If the force on a rocket moving with a velocity of 300m/s is 345 N, then the rate of combustion of the fuel is

Options

A. 2.25kg/s
B. 0.75kg/s
C. 1.15 kg/s
D. 0.55kg/s

Explanation

To solve the problem of finding the rate of combustion of fuel for a rocket moving with a velocity of 300 m/s and experiencing a force of 345 N, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
We need to find the rate of combustion of fuel (dm/dt) for a rocket that is moving at a constant velocity (300 m/s) and is subjected to a force (345 N).

2. **Use the Formula for Force**:
The force acting on the rocket can be expressed in terms of the change in momentum:
\[ F = \frac{d(mv)}{dt} \]
where \( F \) is the force, \( m \) is the mass of the rocket, and \( v \) is its velocity.

3. **Apply the Product Rule**:
Since the mass of the rocket is changing (due to fuel combustion), we can apply the product rule:
\[ F = m \frac{dv}{dt} + v \frac{dm}{dt} \]

4. **Consider Constant Velocity**:
Given that the rocket is moving at a constant velocity, the acceleration is zero, which means:
\[ \frac{dv}{dt} = 0 \]
Therefore, the equation simplifies to:
\[ F = v \frac{dm}{dt} \]

5. **Rearrange the Equation**:
Rearranging the equation to solve for the rate of mass change (rate of combustion of fuel):
\[ \frac{dm}{dt} = \frac{F}{v} \]

6. **Substitute the Given Values**:
Substitute the values of force and velocity into the equation:
\[ \frac{dm}{dt} = \frac{345 \, \text{N}}{300 \, \text{m/s}} \]

7. **Calculate the Rate of Combustion**:
Performing the calculation:
\[ \frac{dm}{dt} = \frac{345}{300} = 1.15 \, \text{kg/s} \]

8. **Final Answer**:
The rate of combustion of the fuel is:
\[ \frac{dm}{dt} = 1.15 \, \text{kg/s} \]