of Results
- Final velocity after collision: \( \frac{2}{3} \, \text{m/s} \)
- Initial momentum: \( 2 \, \text{kg m/s} \)
- Final momentum: \( 2 \, \text{kg m/s} \)
- Loss of kinetic energy: \( \frac{49}{3} \, \text{J} \)