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 will follow these steps:

### Step 1: Understand the given information
We have a rocket in free space with a varying mass due to fuel being exhausted. The rate of change of mass is given by:
\[ \frac{dm(t)}{dt} = -bv^2(t) \]
where \( m(t) \) is the instantaneous mass, \( b \) is a constant, and \( v(t) \) is the instantaneous velocity of the rocket.

### Step 2: Apply Newton's second law
According to Newton's second law, the force acting on an object is equal to the rate of change of momentum:
\[ F = \frac{dp}{dt} \]
where \( p \) is the momentum of the rocket, given by \( p = m(t) v(t) \).

### Step 3: Differentiate momentum with respect to time
Using the product rule for differentiation, we can express the rate of change of momentum as:
\[ \frac{dp}{dt} = \frac{d(m(t)v(t))}{dt} = m(t) \frac{dv(t)}{dt} + v(t) \frac{dm(t)}{dt} \]

### Step 4: Substitute the rate of change of mass
From the problem, we know:
\[ \frac{dm(t)}{dt} = -bv^2(t) \]
Substituting this into the momentum equation gives:
\[ \frac{dp}{dt} = m(t) \frac{dv(t)}{dt} + v(t)(-bv^2(t)) \]
This simplifies to:
\[ \frac{dp}{dt} = m(t) \frac{dv(t)}{dt} - bv^3(t) \]

### Step 5: Set the thrust force equal to the change in momentum
The thrust force \( F \) exerted by the rocket is related to the velocity of the gases being ejected \( u \) (with respect to the rocket):
\[ F = u \frac{dm(t)}{dt} \]
Substituting the expression for \( \frac{dm(t)}{dt} \):
\[ F = u(-bv^2(t)) = -ubv^2(t) \]

### Step 6: Relate thrust force to acceleration
According to Newton's second law, the thrust force is also equal to the mass of the rocket times its acceleration \( a \):
\[ F = m(t)a \]
Setting the two expressions for force equal gives:
\[ m(t)a = -ubv^2(t) \]

### Step 7: Solve for acceleration
Rearranging the equation to solve for acceleration \( a \):
\[ a = \frac{-ubv^2(t)}{m(t)} \]
Since we are interested in the magnitude of acceleration, we can express it as:
\[ a = \frac{ubv^2(t)}{m(t)} \]

### Final Answer
The instantaneous acceleration of the rocket is:
\[ a = \frac{ubv^2(t)}{m(t)} \]