Book edition 4th edition

Author(s) Ruth W. Chabay, Bruce A. Sherwood

Pages 1135 pages

ISBN 9781118875865

Answers without the blur.

Just sign up for free and you're in.

Short Answer

A space station has the form of a hoop of radius R, with mass M. Initially its center of mass is not moving, but it is spinning. Then a small package of mass m is thrown by a spring-loaded gun toward a nearby spacecraft as shown in Figure 3.66; the package has a speed v after launch. Calculate the center-of-mass velocity (a vector) of the space station after the launch.

The horizontal and vertical components of the center of the mass of the satellite are $v_{x} = \frac{m}{M} (vcosθ)$ and $v_{y} = \frac{m}{M} (vsinθ)$ , respectively.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

The mass of the launched package is m

The mass of the space station is M

The radius of the space station is M

The initial speed of the space station is 0

The initial speed of the launched package is $\vec{v}$

Step 2: Conservation of linear momentum

The linear momentum remains conserved in an elastic collision, therefore, if two masses $m_{1}, and m_{2}$ have the initial velocities of $v_{1}, and v_{2}$ and final velocities of $v_{3}, and v_{4}$ , then according to the law of conservation of momentum, we will have,

$m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{3} + m_{2} v_{4}$

Step 3: Conservation of momentum in the x-direction

Horizontal components of the launched package are $vcosθ$ , while the space station has the speeds before and after the launch $0 and v_{x}$ , respectively.

Therefore, according to the law of conservation of linear momentum, we will have,

$mvcosθ = {Mv}_{x} {Mv}_{x} = mv cosθ v_{x} = \frac{m}{M} (vcosθ)$

Step 4: Conservation of momentum in the y-direction

Vertical components of the launched package are $vsinθ$ , while the space station has the speeds before and after the launch $0 and v_{y}$ , respectively.

Therefore, according to the law of conservation of linear momentum, we will have,

$mvsinθ = {Mv}_{y} {Mv}_{y} = mv sinθ v_{y} = \frac{m}{M} (vsinθ)$

Thus, the horizontal and vertical components of the center of the mass of the satellite are $v_{x} = \frac{m}{M} (vcosθ)$ and $v_{y} = \frac{m}{M} (vsinθ)$ , respectively.

Find Study Materials for

Create Study Materials

Select your language

Matter & Interactions

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given

Step 2: Conservation of linear momentum

Step 3: Conservation of momentum in the x-direction

Step 4: Conservation of momentum in the y-direction

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Matter & Interactions

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given

Step 2: Conservation of linear momentum

Step 3: Conservation of momentum in the x-direction

Step 4: Conservation of momentum in the y-direction

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics