/*! 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 19 (II) A 125-kg astronaut (includi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule. \((a)\) What is the change in speed of the space capsule? \((b)\) If the push lasts 0.600 s, what is the average force exerted by each on the other? As the reference frame, use the position of the capsule before the push. \((c)\) What is the kinetic energy of each after the push?

Short Answer

Expert verified
(a) The capsule's speed changes by 0.1645 m/s. (b) Average force exerted is 520.83 N. (c) Astronaut: 390.63 J; Capsule: 25.70 J.

Step by step solution

01

Analyze the problem setup

We have two objects, the astronaut and the space capsule, initially at rest. The astronaut pushes off from the space capsule and gets a speed of 2.50 m/s. We need to apply the law of conservation of momentum and calculate forces and energies.
02

Apply conservation of momentum

According to the conservation of momentum, the initial momentum of the system (astronaut + capsule) is zero because both are initially at rest. After the push, their momenta must also add up to zero. Therefore, \( m_1v_1 + m_2v_2 = 0 \), where \( m_1 = 125 \, \text{kg} \), \( v_1 = 2.50 \, \text{m/s} \), and \( m_2 = 1900 \, \text{kg} \). Solving for \( v_2 \), the velocity of the space capsule, gives us \( 125 \times 2.50 + 1900v_2 = 0 \).
03

Calculate the velocity of the space capsule

Solving the equation from the previous step: \[1900v_2 = - 125 \times 2.50 \v_2 = \frac{- 312.5}{1900} \v_2 = -0.1645 \, \text{m/s}\]The negative sign indicates the capsule moves in the opposite direction to the astronaut.
04

Calculate the average force exerted

Using the impulse-momentum theorem, the change in momentum equals the impulse: \[ F \times \Delta t = m \Delta v \]For the astronaut, \[ F \times 0.600 = 125 \times 2.50 \F = \frac{312.5}{0.600} = 520.83 \, \text{N} \]For the capsule, the same magnitude of force acts in the opposite direction.
05

Calculate kinetic energies

Kinetic energy is given by \( KE = \frac{1}{2}mv^2 \). For the astronaut:\[KE_\text{astronaut} = \frac{1}{2} \times 125 \times (2.50)^2 = 390.625 \, \text{J}\]For the space capsule:\[KE_\text{capsule} = \frac{1}{2} \times 1900 \times (0.1645)^2 = 25.70 \, \text{J}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Impulse-Momentum Theorem
The impulse-momentum theorem is a fundamental idea in physics that explains how the momentum of an object changes when a force is applied over a period of time. This concept is crucial in our understanding of how the astronaut and the space capsule interact.

When the astronaut pushes away from the capsule, she experiences a change in momentum. This happens because her velocity changes from 0 to 2.50 m/s. According to the theorem, this change in momentum equals the impulse applied by the force over time. The formula for this is \[ F \times \Delta t = m \Delta v \]where:
  • \( F \) is the force applied
  • \( \Delta t \) is the time period over which the force acts
  • \( m \Delta v \) is the change in momentum
In our scenario, both the astronaut and the capsule experience this force, but in opposite directions, in line with Newton's Third Law. By calculating the impulse, we can determine the average force each applies to the other over the given time frame of 0.600 s. This gives us valuable insights into how the astronaut and capsule set their motion.
Kinetic Energy Calculation
Kinetic energy is the energy of motion. For any moving object, we calculate it using the formula:

\[ KE = \frac{1}{2}mv^2 \]Here, \( m \) is the mass and \( v \) is the velocity of the object. In our exercise, both the astronaut and the capsule have kinetic energy once they start moving after the push.

For the astronaut, the mass is 125 kg, moving at a speed of 2.50 m/s. Plugging into the formula, we calculate:\[ KE_{\text{astronaut}} = \frac{1}{2} \times 125 \times (2.50)^2 = 390.625 \text{ J} \]
The space capsule, significantly heavier at 1900 kg, moves slower; its velocity is -0.1645 m/s due to the push:\[ KE_{\text{capsule}} = \frac{1}{2} \times 1900 \times (0.1645)^2 = 25.70 \text{ J} \]
The differences here illustrate that a small object with high velocity can possess more kinetic energy than a large object moving slowly. These calculations help us understand how energy is split between the two objects after the push.
Newton's Third Law
Newton's Third Law of Motion is pivotal in understanding interactions like the push-off between the astronaut and the capsule. It states: "For every action, there is an equal and opposite reaction."

This means whenever the astronaut applies a force on the space capsule, the capsule applies an equal force back on her, but in the opposite direction.

In practical terms, this is why the astronaut gains speed in one direction, while the capsule moves in the opposite direction.
  • The force exerted by the astronaut on the capsule causes the capsule to accelerate in the opposite direction of her push.
  • Similarly, the force by the capsule ensures that the astronaut moves forward at a calculated 2.50 m/s.
This law not only helps explain the motions post-impact but also ensures that the total momentum remains conserved. Thus, the astronaut and capsule's combined effects don't just highlight their individual actions but underline a balanced interaction essential in zero-gravity environments.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(II) An object at rest is suddenly broken apart into two fragments by an explosion. One fragment acquires twice the kinetic energy of the other. What is the ratio of their masses?

(II) A 0.280-kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball. \((a)\) What is the mass of the second ball? \((b)\) What fraction of the original kinetic energy \((\Delta KE/KE)\) gets transferred to the second ball?

(III) An atomic nucleus of mass \(m\) traveling with speed \(\upsilon\) collides elastically with a target particle of mass \(2m\) (initially at rest) and is scattered at 90\(^{\circ}\). \((a)\) At what angle does the target particle move after the collision? \((b)\) What are the final speeds of the two particles? \((c)\) What fraction of the initial kinetic energy is transferred to the target particle?

Two astronauts, one of mass 55 kg and the other 85 kg, are initially at rest together in outer space. They then push each other apart. How far apart are they when the lighter astronaut has moved 12 m?

Astronomers estimate that a 2.0-km-diameter asteroid collides with the Earth once every million years. The collision could pose a threat to life on Earth. \((a)\) Assume a spherical asteroid has a mass of 3200 kg for each cubic meter of volume and moves toward the Earth at 15 km/s. How much destructive energy could be released when it embeds itself in the Earth? \((b)\) For comparison, a nuclear bomb could release about \(4.0 \times 10^{16} J\). How many such bombs would have to explode simultaneously to release the destructive energy of the asteroid collision with the Earth?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.