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At illustrates the physics principles in this problem. An astronaut in his space suit and with a propulsion unit (empty of its gas propellant) strapped to his back has a mass of \(146 \mathrm{~kg}\). During a space-walk, the unit, which has been completely filled with propellant gas, ejects some gas with a velocity of \(+32 \mathrm{~m} / \mathrm{s}\). As a result, the astronaut recoils with a velocity of \(-0.39 \mathrm{~m} / \mathrm{s}\). After the gas is ejected, the mass of the astronaut (now wearing a partially empty propulsion unit) is \(165 \mathrm{~kg}\). What percentage of the gas propellant in the completely filled propulsion unit was depleted?

Short Answer

Expert verified
The astronaut depleted approximately 9.36% of the gas propellant.

Step by step solution

01

Understand the conservation of momentum

The principle of conservation of momentum states that the total momentum of a system remains constant as long as no external forces are acting on it. In this problem, before the gas ejection, the total momentum is zero because the astronaut is stationary. When gas is ejected, the momentum of the gas and the astronaut's momentum in the opposite direction must add up to zero.
02

Set up the momentum equation

Let \( m_g \) represent the mass of the gas that was ejected. The momentum scenario can be described by the equation: \( m_a \cdot v_a + m_g \cdot v_g = 0 \), where \( m_a = 146 \mathrm{~kg} \) and \( v_a = -0.39 \mathrm{~m/s} \) (the astronaut's velocity after ejection), and \( v_g = 32 \mathrm{~m/s} \) (the velocity of the gas).
03

Solve for mass of the ejected gas

Rearranging the equation from Step 2, \( m_g = \frac{-m_a \cdot v_a}{v_g} \). Substituting in the values: \( m_g = \frac{-(146 \, \mathrm{kg}) \times (-0.39 \, \mathrm{m/s})}{32 \, \mathrm{m/s}} \approx 1.78 \, \mathrm{kg} \).
04

Calculate the initial and remaining mass of propellant

Initially, the astronaut's total mass was 165 kg after ejection, which includes the astronaut, propulsion unit (without propellant), and the remaining propellant. The difference in mass before ejection (146 kg when unit was empty) and after ejection is the ejected gas plus the remaining gas, which is \(165 - 146 = 19 \mathrm{~kg}\).
05

Determine percentage of gas depleted

The total initial mass of the gas was \(19 \mathrm{~kg} - m_g \). Calculate the percentage of gas used by dividing the mass of the gas ejected \(m_g = 1.78 \, \mathrm{kg}\) by the total initial gas mass and multiplying by 100 to get the percentage.

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Key Concepts

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

Momentum Equation
Momentum is a key concept in physics that helps explain how objects move and interact. It is represented by the equation: \[ p = m \cdot v \]where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity of an object. In this problem involving an astronaut and propulsion in space, the conservation of momentum comes into play.

Before the astronaut ejects the gas propellant, the system is stationary, meaning the total momentum is zero. When the gas is expelled, it gains momentum in one direction, while the astronaut recoils, gaining momentum in the opposite direction. These momenta must add up to zero, as per the principle of momentum conservation:
  • Initial total momentum = 0
  • Momentum after gas ejection: \( m_a \cdot v_a + m_g \cdot v_g = 0 \)
  • Here, \( m_a \) is the mass of the astronaut after ejection, \( v_a \) is the recoil velocity, \( m_g \) is the mass of the gas, and \( v_g \) is the velocity of the gas ejected.

This equation forms the basis for calculating how much gas was needed to propel the astronaut backward, solving for the mass of the ejected gas propellant.
Recoil Velocity
Recoil velocity is an important concept related to the reaction of an object when a force is applied in the opposite direction. It is essentially the speed and direction at which an object moves back after another object or force pushes against it.

In this scenario, as the gas propellant is ejected at a high velocity, it creates an equal and opposite reaction causing the astronaut and his propulsion unit to move in the reverse direction. This backward movement is referred to as recoil velocity.
  • The astronaut moved with a recoil velocity of \(-0.39 \, \text{m/s}\).
  • This movement is dictated by Newton's third law, which states every action has an equal and opposite reaction.
  • The calculated recoil velocity helps determine how much propulsion the ejected gas provides.

Understanding recoil velocity aids in analyzing how interactions between different masses affect their motion in a frictionless environment such as space.
Gas Propulsion
Gas propulsion is a method of moving an object by expelling gas in a specific direction, which creates an equal and opposite movement in the opposite direction of the expelled gas. This principle is especially useful in spaces like outer space, where traditional movement methods, like wheels, don't work.

