To solve the problem step by step, we will follow the principles of conservation of momentum and kinetic energy.
### Step 1: Determine the initial momentum of the system.
The initial momentum (p_initial) of the object can be calculated using the formula:
\[
p_{\text{initial}} = m \cdot v
\]
Where:
- \( m = 5 \, \text{kg} \) (mass of the object)
- \( v = 10 \, \text{m/s} \) (initial speed)
Calculating:
\[
p_{\text{initial}} = 5 \, \text{kg} \cdot 10 \, \text{m/s} = 50 \, \text{kg m/s}
\]
### Step 2: Determine the final momentum of the system.
After the explosion, the object splits into two pieces of equal mass (2.5 kg each). One piece comes to rest, which means its velocity is 0 m/s. Let the velocity of the other piece be \( v' \).
Using conservation of momentum:
\[
p_{\text{initial}} = p_{\text{final}}
\]
\[
50 \, \text{kg m/s} = (2.5 \, \text{kg} \cdot 0) + (2.5 \, \text{kg} \cdot v')
\]
This simplifies to:
\[
50 = 2.5 \cdot v'
\]
Solving for \( v' \):
\[
v' = \frac{50}{2.5} = 20 \, \text{m/s}
\]
### Step 3: Calculate the initial kinetic energy of the system.
The initial kinetic energy (KE_initial) can be calculated using the formula:
\[
KE_{\text{initial}} = \frac{1}{2} m v^2
\]
Calculating:
\[
KE_{\text{initial}} = \frac{1}{2} \cdot 5 \, \text{kg} \cdot (10 \, \text{m/s})^2
\]
\[
KE_{\text{initial}} = \frac{1}{2} \cdot 5 \cdot 100 = 250 \, \text{J}
\]
### Step 4: Calculate the final kinetic energy of the system.
The final kinetic energy (KE_final) consists of the kinetic energy of the moving piece (2.5 kg at 20 m/s) since the other piece is at rest:
\[
KE_{\text{final}} = \frac{1}{2} m' (v')^2
\]
Where \( m' = 2.5 \, \text{kg} \) and \( v' = 20 \, \text{m/s} \):
\[
KE_{\text{final}} = \frac{1}{2} \cdot 2.5 \, \text{kg} \cdot (20 \, \text{m/s})^2
\]
Calculating:
\[
KE_{\text{final}} = \frac{1}{2} \cdot 2.5 \cdot 400 = 500 \, \text{J}
\]
### Step 5: Calculate the kinetic energy added to the system during the explosion.
The kinetic energy added (KE_added) can be found by subtracting the initial kinetic energy from the final kinetic energy:
\[
KE_{\text{added}} = KE_{\text{final}} - KE_{\text{initial}}
\]
Calculating:
\[
KE_{\text{added}} = 500 \, \text{J} - 250 \, \text{J} = 250 \, \text{J}
\]
### Final Answer:
The kinetic energy added to the system during the explosion is **250 J**.
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