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 formula for acceleration. Here’s a step-by-step breakdown of how to find the increase in speed of the rocket:

### Step 1: Identify the given values
- Mass of the rocket, \( m_0 = 2000 \, \text{kg} \)
- Rate of mass expulsion, \( \frac{dm}{dt} = R = 0.5 \, \text{kg/s} \)
- Exhaust velocity, \( V_r = 400 \, \text{m/s} \)
- Time of expulsion, \( t = 5 \, \text{s} \)

### Step 2: Calculate the acceleration of the rocket
The formula for acceleration \( a \) in rocket propulsion is given by:
\[ a = \frac{R \cdot V_r}{m_0} - \frac{R}{m_0} \cdot t \]
Where:
- \( R \) is the rate of mass expulsion
- \( V_r \) is the exhaust velocity
- \( m_0 \) is the initial mass of the rocket

Substituting the values:
\[ a = \frac{0.5 \, \text{kg/s} \cdot 400 \, \text{m/s}}{2000 \, \text{kg}} - \frac{0.5 \, \text{kg/s}}{2000 \, \text{kg}} \cdot 5 \, \text{s} \]

Calculating the first term:
\[ \frac{0.5 \cdot 400}{2000} = \frac{200}{2000} = 0.1 \, \text{m/s}^2 \]

Calculating the second term:
\[ \frac{0.5}{2000} \cdot 5 = \frac{2.5}{2000} = 0.00125 \, \text{m/s}^2 \]

Now substituting back into the acceleration formula:
\[ a = 0.1 - 0.00125 = 0.09875 \, \text{m/s}^2 \]

### Step 3: Calculate the increase in speed
Using the formula for velocity, \( v = u + at \), where \( u \) is the initial velocity (which is 0 since the rocket starts from rest):
\[ v = 0 + (0.09875 \, \text{m/s}^2 \cdot 5 \, \text{s}) = 0.49375 \, \text{m/s} \]

### Step 4: Round the result
Rounding the result gives:
\[ v \approx 0.5 \, \text{m/s} \]

### Final Answer
The increase in the speed of the rocket in this time is approximately **0.5 m/s**.

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