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A rocket is fired in deep space, where gravity is negligible. If the rocket has an initial mass of 6000 \(\mathrm{kg}\) and ejects gas at a relative velocity of magnitude 2000 \(\mathrm{m} / \mathrm{s}\) , how much gas must it eject in the first second to have an initial acceleration of 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) .

Short Answer

Expert verified
The rocket must eject 75 kg of gas in the first second.

Step by step solution

01

Define the Rocket Equation

By Newton's second law, the force on the rocket is equal to the mass times acceleration: \[ F = ma \]The thrust force due to the ejected gas is given by the momentum equation: \[ F = v \cdot \frac{dm}{dt} \]where \( v = 2000 \, \mathrm{m/s} \) is the velocity of the ejected gas, and \( \frac{dm}{dt} \) is the rate of change of mass.
02

Set Up the Equation

We want the acceleration \( a = 25.0 \, \mathrm{m/s^2} \). The initial mass \( m = 6000 \, \mathrm{kg} \). According to the rocket equation, the thrust force needed for this acceleration is:\[ m \cdot a = v \cdot \frac{dm}{dt} \]
03

Substitute Known Values

Substitute the known values into the equation: \[ 6000 \, \mathrm{kg} \times 25.0 \, \mathrm{m/s^2} = 2000 \, \mathrm{m/s} \times \frac{dm}{dt} \]
04

Solve for Mass Ejection Rate

Solve the equation for the mass ejection rate \( \frac{dm}{dt} \):\[ \frac{dm}{dt} = \frac{6000 \times 25}{2000} \]\[ \frac{dm}{dt} = 75 \, \mathrm{kg/s} \]
05

Find Total Ejected Mass in the First Second

Since we're interested in the amount of gas ejected in the first second, we multiply the rate by the time duration (1 second):\[ \Delta m = 75 \, \mathrm{kg} \times 1 \, \mathrm{s} = 75 \, \mathrm{kg} \]

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle that helps us understand how forces affect motion. It states that the force acting on an object is equal to the mass of the object multiplied by the acceleration it experiences. Mathematically, it is expressed as:
  • \[ F = ma \]
In the context of a rocket, the thrust force needed to propel it in space is derived from this principle.
This force results from the rocket's engine ejecting gas at high speed.
The resulting acceleration depends both on the mass of the rocket and the strength of the force exerted by the gas. In our rocket example, knowing the desired acceleration and mass allows us to calculate the required force through this law.
Acceleration
Acceleration is the rate at which an object changes its velocity. In simple terms, it's how quickly something speeds up or slows down. It is crucial in understanding rocket motion because the goal is to achieve a specific acceleration to control speed and direction.
Acceleration is calculated by dividing the force by mass, as shown in Newton’s second law:
  • \[ a = \frac{F}{m} \]
For the rocket, an initial acceleration of 25.0 \( \, \mathrm{m/s^2} \) was desired.
This tells us how fast the rocket needs to increase its velocity in the first second after ignition. Knowing this, we can adjust the force applied by manipulating the mass ejection rate.
Mass Ejection Rate
The mass ejection rate is how quickly mass is expelled from the rocket. This rate is crucial for generating the thrust needed to propel the rocket.
Ejecting mass results in a change of momentum which, in turn, generates a force as per Newton's laws. In equations, this is expressed as:
  • \[ F = v \cdot \frac{dm}{dt} \]
Here, \( \frac{dm}{dt} \) represents the mass ejection rate, and \( v \) is the velocity of the ejected gas.
In our scenario, to achieve an acceleration of 25.0 \( \, \mathrm{m/s^2} \), we found that the mass ejection rate must be 75 \( \, \mathrm{kg/s} \).
This means that for the rocket to accelerate as desired, 75 kilograms of gas must be expelled per second.
Momentum
Momentum is a measure of the quantity of motion of a moving object, which is determined by the product of its mass and velocity. For a rocket, altering momentum is critical because the ejection of gas changes the rocket's speed and direction. The key concept here is the conservation of momentum.
When gas is ejected backward, the rocket moves forward to conserve the total momentum. This relationship forms the foundation of the rocket equation:
  • \[ v \cdot \frac{dm}{dt} = m \cdot a \]
This principle ensures that the momentum change due to the gas ejection matches the needed force for the desired acceleration.
Through managing the momentum, rockets are able to achieve controlled flight and reach high speeds necessary for space travel.

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Most popular questions from this chapter

\(\cdot\) Two figure skaters, one weighing 625 \(\mathrm{N}\) and the other \(725 \mathrm{N},\) push off against each other on frictionless ice. (a) If the heavier skater travels at 1.50 \(\mathrm{m} / \mathrm{s}\) , how fast will the lighter one travel? (b) How much kinetic energy is "created" during the skaters' maneuver, and where does this energy come from?

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