/*! 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 9 A space probe is traveling in ou... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A space probe is traveling in outer space with a momentum that has a magnitude of 7.5 107 kg m/s. A retrorocket is fired to slow down the probe. It applies a force to the probe that has a magnitude of 2.0 106 N and a direction opposite to the probe’s motion. It fires for a period of 12 s. Determine the momentum of the probe after the retrorocket ceases to fire.

Short Answer

Expert verified
The final momentum of the probe is \(5.1 \times 10^7\, \text{kg m/s}\).

Step by step solution

01

Identify Initial Momentum

The initial momentum of the probe is given as \(7.5 \times 10^7\, \text{kg m/s}\).
02

Determine Change in Momentum

The change in momentum is equal to the force times the time interval over which the force is applied. Here, force \(F = 2.0 \times 10^6\, \text{N}\) and time \(t = 12\, \text{s}\). Calculate the change in momentum \(\Delta p = F \times t\).
03

Calculate Change in Momentum

\[ \Delta p = 2.0 \times 10^6\, \text{N} \times 12\, \text{s} = 2.4 \times 10^7\, \text{kg m/s} \]
04

Determine Final Momentum

Since the force is applied in the opposite direction of the initial momentum, subtract the change in momentum from the initial momentum: \[ p_{\text{final}} = 7.5 \times 10^7\, \text{kg m/s} - 2.4 \times 10^7\, \text{kg m/s} \]
05

Calculate Final Momentum

\[ p_{\text{final}} = 5.1 \times 10^7\, \text{kg m/s} \]

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.

Force Application
Imagine you're on a space probe floating through the vast emptiness of space. Everything is calm until a retrorocket fires, creating a force that slows the probe down. Force application is a concept that explains how force is used to change the motion of an object. In this case, the retrorocket exerts a specific amount of force, described in Newtons, opposite the direction of the probe's original motion.

Think of Newton's second law of motion, which helps us understand how force is linked to acceleration and mass. This principle tells us that when you apply a force to an object, it accelerates. For the space probe, the force causes negative acceleration, or deceleration, because it's moving in the opposite direction of the original motion.

Here's how to breakdown the impact of force application:
  • Magnitude of Force: This is the strength of the force, measured in Newtons (N). For the probe, the force is 2.0 million Newtons.
  • Direction of Force: This is crucial because force in the opposite direction reduces the speed. It's like pulling your bicycle handlebars backward to stop.
Momentum Change
Momentum change is central to this problem. Momentum, the product of an object's mass and velocity, tells us how much motion the object has. The retrorocket firing changes the probe's momentum. But how, exactly?

To calculate this momentum change, multiply the force applied by the time it acts on the object. In our situation, the formula \[\Delta p = F \times t\] applies, where \(\Delta p\) stands for change in momentum, \(F\) is the force, and \(t\) is the time. For the probe, \[\Delta p = 2.0 \times 10^6 \ \text{N} \times 12 \ \text{s} = 2.4 \times 10^7 \ \text{kg m/s}\]Understand the steps to see how momentum change influences the object's motion:

  • Initial Momentum: The probe starts with a momentum of 7.5 million kg m/s.
  • Change in Momentum: Calculated using the force and the time (2.4 million kg m/s in this problem).
  • Subtract Change: Reduces the initial momentum due to the opposite force direction, leading to a final momentum of 5.1 million kg m/s.
Momentum change elegantly reveals how forces, over time, alter an object's motion.
Outer Space Physics
Outer space presents a unique environment where everyday forces, like friction, aren't present. This is key to understanding how physics operates in the cosmos. Let's explore a few fascinating aspects of space physics that are relevant to our problem.

In space, objects continue moving in a straight line at a constant speed unless acted upon. This is due to the lack of opposing forces such as air resistance or friction, which you encounter on Earth. Hence, when the retrorocket was not firing, the probe maintained its motion due to inertia.

Consider these unique conditions of space:
  • Lack of Friction: With no atmosphere, there's no air resistance. Once you start moving, nothing slows you down unless another force acts.
  • Newton's First Law: An object in motion stays in motion with the same speed and direction unless acted upon by an unbalanced force. The retrorocket firing is that unbalanced force causing change.
In essence, mastering the forces and movements in outer space requires understanding how momentum conservation laws apply without traditional forces at play. This allows us to predict and analyze the motion of objects like our space probe precisely.

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

@ Object A is moving due east, while object \(\mathrm{B}\) is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of \(m_{\mathrm{A}}=17.0 \mathrm{kg}\) and an initial velocity of \(\vec{v}_{0}=8.00 \mathrm{m} / \mathrm{s}\) , due east. Object \(\mathrm{B}\) , however, has a mass of \(m_{\mathrm{B}}=29.0 \mathrm{kg}\) and an initial velocity of \(\overrightarrow{\mathrm{v}}_{\mathrm{oB}}=5.00 \mathrm{m} / \mathrm{s},\) due north. Find the magnitude and direction of the total momentum of the two-object system after the collision.

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is 95.0 \(\mathrm{kg}\) . The mass of the rock is 0.300 \(\mathrm{kg}\) . Initially the wagon is rolling forward at a speed of 0.500 \(\mathrm{m} / \mathrm{s}\) . Then the person throws the rock with a speed of 16.0 \(\mathrm{m} / \mathrm{s}\) . Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown directly forward in one case and directly backward in another.

The drawing shows a human figure 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 whole-body mass) has been ignored to simplify the drawing.

ssm A dump truck is being filled with sand. The sand falls straight downward from rest from a height of 2.00 m above the truck bed, and the mass of sand that hits the truck per second is 55.0 kg/s. The truck is parked on the platform of a weight scale. By how much does the scale reading exceed the weight of the truck and sand?

ssm \(\mathrm{A} 50.0\) -kg skater is traveling due east at a speed of 3.00 \(\mathrm{m} / \mathrm{s} . \mathrm{A}\) 70.0-kg skater is moving due south at a speed of 7.00 \(\mathrm{m} / \mathrm{s}\) . They collide and hold on to each other after the collision, managing to move off at an angle \(\theta\) south of east, with a speed of \(v_{\mathrm{f}}\) . Find \(\quad\) (a) the angle \(\theta\) and (b) the speed \(v_{\mathrm{f}},\) assuming that friction can be ignored.

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.