In the astronaut’s case, gas is ejected to create a propulsion force. This force makes the astronaut move backward, and it's this specific movement that defines the essence of gas propulsion.
  • The propulsion unit was initially full of gas propellant.
  • The expelled gas created a velocity of \(+32 \, \text{m/s}\).
  • This jet action demonstrated by the propulsion unit is crucial in guiding motion in space.

Gas propulsion exploits Newton’s third law of motion, providing consistent and calculable movement through the ejection of mass. It's a foundational concept in astronaut dynamics as it applies directly to maneuvering without atmospheric resistance.
Astronaut Dynamics
Astronaut dynamics explore how astronauts move and function in space environments where unique forces come into play. In zero gravity, the rules change compared to when they're on Earth.

This exercise demonstrates several important aspects of astronaut dynamics:
  • Mass and motion: The astronaut's movement is influenced by their total mass, which includes their gear and any gas interacting with their system.
  • Understanding mass change: Before and after ejecting gas, the mass of the astronaut and equipment changes, which affects how they move.
  • Velocity adjustments: After gas ejection, the astronaut adjusts to a new velocity, showcasing how small actions can alter an astronaut's path drastically.
Learning these dynamics helps in designing better space missions and equipment, ensuring astronaut safety and mission success. Their ability to navigate effectively using available resources is a testament to the thorough understanding of both physics and engineering in space operations.

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Most popular questions from this chapter

At illustrates how to model a problem similar to this one. An automobile has a mass of \(2100 \mathrm{~kg}\) and a velocity of \(+17 \mathrm{~m} / \mathrm{s}\). It makes a rear-end collision with a stationary car whose mass is \(1900 \mathrm{~kg} .\) The cars lock bumpers and skid off together with the wheels locked. (a) What is the velocity of the two cars just after the collision? (b) Find the impulse (magnitude and direction) that acts on the skidding cars from just after the collision until they come to a halt. (c) If the coefficient of kinetic friction between the wheels of the cars and the pavement is \(\mu_{k}=0.68,\) determine how far the cars skid before coming to rest.

The drawing shows a human figure approximately in a sitting position. For purposes of this problem, there are three parts to the figure, and the center of mass of each one is shown in the drawing. These parts are: (1) the torso, neck, and head (total mass \(=41 \mathrm{~kg}\) ) with a center of mass located on the \(y\) axis at a point \(0.39 \mathrm{~m}\) above the origin, (2) the upper legs (mass \(=17 \mathrm{~kg}\) ) with a center of mass located on the \(x\) axis at a point \(0.17 \mathrm{~m}\) to the right of the origin, and (3) the lower legs and feet (total mass \(=9.9 \mathrm{~kg}\) ) with a center of mass located \(0.43 \mathrm{~m}\) to the right of and \(0.26 \mathrm{~m}\) below the origin. Find the \(x\) and \(y\) coordinates of the center of mass of the human figure. Note that the mass of the arms and hands (approximately \(12 \%\) of the wholebody mass) has been ignored to simplify the drawing.

Part \(a\) of the drawing shows a bullet approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part \(b\). Note that both blocks are moving after the collision with the bullet. (a) Can the conservation of linear momentum be applied to this three-object system, even though the second collision occurs a bit later than the first one? Justify your answer. Neglect any mass removed from the first block by the bullet. (b) Is the total kinetic energy of this three-body system conserved? If not, would the total kinetic energy after the collisions be greater than or smaller than that before the collisions? Justify your answer.

In a performance test two cars take the same time to accelerate from rest up to the same speed. Car A has a mass of \(1400 \mathrm{~kg}\), and car \(\mathrm{B}\) has a mass of \(1900 \mathrm{~kg}\). During the test, which car (a) has the greater change in momentum, (b) experiences the greater impulse, and (c) is acted upon by the greater net average force? In each case, give your reasoning.

Consult Concept Simulation \(7.1\) at in preparation for this problem. Two friends, \(\mathrm{Al}\) and Jo, have a combined mass of \(168 \mathrm{~kg}\). At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of \(0.90 \mathrm{~m} / \mathrm{s}\), while Jo moves off in the opposite direction at a speed of \(1.2 \mathrm{~m} / \mathrm{s}\). Assuming that friction is negligible, find Al's mass.

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