/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 The first stage of a Saturn \(V\... [FREE SOLUTION] | 91Ó°ÊÓ

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The first stage of a Saturn \(V\) space vehicle consumed fuel and oxidizer at the rate of \(1.50 \times 10^{4} \mathrm{kg} / \mathrm{s}\), with an exhaust speed of \(2.60 \times 10^{3} \mathrm{m} / \mathrm{s} .\) (a) Calculate the thrust produced by these engines. (b) Find the acceleration of the vehicle just as it lifted off the launch pad on the Earth if the vehicle's initial mass was \(3.00 \times 10^{6} \mathrm{kg} .\) Note: You must include the gravitational force to solve part (b).

Short Answer

Expert verified
The thrust produced by the engines is \(3.90 \times 10^{7} N\) and the initial acceleration of the vehicle as it lifted off is \(3.2 m/s^2\).

Step by step solution

01

Force Calculation

Firstly, calculate the force which gives the thrust of the rocket. The momentum carried away by the propellant each second is the rate of mass loss of the propellant times its exhaust speed \(v_e\). Conservation of momentum means that the force on the rocket is equal to the force on the propellant coming out. Hence, the force can be calculated using the formula \(F = \Delta p/ \Delta t = v_e (dm/dt)\). Use the given values for exhaust speed \(v_e = 2.60 \times 10^{3} \mathrm{m} / \mathrm{s}\) and rate of mass loss\(dm/dt = 1.50 \times 10^{4} \mathrm{kg} / \mathrm{s}\) and substitute into the equation to find F = \(2.60 \times 10^{3} \times 1.50 \times 10^{4} = 3.90 \times 10^{7} N\).
02

Acceleration Calculation

Now, calculate the acceleration of the rocket. We can start by finding the force of gravity on the rocket which can be given by \(F_g = mg\), where m is mass of rocket and g is acceleration due to gravity. Substituting given values, \(F_g = 3.00 \times 10^{6} kg \times 9.8 m/s^2 = 2.94 \times 10^{7} N\). To find the acceleration, apply Newton’s \(2^{nd}\) law, using the net force which is the difference of the thrust force calculated in step 1 and the force of gravity calculated in this step. So, \(a = F_net/m\) gives \( a = (3.90 \times 10^{7} N - 2.94 \times 10^{7} N) / 3.00 \times 10^{6} kg = 3.2 m/s^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum
Understanding the conservation of momentum is essential when studying rocket propulsion. In the context of rockets, this principle states that if no external force acts on a system, the total momentum of that system remains constant over time. When a rocket burns fuel, the propellants are ejected out of the back at high speed, creating a backward momentum. To conserve momentum, the rocket gains forward momentum. This relationship can be represented by the formula \( F = \frac{\Delta p}{\Delta t} = v_e (\frac{dm}{dt}) \), where \( F \) is the force (thrust) produced, \( v_e \) is the exhaust velocity, and \( \frac{dm}{dt} \) is the rate of change of mass (mass flow rate).For the Saturn V vehicle, this is illustrated by the engine expelling fuel and oxidizer at a certain rate, generating thrust equal in magnitude but opposite in direction to the momentum carried away by the exhaust gases. This ensures the rocket accelerates upwards, away from Earth.
Rocket Propulsion
Rocket propulsion is a clear demonstration of Newton’s third law of motion, which states that for every action, there is an equal and opposite reaction. The engines of a rocket produce propulsion by expelling exhaust gases at high speed. The reaction to this action is that the rocket is pushed in the opposite direction. The thrust produced by the rocket's engines propels the rocket forward. This thrust can be calculated using the relationship between force and momentum previously mentioned, taking into account the rate at which the fuel is consumed and the speed at which it is expelled from the rocket.

