Derivation of Equations for a Mass Bouncing on a Spring

Just for the fun of it, I’d like to show the derivation of equations for a block of mass, m, oscillating on a vertical spring with constant, k. I’m sure there are several ways to go about doing this, but I haven’t found any on the internet that do so for a general situation algebraically. So here’s the set-up we will analyze.

We will assume that a motion detector has been placed below this system, with the bottom of the block attached to the unstretched spring at position 0, the equilibrium position is at position 2 and that the origin is at the motion detector on the table below position 3. If the block is pulled up a distance A to position 1 and released, we get a graph that looks like this:

Which we could represent with the general equation:

y = A cos \left( Bt+C \right) + D

Since the amplitude is A, that term can stay A. We know from math that the vertical compression term, B, is the angular velocity of the system, \omega. Since the graph is in phase starting from the maximum value, C is zero and can be ignored. The value of D would be how high position 2 is above the origin, which we designate as y_{eq}. Therefore our equation for the vertical position as function of time would be:

y = A cos \left(\omega t\right) + y_{eq}

If we display the graph of velocity vs. time, we see a graph that looks like this:

We can represent this equation as:

v = Asin\left(Bt+C\right) + D

For us, the amplitude is v_{max}, the period is the same as above, is the B term would be the same as for position. C and D are both zero. So the equation begins to simplify to:

v = v_{max} sin \left(\omega t\right)

From calculus, we know that:

\dfrac{d}{dx} Acos\left(Bx\right) = AB sin\left(Bx\right)

So our final equation for velocity is:

v = A\omega sin\left( \omega t \right)

I’m well aware that we could have done the analysis of velocity using a phase shift instead of switching from cosine to sine, as I had a professor in college that would not allow us to use the “evil” cosine function at all. However, most calculus texts show the derivative of sine as switching to cosine {and cosine to sine} over using phase shifts.

The graph of acceleration vs. time looks like this

We can represent the amplitude with the maximum acceleration, a_{max}, and since it has the same phase as the other graph, can be represented as:

a = -a_{max}cos\left(\omega t\right)

Since we know from calculus:

\dfrac{d}{dx} A sin\left( B x \right) = -AB cos \left(Bx\right)

Our final equation for acceleration vs. time is:

a = -A\omega^2 cos \left(\omega t \right)

Now to the fun of deriving the equations for energy as a function of time. We know that kinetic energy is:

K = \frac{1}{2} mv^2

If we substitute our equation for velocity from above, we get:

K = \frac{1}{2} m\left[A\omega sin\left(\omega t\right)\right]^2

If we multiply out, we get:

K = \frac{1}{2}mA^2\omega^2sin^2\left(\omega t\right)

We know from trigonometry:

sin^2\left(\theta\right) = \dfrac{1 - cos\left(2\theta\right)}{2}

If we substitute this trig identity into our equation, we get:

K = \frac{1}{2} mA^2\omega^2\left[\dfrac{1 - cos\left(2\omega t\right)}{2}\right]

After multiplying through to simplify, we get:

K = \frac{1}{4}m\omega^2A^2 -\frac{1}{4}m\omega^2 A^2 cos\left(2\omega t\right)

We know the period of a mass on a spring from our experiment is:

T = 2\pi\sqrt{\dfrac{m}{k}}

From Unit 7, we know that at object moving in a circle at constant speed has a speed of:

v = \dfrac{2\pi r}{T}

From Unit 8, we learn that we can relate linear velocity to angular velocity by:

v = \omega r

Therefore, we can say that:

\dfrac{2\pi r}{T} = \omega r

Which we can simplify to:

\omega = \dfrac{2 \pi}{T}

If we now plug in our equation from the experiment in Unit 9, we get:

\omega = \dfrac{2 \pi}{2\pi \sqrt{\dfrac{m}{k}}}

Which simplifies to:

\omega = \sqrt{\dfrac{k}{m}}

If we plug this into our equation for velocity, we get:

K = \frac{1}{4}m\left( \sqrt{\dfrac{k}{m}} \right)^2 A^2 - \frac{1}{4}m\left( \sqrt{\dfrac{k}{m}} \right)^2 A^2 cos\left(2\omega t\right)

Which simplifies to:

K = \frac{1}{4}kA^2 - \frac{1}{4}kA^2 cos\left(2\omega t\right)

The graph of this would look like this:

We know that gravitational energy is:

U_G = mg\Delta y

Where \Delta y represents the height above our zero position. Since we using the table as our zero, y_o is at zero, so:

U_G = mgy

If we plug in our equation for position vs. time from above, we get:

U_G = mg\left[A cos\left(\omega t\right)+y_{eq}\right]

Which can be rewritten to give us the final equation for gravitational energy:

U_G = mgA cos\left(\omega t\right) + mgy_{eq}

If we make a graph of gravitational energy vs. time we get:

We know that spring energy is:

U_S = \frac{1}{2} k\left(\Delta x\right)^2

where \Delta x represents how much the spring has been deformed. For this event, that would mean:

U_S = \frac{1}{2}k\left(y - y _o \right)

If we substitute our equation for position vs. time, we get:

U_S = \frac{1}{2} k\{ \left[A cos\left( \omega t \right) + y_{eq} \right] -y_o\}^2

If we rearrange by grouping the position terms together, we see:

U_S = \frac{1}{2} k\left[\left(y_{eq} -  y_o\right) + A cos\left( \omega t \right)  \right]^2

If we multiply out the squared quantity inside the brackets, we get:

U_S = \frac{1}{2}k\left[\left(y_ {eq}  - y_o\right)^2 +2\left(y_ {eq}  - y_o\right) A cos\left(\omega t\right) + A^2 cos^2\left(\omega t\right)   \right]

