A ball of mass 5 kg moving with speed 8 m/s collides head on with another stationary ball of mass 15 kg. If collision is perfectly inelastic, then loss in kinetic energ is

Options

A. 160 J
B. 80 J
C. 40 J
D. 120 J

Answer

To solve the problem, we need to follow these steps:

### Step 1: Identify the given data
- Mass of ball 1 (m1) = 5 kg
- Initial velocity of ball 1 (u1) = 8 m/s
- Mass of ball 2 (m2) = 15 kg
- Initial velocity of ball 2 (u2) = 0 m/s (stationary)

### Step 2: Apply the law of conservation of momentum
In a perfectly inelastic collision, the two objects stick together after colliding. The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

The equation for conservation of momentum is:
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \]

Substituting the known values:
\[ 5 \times 8 + 15 \times 0 = (5 + 15) v \]
\[ 40 = 20v \]

### Step 3: Solve for the final velocity (v)
Now, we can solve for v:
\[ v = \frac{40}{20} = 2 \, \text{m/s} \]

### Step 4: Calculate the initial kinetic energy (Ki)
The initial kinetic energy (Ki) is given by the formula:
\[ K_i = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]

Substituting the values:
\[ K_i = \frac{1}{2} \times 5 \times 8^2 + \frac{1}{2} \times 15 \times 0^2 \]
\[ K_i = \frac{1}{2} \times 5 \times 64 + 0 \]
\[ K_i = \frac{1}{2} \times 320 = 160 \, \text{J} \]

### Step 5: Calculate the final kinetic energy (Kf)
The final kinetic energy (Kf) after the collision is given by:
\[ K_f = \frac{1}{2} (m_1 + m_2) v^2 \]

Substituting the values:
\[ K_f = \frac{1}{2} (5 + 15) \times 2^2 \]
\[ K_f = \frac{1}{2} \times 20 \times 4 \]
\[ K_f = \frac{1}{2} \times 80 = 40 \, \text{J} \]

### Step 6: Calculate the loss in kinetic energy (ΔK)
The loss in kinetic energy is calculated as:
\[ \Delta K = K_i - K_f \]
\[ \Delta K = 160 - 40 = 120 \, \text{J} \]

### Conclusion
The loss in kinetic energy during the perfectly inelastic collision is **120 Joules**.

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