A rocket moving in free space has varying mass due to fuel exhausted `d m(t)/dt = -bv^2(t)` where m(t) = instantaneous mass, b = constant, v(t) = instantaneous velocity . If gases are ejected with velocity u, with respect to rocket, instantaneous acceleration of should be

Options

A. `ubv^2(t)/m(t)`
B. `ubv^2(t)/2 m(t)`
C. ubv(t)/ m(t)`
D. `ub/m(t)`

Answer

To solve the problem, we need to analyze the motion of a rocket with varying mass due to fuel exhaustion. The given information is that the rate of change of mass with respect to time is given by:

\[ \frac{dm(t)}{dt} = -bv^2(t) \]

where:
- \( m(t) \) is the instantaneous mass of the rocket,
- \( b \) is a constant,
- \( v(t) \) is the instantaneous velocity of the rocket.

Additionally, the gases are ejected with a velocity \( u \) with respect to the rocket. We need to find the instantaneous acceleration of the rocket.

### Step-by-Step Solution:

1. **Understanding the System**:
The rocket is a system with a varying mass. As the rocket burns fuel, its mass decreases. The thrust generated by the rocket is due to the ejection of gases.

2. **Applying Newton's Second Law**:
According to Newton's second law, the force acting on the rocket can be expressed as the rate of change of momentum:

\[ F = \frac{dp}{dt} \]

where \( p \) is the momentum of the rocket.

3. **Expressing Momentum**:
The momentum \( p \) of the rocket is given by:

\[ p = m(t) v(t) \]

Thus, the rate of change of momentum can be expressed using the product rule:

\[ \frac{dp}{dt} = \frac{dm}{dt} v + m \frac{dv}{dt} \]

4. **Substituting the Rate of Change of Mass**:
From the problem statement, we know:

\[ \frac{dm}{dt} = -bv^2 \]

Substituting this into the momentum equation gives:

\[ \frac{dp}{dt} = (-bv^2)v + m \frac{dv}{dt} \]

5. **Identifying Thrust Force**:
The thrust force \( F \) can also be expressed in terms of the ejected gases. The thrust force is given by:

\[ F = u \frac{dm}{dt} \]

Substituting for \( \frac{dm}{dt} \):

\[ F = u(-bv^2) \]

6. **Equating Forces**:
The thrust force is equal to the rate of change of momentum:

\[ u(-bv^2) = -bv^2 v + m \frac{dv}{dt} \]

Rearranging gives:

\[ m \frac{dv}{dt} = u(-bv^2) + bv^3 \]

7. **Finding Instantaneous Acceleration**:
The instantaneous acceleration \( a \) of the rocket is given by:

\[ a = \frac{dv}{dt} \]

Thus, we can express it as:

\[ a = \frac{u(-bv^2) + bv^3}{m} \]

Simplifying this, we get:

\[ a = \frac{u b v^2}{m} - \frac{b v^3}{m} \]

8. **Final Expression**:
The final expression for the instantaneous acceleration of the rocket is:

\[ a = \frac{ubv^2}{m} \]