A comet fragment of mass `1.96 xx 10^13` kg is moving at `6.50 xx 10^4` m/s when it crashes into Callisto, a moon of Jupiter. The mass of Callisto is `1.08 xx 10^23` kg. The collision is completely inelastic.
How much kinetic energy was released in the collision ?

Options

A. `8.28xx10^22` J
B. `3.51xx10^27` J
C. `7.02xx10^27` J
D. `4.14xx10^22` J

Answer

To find the kinetic energy released in the collision between the comet fragment and Callisto, we will follow these steps:

### Step 1: Calculate the initial kinetic energy of the comet fragment.
The initial kinetic energy (KE_initial) of the comet fragment can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 \]
where:
- \( m = 1.96 \times 10^{13} \, \text{kg} \) (mass of the comet fragment)
- \( u = 6.50 \times 10^{4} \, \text{m/s} \) (initial velocity of the comet fragment)

Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times (1.96 \times 10^{13}) \times (6.50 \times 10^{4})^2 \]

Calculating \( (6.50 \times 10^{4})^2 \):
\[ (6.50 \times 10^{4})^2 = 4.225 \times 10^{9} \]

Now substituting this back into the kinetic energy formula:
\[ KE_{\text{initial}} = \frac{1}{2} \times (1.96 \times 10^{13}) \times (4.225 \times 10^{9}) \]
\[ KE_{\text{initial}} = 4.14 \times 10^{22} \, \text{J} \]

### Step 2: Calculate the final velocity after the collision.
Since the collision is completely inelastic, both bodies will move together after the collision. We can use the conservation of momentum to find the final velocity \( V \).

The total momentum before the collision is equal to the total momentum after the collision:
\[ m u + M \cdot 0 = (M + m) V \]
where:
- \( M = 1.08 \times 10^{23} \, \text{kg} \) (mass of Callisto)

Rearranging gives:
\[ V = \frac{m u}{M + m} \]

Substituting the values:
\[ V = \frac{(1.96 \times 10^{13}) \times (6.50 \times 10^{4})}{(1.08 \times 10^{23} + 1.96 \times 10^{13})} \]

Since \( M \) is much larger than \( m \), we can approximate:
\[ V \approx \frac{(1.96 \times 10^{13}) \times (6.50 \times 10^{4})}{1.08 \times 10^{23}} \]

Calculating:
\[ V \approx \frac{1.274 \times 10^{18}}{1.08 \times 10^{23}} \approx 1.18 \times 10^{-5} \, \text{m/s} \]

### Step 3: Calculate the final kinetic energy after the collision.
The final kinetic energy (KE_final) can be calculated using:
\[ KE_{\text{final}} = \frac{1}{2} (M + m) V^2 \]

Since \( M \) is much larger than \( m \), we can approximate:
\[ KE_{\text{final}} \approx \frac{1}{2} M V^2 \]

Substituting the values:
\[ KE_{\text{final}} = \frac{1}{2} (1.08 \times 10^{23}) (1.18 \times 10^{-5})^2 \]
Calculating \( (1.18 \times 10^{-5})^2 \):
\[ (1.18 \times 10^{-5})^2 = 1.3924 \times 10^{-10} \]

Now substituting back:
\[ KE_{\text{final}} = \frac{1}{2} (1.08 \times 10^{23}) \times (1.3924 \times 10^{-10}) \approx 7.51 \times 10^{12} \, \text{J} \]

### Step 4: Calculate the kinetic energy released in the collision.
The kinetic energy released (ΔKE) is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} \]
Substituting the values:
\[ \Delta KE = (4.14 \times 10^{22}) - (7.51 \times 10^{12}) \approx 4.14 \times 10^{22} \, \text{J} \]

### Final Answer:
The kinetic energy released in the collision is approximately:
\[ \Delta KE \approx 4.14 \times 10^{22} \, \text{J} \]

---