/*! 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 A rocket is fired in deep space,... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Step by step solution

01

Understand the Problem

We need to calculate the mass of gas the rocket must eject to achieve a specified acceleration. The rocket's initial mass is \(6000\, \text{kg}\), the relative velocity of the ejected gas is \(2000\, \text{m/s}\), and the desired acceleration is \(25.0\, \text{m/s}^2\).
02

Use the Rocket Equation

The rocket equation is given by \(F = m_{eject} \times v_{rel}\), where \(m_{eject}\) is the mass of gas ejected per second, and \(v_{rel}\) is the relative velocity of the ejected gas. The force \(F\) can also be expressed as \(F = m_{initial} \times a\), where \(a\) is the acceleration.
03

Set Up the Equation

Equating the force from the rocket to the product of its mass and acceleration, we have:\[m_{eject} \times v_{rel} = m_{initial} \times a\]\[m_{eject} \times 2000 = 6000 \times 25.0\]
04

Solve for the Mass Ejected \(m_{eject}\)

Rearrange the equation to solve for \(m_{eject}\):\[m_{eject} = \frac{6000 \times 25.0}{2000}\]\[m_{eject} = \frac{150000}{2000}\]\[m_{eject} = 75\, \text{kg}\]
05

Conclusion

The rocket must eject \(75\, \text{kg}\) of gas in the first second to achieve an acceleration of \(25.0\, \text{m/s}^2\).

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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 fundamental in understanding motion and forces. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. This is expressed in the formula: \( F = ma \). Here, \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
This law is crucial in comprehending rocket motion because it lays down the principle that any change in motion (or acceleration) requires a force.
In our problem, Newton's Second Law guides us to find how much force is needed to propel the rocket by a specified acceleration amount. This force comes from the ejection of gas at high velocity.
Rocket equation
The Rocket Equation is a pivotal concept that explains how rockets can accelerate by expelling some of their mass. Formally, it is given by the equation:
  • \( F = m_{eject} \times v_{rel} \)
The equation states that the force \( F \) exerted by the rocket is equal to the mass of the ejected gas \( m_{eject} \) times the relative velocity \( v_{rel} \) of the gas.
In essence, this equation describes how the rocket generates thrust, using the mass it ejects.
It is the foundational equation of our exercise, connecting the velocity of the expelled gases to the force required to move the rocket with certain acceleration.
Mass ejection
Mass ejection in rocketry refers to the process of expelling mass to produce thrust. The mass of the rocket decreases over time as fuel and gases are expelled.
This principle is vital because rockets rely on the ejected mass to change their momentum, as described by the Rocket Equation.
In our exercise, the rocket must eject a specific amount of mass in the form of gas to achieve the desired acceleration. The calculation was based on relating the ejected mass and the speed at which it's expelled.
In this case, we calculated that the rocket has to eject \(75 \text{ kg}\) of gas per second to achieve an acceleration of \(25 \text{ m/s}^2\).
Acceleration calculation
Acceleration calculation in this context involves determining how much the velocity of the rocket changes per unit of time.
To achieve this, the Rocket Equation was used to equate the thrust produced by ejected gases to the necessary force needed for a specific acceleration.
The equation \( m_{eject} \times v_{rel} = m_{initial} \times a \) allows us to calculate the mass of gas that needs to be ejected by considering the initial mass of the rocket and the desired acceleration.
Solving for the ejected mass \( m_{eject} \), the exercise shows that the required rate of fuel or gas expulsion is correctly calculated to meet the acceleration target.
In summary, this approach demonstrates how calculations are used for precise adjustments in rocket velocity behavior.

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Most popular questions from this chapter

You (mass \(55 \mathrm{~kg}\) ) are riding your frictionless skateboard (mass \(5.0 \mathrm{~kg}\) ) in a straight line at a speed of \(4.5 \mathrm{~m} / \mathrm{s}\) when a friend standing on a balcony above you drops a \(2.5 \mathrm{~kg}\) sack of flour straight down into your arms. (a) What is your new speed, while holding the flour sack? (b) since the sack was dropped vertically, how can it affect your horizontal motion? Explain. (c) Suppose you now try to rid yourself of the extra weight by throwing the flour sack straight up. What will be your speed while the sack is in the air? Explain.

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