A `5.0xx10^(4)` kg space probe is traveling at a speed of 11000 m/s through deep space. Auxiliary rockets are fired along the line of motion of reduce the probe's speed. The auxiliary rockets generate a force of `4.0xx10^(5)N` over a distance of 2500 km. What is the final speed of the probe ?

Options

A. 10000m/s
B. 8000m/s
C. 9000m/s
D. 7000m/s

Answer

To find the final speed of the space probe after the auxiliary rockets have fired, we can use the work-energy principle. The work done by the rockets will change the kinetic energy of the probe.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the space probe, \( m = 5.0 \times 10^4 \) kg
- Initial speed of the probe, \( u = 11000 \) m/s
- Force exerted by the rockets, \( F = 4.0 \times 10^5 \) N
- Distance over which the force is applied, \( d = 2500 \) km = \( 2500 \times 1000 = 2.5 \times 10^6 \) m

2. **Calculate the Work Done by the Rockets:**
The work done \( W \) by the rockets can be calculated using the formula:
\[ W = F \times d \]
Substituting the values:
\[ W = 4.0 \times 10^5 \, \text{N} \times 2.5 \times 10^6 \, \text{m} = 1.0 \times 10^{12} \, \text{J} \]

3. **Calculate the Initial Kinetic Energy:**
The initial kinetic energy \( KE_i \) of the probe is given by:
\[ KE_i = \frac{1}{2} m u^2 \]
Substituting the values:
\[ KE_i = \frac{1}{2} \times 5.0 \times 10^4 \, \text{kg} \times (11000 \, \text{m/s})^2 \]
\[ KE_i = \frac{1}{2} \times 5.0 \times 10^4 \times 1.21 \times 10^8 = 3.025 \times 10^{12} \, \text{J} \]

4. **Calculate the Final Kinetic Energy:**
The work done by the rockets will reduce the kinetic energy of the probe. Therefore, the final kinetic energy \( KE_f \) can be expressed as:
\[ KE_f = KE_i - W \]
Substituting the values:
\[ KE_f = 3.025 \times 10^{12} \, \text{J} - 1.0 \times 10^{12} \, \text{J} = 2.025 \times 10^{12} \, \text{J} \]

5. **Calculate the Final Speed:**
The final kinetic energy can also be expressed in terms of the final speed \( v \):
\[ KE_f = \frac{1}{2} m v^2 \]
Setting the two expressions for kinetic energy equal:
\[ 2.025 \times 10^{12} = \frac{1}{2} \times 5.0 \times 10^4 \times v^2 \]
Rearranging to solve for \( v^2 \):
\[ v^2 = \frac{2 \times 2.025 \times 10^{12}}{5.0 \times 10^4} \]
\[ v^2 = \frac{4.05 \times 10^{12}}{5.0 \times 10^4} = 8.1 \times 10^7 \]
Taking the square root:
\[ v = \sqrt{8.1 \times 10^7} \approx 9000 \, \text{m/s} \]

### Final Answer:
The final speed of the probe is approximately **9000 m/s**.