An object of mass 80 kg moving with velocity `2ms^(-1)` hit by collides with another object of mass 20 kg moving with velocity `4 ms ^(-1)` Find the loss of energy assuming a perfectly, inelastic collision

Options

A. `12J`
B. `24J`
C. `30J`
D. `32J`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the masses and velocities
- Mass of object 1 (m1) = 80 kg, velocity (u1) = 2 m/s
- Mass of object 2 (m2) = 20 kg, velocity (u2) = 4 m/s

### Step 2: Calculate the initial momentum
The initial momentum (p_initial) of the system can be calculated using the formula:
\[ p_{\text{initial}} = m_1 \cdot u_1 + m_2 \cdot u_2 \]
Substituting the values:
\[ p_{\text{initial}} = (80 \, \text{kg} \cdot 2 \, \text{m/s}) + (20 \, \text{kg} \cdot 4 \, \text{m/s}) \]
\[ p_{\text{initial}} = 160 \, \text{kg m/s} + 80 \, \text{kg m/s} = 240 \, \text{kg m/s} \]

### Step 3: Calculate the final velocity after the collision
In a perfectly inelastic collision, the two objects stick together. The total mass after the collision (m_total) is:
\[ m_{\text{total}} = m_1 + m_2 = 80 \, \text{kg} + 20 \, \text{kg} = 100 \, \text{kg} \]

Using conservation of momentum:
\[ p_{\text{initial}} = p_{\text{final}} \]
\[ 240 \, \text{kg m/s} = m_{\text{total}} \cdot V_f \]
Where \( V_f \) is the final velocity after the collision.
\[ 240 \, \text{kg m/s} = 100 \, \text{kg} \cdot V_f \]
Solving for \( V_f \):
\[ V_f = \frac{240 \, \text{kg m/s}}{100 \, \text{kg}} = 2.4 \, \text{m/s} \]

### Step 4: Calculate the initial kinetic energy
The initial kinetic energy (KE_initial) is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]
Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} (80 \, \text{kg}) (2 \, \text{m/s})^2 + \frac{1}{2} (20 \, \text{kg}) (4 \, \text{m/s})^2 \]
\[ KE_{\text{initial}} = \frac{1}{2} (80 \cdot 4) + \frac{1}{2} (20 \cdot 16) \]
\[ KE_{\text{initial}} = 160 \, \text{J} + 160 \, \text{J} = 320 \, \text{J} \]

### Step 5: Calculate the final kinetic energy
The final kinetic energy (KE_final) after the collision is:
\[ KE_{\text{final}} = \frac{1}{2} m_{\text{total}} V_f^2 \]
Substituting the values:
\[ KE_{\text{final}} = \frac{1}{2} (100 \, \text{kg}) (2.4 \, \text{m/s})^2 \]
\[ KE_{\text{final}} = \frac{1}{2} (100) (5.76) \]
\[ KE_{\text{final}} = 288 \, \text{J} \]

### Step 6: Calculate the loss of kinetic energy
The loss of kinetic energy (ΔKE) is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} \]
Substituting the values:
\[ \Delta KE = 320 \, \text{J} - 288 \, \text{J} = 32 \, \text{J} \]

### Conclusion
The loss of energy in the perfectly inelastic collision is **32 Joules**.

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