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. `4 xx 10^(-7) J`
C. 4 J
D. 0 J

Explanation

To solve the problem step-by-step, we will calculate the loss in kinetic energy (KE) of the system after the collision.

### Step 1: Calculate the initial kinetic energy of the system
The initial kinetic energy (KE_initial) of the system can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
For the 1 kg block moving at 4 m/s:
\[ KE_{initial} = \frac{1}{2} \times 1 \, \text{kg} \times (4 \, \text{m/s})^2 = \frac{1}{2} \times 1 \times 16 = 8 \, \text{J} \]
The 2 kg block is at rest, so its kinetic energy is:
\[ KE_{2kg} = \frac{1}{2} \times 2 \, \text{kg} \times (0 \, \text{m/s})^2 = 0 \, \text{J} \]
Thus, the total initial kinetic energy of the system is:
\[ KE_{initial\_total} = KE_{initial} + KE_{2kg} = 8 \, \text{J} + 0 \, \text{J} = 8 \, \text{J} \]

### Step 2: Determine the final kinetic energy of the system
After the collision, the 1 kg block comes to rest, and we need to find the velocity of the 2 kg block. We can use the conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]
The initial momentum is:
\[ p_{initial} = (1 \, \text{kg} \times 4 \, \text{m/s}) + (2 \, \text{kg} \times 0) = 4 \, \text{kg m/s} \]
Let \( v_2 \) be the final velocity of the 2 kg block. The final momentum is:
\[ p_{final} = (1 \, \text{kg} \times 0) + (2 \, \text{kg} \times v_2) = 2v_2 \]
Setting the initial momentum equal to the final momentum gives:
\[ 4 = 2v_2 \implies v_2 = 2 \, \text{m/s} \]
Now we can calculate the final kinetic energy of the system:
\[ KE_{final} = KE_{1kg} + KE_{2kg} \]
The kinetic energy of the 1 kg block is:
\[ KE_{1kg} = \frac{1}{2} \times 1 \times (0)^2 = 0 \, \text{J} \]
The kinetic energy of the 2 kg block is:
\[ KE_{2kg} = \frac{1}{2} \times 2 \times (2)^2 = \frac{1}{2} \times 2 \times 4 = 4 \, \text{J} \]
Thus, the total final kinetic energy of the system is:
\[ KE_{final\_total} = 0 \, \text{J} + 4 \, \text{J} = 4 \, \text{J} \]

### Step 3: Calculate the loss in kinetic energy
The loss in kinetic energy (ΔKE) is given by:
\[ \Delta KE = KE_{initial\_total} - KE_{final\_total} \]
Substituting the values we found:
\[ \Delta KE = 8 \, \text{J} - 4 \, \text{J} = 4 \, \text{J} \]

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

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