Suppose that a rocket is launched vertically and it is known that the exaust gases are emitted at a constant velocity of 18.5 m/s relative to the rocket,

Question

Suppose that a rocket is launched vertically and it is known that the exhaust gases are emitted at a constant velocity of 18.5 m/s relative to the rocket, the initial mass is 1.4 kg and we take the acceleration due to gravity to be 9.81 ms^−2

(a) If it is initially at rest, and after 0.6 seconds the vertical velocity is 3.79 m/s, then what is α, the rate at which it burns fuel, in kg/s?

Enter your answer to 2 decimal places.

(b) How long does it take until the fuel is all used up? Enter in seconds correct to 2 decimal places.

(c) If we assume that the mass of the shell is negligible, then what height would we expect the rocket to attain when all of the fuel is used up? Enter an answer in meters to decimal places.

(Hint: the solution of the DE doesn't apply when m(t)=0 but you can look at what happens as m(t)→0 The limit lim x -> 0^+ x ln x = 0 may be useful).



Enter in metres (to the nearest metre)

Answer 1

To solve this problem, we can use the momentum conservation principles and the rocket equation.

(a) To find the rate at which the rocket burns fuel (\( \alpha \)), we can use the rocket equation in the form:

\[ v(t) = v_e \ln\left(\frac{m_0}{m(t)}\right) - g t \]

Where:

- \( v(t) \) is the velocity of the rocket at time \( t \),

- \( v_e \) is the exhaust velocity relative to the rocket,

- \( m_0 \) is the initial mass of the rocket,

- \( m(t) \) is the mass of the rocket at time \( t \),

- \( g \) is the acceleration due to gravity.

Given that the rocket's velocity after 0.6 seconds is 3.79 m/s, we can rearrange the equation to solve for \( m(t) \) and subsequently find \( \alpha \) using the difference in mass over time.

(b) To find the time it takes until the fuel is used up, we can integrate the rate of fuel consumption over time to find when the mass of the rocket reaches the mass of the rocket's shell (assuming it's negligible, we're looking for when the mass of the fuel goes to zero).

(c) To find the maximum height attained by the rocket, we can use the energy principles or continue integrating the velocity over time until the fuel is depleted. The hint suggests using the limit \( \lim_{x \to 0^+} x \ln(x) = 0 \), which might come into play when evaluating the final height as the mass of the fuel approaches zero.

Let's start with part (a) by calculating \( \alpha \).

For part (a), the rate at which the rocket burns fuel (\( \alpha \)) is approximately \( 0.95 \) kg/s.

Now, let's proceed to part (b) and calculate the time until the fuel is used. We'll use the rate of fuel consumption (\( \alpha \)) to find when the mass of the rocket equals the mass of the shell (assuming it's negligible, we're considering the fuel to be the only mass that's being depleted).

Part (b) takes approximately \( 1.47 \) seconds until all the fuel is used up.

For part (c), the height that we expect the rocket to attain when all the fuel is used up is approximately \( 16.61 \) meters.

To summarize:

- (a) The rate at which the rocket burns fuel (\( \alpha \)) is \( 0.95 \) kg/s.

- (b) It takes \( 1.47 \) seconds until all the fuel is used up.

- (c) The expected height attained by the rocket when all of the fuel is used up is approximately \( 17 \) meters (rounded to the nearest meter).