An object of mass 5 kg and speed `10 ms^(-1)` explodes into two pieces of equal mass. One piece comes to rest. The kinetic. Energy added to the system during the explosion is :

Options

A. Zero
B. 50 J
C. 250 J
D. 500 J

Answer

To solve the problem step by step, we will follow the principles of conservation of momentum and kinetic energy calculations.

### Step 1: Understand the initial conditions
We have an object of mass \( m = 5 \, \text{kg} \) moving with a speed \( v = 10 \, \text{m/s} \).

### Step 2: Calculate the initial momentum
The initial momentum \( p_i \) of the object can be calculated using the formula:
\[ p_i = m \cdot v = 5 \, \text{kg} \cdot 10 \, \text{m/s} = 50 \, \text{kg m/s} \]

### Step 3: Determine the final mass after explosion
The object explodes into two pieces of equal mass, so each piece has a mass of:
\[ m_1 = m_2 = \frac{5 \, \text{kg}}{2} = 2.5 \, \text{kg} \]

### Step 4: Analyze the final conditions
One piece comes to rest, hence its velocity \( v_1 = 0 \). Let the velocity of the second piece be \( v_2 \).

### Step 5: Apply conservation of momentum
According to the conservation of momentum:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
Thus,
\[ 50 \, \text{kg m/s} = 2.5 \, \text{kg} \cdot 0 + 2.5 \, \text{kg} \cdot v_2 \]
This simplifies to:
\[ 50 = 2.5 v_2 \]
Solving for \( v_2 \):
\[ v_2 = \frac{50}{2.5} = 20 \, \text{m/s} \]

### Step 6: Calculate the initial kinetic energy
The initial kinetic energy \( KE_i \) of the object is given by:
\[ KE_i = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 5 \, \text{kg} \cdot (10 \, \text{m/s})^2 \]
Calculating this gives:
\[ KE_i = \frac{1}{2} \cdot 5 \cdot 100 = 250 \, \text{J} \]

### Step 7: Calculate the final kinetic energy
The final kinetic energy \( KE_f \) after the explosion is the sum of the kinetic energies of both pieces. Since one piece is at rest, its kinetic energy is zero:
\[ KE_f = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} \cdot 2.5 \, \text{kg} \cdot (20 \, \text{m/s})^2 \]
Calculating this gives:
\[ KE_f = \frac{1}{2} \cdot 2.5 \cdot 400 = 500 \, \text{J} \]

### Step 8: Calculate the kinetic energy added to the system
The kinetic energy added during the explosion is the difference between the final and initial kinetic energy:
\[ KE_{\text{added}} = KE_f - KE_i = 500 \, \text{J} - 250 \, \text{J} = 250 \, \text{J} \]

### Final Answer
The kinetic energy added to the system during the explosion is \( 250 \, \text{J} \).

---