A 500kg rocket has to be fired vertically. Exhaust velocity of the gases is 1.96 km/s. Minimum mass of the fuel to be released in kg per second is

Options

A. 250
B. 25
C. 2.5
D. 50

Answer

To solve the problem of determining the minimum mass of fuel to be released per second for a 500 kg rocket with an exhaust velocity of 1.96 km/s, we can follow these steps:

### Step 1: Understand the Forces Acting on the Rocket
The rocket must overcome its weight (mg) to ascend. The weight of the rocket can be calculated using the formula:
\[ \text{Weight} = mg \]
where \( m = 500 \, \text{kg} \) (mass of the rocket) and \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

### Step 2: Calculate the Weight of the Rocket
Substituting the values:
\[ \text{Weight} = 500 \, \text{kg} \times 10 \, \text{m/s}^2 = 5000 \, \text{N} \]

### Step 3: Apply Newton's Third Law
According to Newton's third law, the force exerted by the exhaust gases must equal the weight of the rocket for it to ascend. The force due to the exhaust can be expressed as:
\[ F = v \cdot \frac{dm}{dt} \]
where:
- \( F \) is the force exerted by the exhaust,
- \( v \) is the exhaust velocity (1.96 km/s converted to m/s),
- \( \frac{dm}{dt} \) is the mass flow rate of the fuel (mass of fuel released per second).

### Step 4: Convert Exhaust Velocity
Convert the exhaust velocity from km/s to m/s:
\[ v = 1.96 \, \text{km/s} = 1.96 \times 1000 \, \text{m/s} = 1960 \, \text{m/s} \]

### Step 5: Set Up the Equation
Setting the force due to the exhaust equal to the weight of the rocket:
\[ mg = v \cdot \frac{dm}{dt} \]
Substituting the known values:
\[ 5000 \, \text{N} = 1960 \, \text{m/s} \cdot \frac{dm}{dt} \]

### Step 6: Solve for Mass Flow Rate
Rearranging the equation to solve for \( \frac{dm}{dt} \):
\[ \frac{dm}{dt} = \frac{5000 \, \text{N}}{1960 \, \text{m/s}} \]

### Step 7: Calculate \( \frac{dm}{dt} \)
Calculating the mass flow rate:
\[ \frac{dm}{dt} \approx \frac{5000}{1960} \approx 2.55 \, \text{kg/s} \]

### Step 8: Round the Result
Rounding to one decimal place, we find:
\[ \frac{dm}{dt} \approx 2.5 \, \text{kg/s} \]

### Conclusion
The minimum mass of the fuel to be released per second is approximately **2.5 kg/s**.

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