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.
Assuming for this calculation that Callisto's initial momentum is zero kg m/s, what is the recoil speed of Callisto immediately after the collision?

Options

A. `3.34xx10^(-18)` m/s
B. `3.58xx10^(-12)` m/s
C. `1.27xx10^(-14)` m/s
D. `1.18xx10^(-5)` m/s

Answer

To solve the problem of finding the recoil speed of Callisto after the inelastic collision with the comet fragment, we can follow these steps:

### Step 1: Understand the Problem
We have a comet fragment with mass \( m = 1.96 \times 10^{13} \) kg moving at a speed \( u = 6.50 \times 10^{4} \) m/s. Callisto, with mass \( M = 1.08 \times 10^{23} \) kg, is initially at rest. The collision is completely inelastic, meaning both bodies will move together after the collision.

### Step 2: Apply Conservation of Momentum
In a completely inelastic collision, the total momentum before the collision is equal to the total momentum after the collision. The momentum before the collision can be expressed as:
\[ \text{Initial Momentum} = m \cdot u + M \cdot 0 = m \cdot u \]
The momentum after the collision, when both bodies move together with a common velocity \( V \), is:
\[ \text{Final Momentum} = (m + M) \cdot V \]
Setting the initial momentum equal to the final momentum gives us:
\[ m \cdot u = (m + M) \cdot V \]

### Step 3: Solve for the Final Velocity \( V \)
Rearranging the equation to solve for \( V \):
\[ V = \frac{m \cdot u}{m + M} \]

### Step 4: Substitute the Values
Now we can substitute the known values into the equation:
\[ V = \frac{(1.96 \times 10^{13} \, \text{kg}) \cdot (6.50 \times 10^{4} \, \text{m/s})}{(1.96 \times 10^{13} \, \text{kg}) + (1.08 \times 10^{23} \, \text{kg})} \]

### Step 5: Simplify the Denominator
Since \( M \) is much larger than \( m \), we can approximate:
\[ m + M \approx M \]
Thus, the equation simplifies to:
\[ V \approx \frac{m \cdot u}{M} \]

### Step 6: Calculate \( V \)
Now we can calculate \( V \):
\[ V \approx \frac{(1.96 \times 10^{13}) \cdot (6.50 \times 10^{4})}{1.08 \times 10^{23}} \]
Calculating the numerator:
\[ 1.96 \times 10^{13} \cdot 6.50 \times 10^{4} = 1.274 \times 10^{18} \]
Now, substituting this back into the equation for \( V \):
\[ V \approx \frac{1.274 \times 10^{18}}{1.08 \times 10^{23}} \approx 1.18 \times 10^{-5} \, \text{m/s} \]

### Final Answer
The recoil speed of Callisto immediately after the collision is approximately:
\[ V \approx 1.18 \times 10^{-5} \, \text{m/s} \]

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