2.3: Conservation Laws

Last updated
Save as PDF

Page ID: 29883

Lumen Learning
Lumen Learning

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

Conservation Laws

Concepts and Principles

The Impulse-Momentum Relation

Just like the kinematic relations and Newton’s second law, the impulse-momentum relation is independently valid in any member of a set of perpendicular directions. Thus, we will typically apply the impulse-momentum relation in its component forms:

The Work-Energy Relation

From Model 1, our expression for the Work-Energy Relation, with gravitational potential energy terms, is:

It’s very important to remember that the work-energy relation is a scalar equation, meaning it cannot be broken into components and “solved” separately in the x-, y-, and z-directions. This is even more important to remember now that we are working in multiple dimensions. This observation results in two important points:

The work-energy relation involves the actual initial and final velocities, not their components. The kinetic energy of an object does not depend on the direction of travel of the object.
In the expression for work, , the product of the magnitude of the force and the magnitude of the displacement is multiplied by cos , where is defined to be the angle between the applied force and the displacement of the object. If the force and displacement are in the same direction = 0°, and the work is positive (the object gains energy). If the force and displacement are in the opposite direction = 180°, and the work is negative (the object loses energy). If the force and displacement are perpendicular, no work is done. Note that the actual directions of the force and the displacement are unimportant, only their directions relative to each other affect the work.

Analysis Tools

Applying the Impulse-Momentum Relation to a Single Object

Let’s investigate the following scenario:

A boy pulls a 30 kg sled, including the mass of his kid brother, along ice. The boy pulls on the tow rope, oriented at 600 above horizontal, with a force of 110 N until his kid brother begins to cry. Like clockwork, his brother always cries upon reaching a speed of 2.0 m/s. The frictional coefficient is (0.20,0.15).

To apply the impulse-momentum relation, you must clearly specify the initial and final events at which you will tabulate the momentum. For example:

Event 1: The instant before the sled begins to move.

P_1x = 0

P_1y = 0

Event 2: The instant the sled reaches 2.0 m/s

P_2x = 30 (2.0) = 60

P_2y = 0

Applying impulse-momentum separately in the x- and y-directions yields:

Substituting the value for the force of the surface into the x-equation,

The kid brother begins to cry after only 2.4 s.

Applying the Work-Energy Relation to a Single Object

What will the work-energy relation tell us about the same scenario?

To apply the work-energy relation, you must clearly specify the initial and final events at which you will tabulate the energy. For example:

Event 1: The instant before the sled begins to move.

KE1 = 0

GE1 = 0

Event 2: The instant the sled reaches 2.0 m/s

KE2 = ½ 30 (2.0)2 = 60

GE2 = 0

Applying the work-energy relation yields:

Notice that the force of the surface does no work, the force of the rope does positive work, and the force of friction does negative work. Each of these terms should make sense if you remember that work is the transfer of energy into (positive) or out of (negative) the system of interest. Also recall that in this form of the work-energy relation we conceptualize gravity as a source of potential energy, not as a force that does work.

Using the result for the force of the surface determined in the first example, F_surface = 199 N, gives:

The kid brother begins to cry after traveling 2.4 m.

Applying Work-Energy with Gravitational Potential Energy

Let’s use the work-energy relation, with gravitational potential energy terms, to analyze the following scenario:

A 30 kg child on his 15 kg sled slides down his parents’ 10 m long, 150 above horizontal driveway after an ice storm. The coefficient of friction between the sled and the driveway is (0.10, 0.08).

To calculate the gravitational energy terms let the bottom of the driveway be zero and up positive. The coordinate system used to calculate gravitational energy does not in general have to be the same as the system you use for the rest of the problem. In fact, since the work-energy relation is a scalar equation, the other portions of the equation should not depend on your choice of coordinate system at all!

Event 1: The instant before the sled begins to move.

KE1 = 0

GE1 = 45 (9.8) (10 sin150) = 1140

Event 2: The instant the sled reaches the bottom of the driveway

KE2 = ½ 45 v2

GE2 = 0

Note:

The only forces that could do work are the force of the ice and the force of friction, since the action of the force of gravity is already incorporated into the gravitational potential energy terms.
The heights in the gravitational potential energy function were measured from the bottom of the driveway, with the positive direction as upward, as required. Notice that the initial height is not the same as the length of the driveway. Since the driveway is 10 m long, at an angle of 15°, the height of the top of the driveway relative to the bottom is (10 m) sin 15°. The height at the bottom of the driveway is defined to be 0 m.

To finish the analysis we need to determine the kinetic frictional force. Since this depends on the force of the ice, apply Newton’s Second Law in the y-direction and find:

Plugging this value into the work-energy relation yields:

A Two-Dimensional Collision

Let’s try a two-dimensional collision.

At a busy intersection, an impatient driver heading south runs a red-light and collides with a delivery truck originally moving at 15 m/s west. The vehicles become entangled and the skid marks from the wreckage are at 22o south of west. The auto mass is 755 kg and the truck mass is 1250 kg.

Partial free-body diagrams (top view) for both the car and the truck during the time interval during the collision are shown below.

These are only partial free-body diagrams because:

Forces perpendicular to the earth’s surface (the force of gravity and the force of the road) are not shown.These are only partial free-body diagrams because:
During a collision, the force between the colliding objects is normally much greater in magnitude than any other forces acting on the objects. Therefore we will often ignore the other forces acting on colliding objects for the duration of a collision. This approximation is termed the impulse approximation. Under the impulse approximation, the frictional forces between the car and truck and the road are ignored.

Also note that the direction of the force acting between the car and truck is unknown. The angle q is not determined from the situation description.

Event 1: The instant before the car and truck collide

Object: Car

P1x = 0

P1y = 755 vcar

Event 2: The instant they reach a common velocity

P2x = 755 (v2 cos 22º) = 700 v2

P2y = 755 (v2 sin 22º) = 283 v2

Object: Truck

P1x = 1250 (15) = 18750

P1y = 0

P2x = 1250 (v2 cos 22º) = 1160 v2

P2y = 1250 (v2 sin 22º) = 468 v2

Applying the impulse-momentum relation to the car and truck yields:

Car

Truck

Since the magnitude of the force on the car due to the truck and the force on the truck due to the car are equal, when the x-equations for the car and truck are added, the impulses cancel!

This is the speed of the wreckage immediately after the collision. Note that this is exactly the same equation we would have written if we had considered the system of the car and truck right from the start. Try it!

Adding the two y-equations yields:

This is the speed of the car immediately before colliding with the truck.

Activities

Below are six different directions in which a baseball can be thrown. In all cases the baseball is thrown at the same initial speed from the same height above the ground. Assume the effects of air resistance are negligible.