A 2000 kg rocket in free space expels 0.5 kg of gas per second at exhaust velocity `"400 m s"^(-1)` for 5 s. The increase in the speed of the rocket in this time is

Options

A. `2000ms^(-1)`
B. `200ms^(-1)`
C. `0.5ms^(-1)`
D. zero

Answer

To solve the problem, we will use the principles of rocket propulsion and the conservation of momentum. Here are the steps to find the increase in the speed of the rocket:

### Step 1: Identify the given values
- Mass of the rocket, \( M_0 = 2000 \, \text{kg} \)
- Rate of gas expulsion, \( \frac{dm}{dt} = 0.5 \, \text{kg/s} \)
- Exhaust velocity, \( V_r = 400 \, \text{m/s} \)
- Time of gas expulsion, \( t = 5 \, \text{s} \)

### Step 2: Calculate the total mass of gas expelled
The total mass of gas expelled over the time period can be calculated as:
\[ \Delta m = \frac{dm}{dt} \times t = 0.5 \, \text{kg/s} \times 5 \, \text{s} = 2.5 \, \text{kg} \]

### Step 3: Calculate the final mass of the rocket
The final mass of the rocket after expelling the gas is:
\[ M_f = M_0 - \Delta m = 2000 \, \text{kg} - 2.5 \, \text{kg} = 1997.5 \, \text{kg} \]

### Step 4: Use the rocket equation to find the change in velocity
Using the rocket equation:
\[ \Delta v = V_r \ln\left(\frac{M_0}{M_f}\right) \]
Substituting the values:
\[ \Delta v = 400 \, \text{m/s} \times \ln\left(\frac{2000}{1997.5}\right) \]

### Step 5: Calculate the natural logarithm
Calculating the logarithm:
\[ \ln\left(\frac{2000}{1997.5}\right) \approx \ln(1.00125) \approx 0.00125 \, \text{(using a calculator)} \]

### Step 6: Calculate the increase in speed
Now substituting back into the equation:
\[ \Delta v \approx 400 \, \text{m/s} \times 0.00125 \approx 0.5 \, \text{m/s} \]

### Final Answer
The increase in the speed of the rocket in this time is approximately \( 0.5 \, \text{m/s} \).

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