Practical Application

For the Saturn V space vehicle, the first stage engines produce thrust by combusting fuel and oxidizer, then expelling the result at high speed. The force generated by these engines was calculated in step 1 of the provided solution, showing the practical application of the rocket propulsion concept.
Newton's Second Law
Going deeper into the mechanics of motion, Newton's second law is pivotal. This law can be stated as \( F = ma \), where \( F \) is net force acting on an object, \( m \) is the object's mass, and \( a \) is the object's acceleration. The law implies that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.In the context of the Saturn V space vehicle, to determine its acceleration at lift-off, one would have to account for the net force acting on it, which is the force of thrust minus the force of gravity pulling it down. The mass of the vehicle is a crucial factor, as the same thrust produces different accelerations on objects with different masses. Step 2 of the solution uses Newton’s second law to find the acceleration by dividing the net force by the rocket’s mass just as it lifts off the launch pad.
Gravitational Force
Gravitational force is an essential consideration for rockets launching from Earth or any celestial body. It is the attraction between objects with mass, and its magnitude can be calculated by the formula \( F_g = mg \), where \( F_g \) is the force of gravity, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity on the surface of the Earth (typically \(9.8 m/s^2\)).When a rocket like the Saturn V lifts off, it must overcome the force of gravity to reach space. The gravitational force acts to pull the vehicle back to Earth and must be subtracted from the thrust to find the net force available to accelerate the rocket upwards. As shown in step 2 of the solution, once the gravitational force is subtracted from the thrust, the result is used along with the mass of the rocket to calculate the initial acceleration off the launch pad using Newton’s second law.

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Most popular questions from this chapter

Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size \(\mathrm{C} 5\) model rocket engine has an average thrust of \(5.26 \mathrm{N},\) a fuel mass of \(12.7 \mathrm{g},\) and an initial mass of \(25.5 \mathrm{g} .\) The duration of its burn is \(1.90 \mathrm{s}\). (a) What is the average exhaust speed of the engine? (b) If this engine is placed in a rocket body of mass \(53.5 \mathrm{g},\) what is the final velocity of the rocket if it is fired in outer space? Assume the fuel burns at a constant rate.

An 80.0 -kg astronaut is working on the engines of his ship, which is drifting through space with a constant velocity. The astronaut, wishing to get a better view of the Universe, pushes against the ship and much later finds himself \(30.0 \mathrm{m}\) behind the ship. Without a thruster, the only way to return to the ship is to throw his \(0.500-\mathrm{kg}\) wrench directly away from the ship. If he throws the wrench with a speed of \(20.0 \mathrm{m} / \mathrm{s}\) relative to the ship, how long does it take the astronaut to reach the ship?

A golf club consists of a shaft connected to a club head. The golf club can be modeled as a uniform rod of length \(\ell\) and mass \(m_{1}\) extending radially from the surface of a sphere of radius \(R\) and mass \(m_{2} .\) Find the location of the club's center of mass, measured from the center of the club head.

There are (one can say) three coequal theories of motion: Newton's second law, stating that the total force on an object causes its acceleration; the work- kinetic energy theorem, stating that the total work on an object causes its change in kinetic energy; and the impulse-momentum theorem, stating that the total impulse on an object causes its change in momentum. In this problem, you compare predictions of the three theories in one particular case. A 3.00-kg object has velocity 7.00 \(\hat{\mathbf{j}}\) m/s. Then, a total force \(12.0 \hat{\mathbf{i}} \mathrm{N}\) acts on the object for 5.00 s. (a) Calculate the object's final velocity, using the impulse-momentum theorem. (b) Calculate its acceleration from \(\mathbf{a}=\left(\mathbf{v}_{f}-\mathbf{v}_{i}\right) / \Delta t\) (c) Calculate its acceleration from \(\mathbf{a}=\Sigma \mathbf{F} / m\) (d) Find the object's vector displacement from \(\Delta \mathbf{r}=\mathbf{v}_{i} t+\frac{1}{2} \mathbf{a} t^{2}\).(e) Find the work done on the object from \(W=\mathbf{F} \cdot \Delta \mathbf{r}\).(f) Find the final kinetic energy from \(\frac{1}{2} m v_{f}\space^{2}$$=\frac{1}{2} m \mathbf{v}_{f} \cdot \mathbf{v}_{f}\).(g) Find the final kinetic energy from \(\frac{1}{2} m v_{i}\space^{2}+W\).

In a slow-pitch softball game, a \(0.200-\mathrm{kg}\) softball crosses the plate at \(15.0 \mathrm{m} / \mathrm{s}\) at an angle of \(45.0^{\circ}\) below the horizontal. The batter hits the ball toward center field, giving it a velocity of \(40.0 \mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal. (a) Determine the impulse delivered to the ball. (b) If the force on the ball increases linearly for \(4.00 \mathrm{ms}\), holds constant for \(20.0 \mathrm{ms},\) and then decreases to zero linearly in another \(4.00 \mathrm{ms},\) what is the maximum force on the ball?

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