In the first second of its flight, rocket ejects 1/60 of its mass with a velocity of `2400 ms^(-1)`. The acceleration of the rocket is

Options

A. `19.6 ms^(-2)`
B. `30.2 ms^(-2)`
C. `40 ms^(-2)`
D. `49.8 ms^(-2)`

Explanation

To find the acceleration of the rocket, we can use the principle of conservation of momentum and the relationship between mass flow rate and thrust. Here’s a step-by-step solution:

### Step 1: Understand the problem
The rocket ejects a portion of its mass and we need to find its acceleration. The mass ejected is \( \frac{1}{60} \) of the total mass \( m \) of the rocket, and the velocity of the ejected mass is \( 2400 \, \text{m/s} \).

### Step 2: Define the variables
Let:
- \( m \) = total mass of the rocket
- \( \Delta m \) = mass ejected in 1 second = \( \frac{m}{60} \)
- \( v_e \) = velocity of the ejected mass = \( 2400 \, \text{m/s} \)

### Step 3: Calculate the thrust force
The thrust force \( F \) produced by the rocket can be calculated using the formula:
\[ F = v_e \cdot \frac{\Delta m}{\Delta t} \]
Since \( \Delta t = 1 \, \text{s} \), we have:
\[ F = v_e \cdot \frac{m}{60} \]
Substituting the values:
\[ F = 2400 \cdot \frac{m}{60} \]
\[ F = 40m \, \text{N} \]

### Step 4: Apply Newton's second law
According to Newton's second law, the net force acting on the rocket is equal to the mass of the rocket multiplied by its acceleration \( a \):
\[ F = m \cdot a \]
Setting the thrust force equal to the mass times acceleration:
\[ 40m = m \cdot a \]

### Step 5: Solve for acceleration
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ a = 40 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the rocket is \( 40 \, \text{m/s}^2 \).

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