A block of mass `1 kg` moving with a speed of `4 ms^(-1)`, collides with another block of mass `2 kg` which is at rest. The lighter block comes to rest after collision. The loss in `KE` of the system is

Options

A. 8 J
B. `4xx10^(-7)J`
C. 4 J
D. 0 J

Answer

To solve the problem, we need to calculate the loss in kinetic energy of the system after the collision between the two blocks. Here are the steps to derive the solution:

### Step 1: Identify the given data
- Mass of the first block (m1) = 1 kg
- Initial velocity of the first block (u1) = 4 m/s
- Mass of the second block (m2) = 2 kg
- Initial velocity of the second block (u2) = 0 m/s (since it is at rest)

### Step 2: Calculate the initial kinetic energy (KE_initial)
The kinetic energy of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
For the first block:
\[ KE_{1\_initial} = \frac{1}{2} \times m_1 \times u_1^2 = \frac{1}{2} \times 1 \times (4)^2 = \frac{1}{2} \times 1 \times 16 = 8 \text{ J} \]
For the second block (since it is at rest):
\[ KE_{2\_initial} = \frac{1}{2} \times m_2 \times u_2^2 = \frac{1}{2} \times 2 \times (0)^2 = 0 \text{ J} \]
Thus, the total initial kinetic energy of the system is:
\[ KE_{initial} = KE_{1\_initial} + KE_{2\_initial} = 8 \text{ J} + 0 \text{ J} = 8 \text{ J} \]

### Step 3: Determine the final velocities after the collision
According to the problem, the lighter block (1 kg) comes to rest after the collision. Therefore:
- Final velocity of the first block (v1) = 0 m/s
- Let the final velocity of the second block (v2) be v.

Using the conservation of momentum:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the values:
\[ 1 \times 4 + 2 \times 0 = 1 \times 0 + 2 \times v \]
This simplifies to:
\[ 4 = 2v \]
Solving for v gives:
\[ v = 2 \text{ m/s} \]

### Step 4: Calculate the final kinetic energy (KE_final)
Now we can calculate the final kinetic energy of the system:
For the first block (which is at rest):
\[ KE_{1\_final} = \frac{1}{2} \times m_1 \times v_1^2 = \frac{1}{2} \times 1 \times (0)^2 = 0 \text{ J} \]
For the second block:
\[ KE_{2\_final} = \frac{1}{2} \times m_2 \times v_2^2 = \frac{1}{2} \times 2 \times (2)^2 = \frac{1}{2} \times 2 \times 4 = 4 \text{ J} \]
Thus, the total final kinetic energy of the system is:
\[ KE_{final} = KE_{1\_final} + KE_{2\_final} = 0 \text{ J} + 4 \text{ J} = 4 \text{ J} \]

### Step 5: Calculate the loss in kinetic energy
The loss in kinetic energy is given by:
\[ \text{Loss in KE} = KE_{initial} - KE_{final} = 8 \text{ J} - 4 \text{ J} = 4 \text{ J} \]

### Final Answer
The loss in kinetic energy of the system is **4 Joules**.

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