Multiplying through by \frac{1}{2} k yields:

U_S = \frac{1}{2}k\left(y_ {eq}  - y_o\right)^2 + k\left( y_ {eq}  - y_o \right)Acos\left(\omega t\right) + \frac{1}{2} kA^2 cos^2\left(\omega t\right)

From trigonometry, we have the identity:

cos^2 \left(\theta\right) = \dfrac{1+cos\left(2\theta\right)}{2}

If we substitute this identity into our equation for spring energy:

U_S = \frac{1}{2}k\left( y_{eq}  - y_o \right)^2 + k\left( y_{eq}  - y_o \right)Acos\left(\omega t\right) + \frac{1}{2} kA^2 \left[  \dfrac{1+cos\left(2\omega t\right)}{2} \right]

If we multiply out the term at the end, and clump similar terms together, we get:

U_S = \frac{1}{2} k  \left(  y_ {eq}  - y_o  \right)^2 +\frac{1}{4}kA^2 + k\left( y_ {eq}  - y_o \right)Acos\left(\omega t \right) +\frac{1}{4}kA^2 cos\left(2\omega t\right)

The graph of spring energy vs. time would look like this:

Since the total energy is:

E = K + U_G + U_S

We can add up the three equations to get our total energy. By looking at the graphs for each energy vs time, the combination may be distorted since they each have a different scale. If we plot all three on the same graph, with the same scale, we get something like this:

If we arrange the three equations strategically, we get:

\frac{1}{4}kA^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~- \frac{1}{4}kA^2 cos\left(2\omega t\right)

+~~~mgy_{eq} ~~~~~~~~~~~~~~~~~~~~+ ~~~~~~~~~~~mgA cos\left(\omega t\right)

+~~~  \frac{1}{4}kA^2 + \frac{1}{2} k  \left(  y_ {eq}  - y_o  \right)^2 + k\left( y_ {eq}  - y_o \right)Acos\left(\omega t \right) +\frac{1}{4}kA^2 cos\left(2\omega t\right)

The first thing that should jump out is the two terms at the far right, - \frac{1}{4}kA^2 cos\left(2\omega t\right) and \frac{1}{4}kA^2 cos\left(2\omega t\right). They are the exact same term. Since the first has a “-” and the last has a “+”, these terms cancel out. The term in the middle, we see:

mgAcos\left(\omega t\right) +  k\left( y_ {eq}  - y_o \right)Acos\left(\omega t \right)

Which can be rewritten as:

\left[ mg + k\left(y_{eq} - y_o\right)\right] A cos\left(\omega t\right)

If we focus in on the terms inside the square brackets we see:

mg + k\left(y_{eq} - y_o\right)

The first term is the force of gravity on the block. The second term is the spring force on the block when it is at the equilibrium position, which is labelled position 2 on the original diagram. At that position, these two forces are equal and opposite in direction, thus:

mg + k\left(y_{eq} - y_o\right)  = 0

What’s left for the total energy is:

E =   \frac{1}{4}kA^2 +  mgy_{eq} +   \frac{1}{4}kA^2 + \frac{1}{2} k  \left(  y_ {eq}  - y_o  \right)^2

We can see that one of the terms repeats, so we can simplify this equation to be:

E =   \frac{1}{2}kA^2 +  mgy_{eq}  + \frac{1}{2} k  \left(  y_ {eq}  - y_o  \right)^2

The second term, mgy_{eq} , represents the gravitational energy at the equilibrium position. The last term, \frac{1}{2} k  \left(  y_ {eq}  - y_o  \right)^2 , represents the spring energy at the equilibrium position.

The first term, \frac{1}{2}kA^2, represents the work done to lift the block from equilibrium by the amplitude A. Since all the cosine and sine terms have cancelled out, we are left with:

E = W + U_{G,equilibrium} + U_{S,equilibrium}

Thus the if we assume no dissipated energy, the total energy would remain constant over time since all three terms are each constant with respect to time.

AP1 Unit 8: Rotating Bodies Model

We begin this unit with a simple paradigm activity similar to Unit 4‘s bowling ball games. We begin with a dot on a pulley. I begin rotating the pulley and tell the students to imagine the disk is spinning at constant speed. I ask them to figure out a way to describe the motion of the dot as a function of time. I give them a few minutes to talk as a group. Since my students math experience ranges from Algebra II to Calculus, I will usually have several students who come up with:

x = r cos \left( \theta\right)

y = r sin \left(\theta\right)

To ensure all the students remember the basic equation, we then discuss the parent function for sine and cosine.

y = Acos \left(Bx + C\right) + D

To do this, I first pull up Desmos and show the effects of changing each of the four variables. Here’s what happens when you increase A:

Sin A

Here’s what happens when you increase B:

Sin B

Here’s an increase in C:

Sin C

And an increase in D:

Sin D

We discuss the meaning of A as the amplitude or vertical stretch of the graph, B as the “frequency” or horizontal stretch of the graph, C as the horizontal shift of the graph, and D as the vertical shift. Since we have discussed frequency last unit, I relate that B is not exactly frequency, but that it is related to frequency, and tell them that we will go into more detail later this unit.

From there, I grab a string with a mass tied at one end, I grab the other end of that string and attach it to the edge of the pulley. I then wind the string around the pulley until the mass is touching the side of the pulley. I then challenge them to come up with a way to describe the dot as a function of time as I let go of the pulley and the mass falls to the table. I give them a few minutes to struggle with coming up with a function. After of few minutes of struggle, I suggest looking at the world from a different perspective. I then ask them what other ways could we describe this beyond using x and y?

I socratically lead them to using polar coordinates. Since the distance of the dot to the center of the pulley doesn’t change, the motion can be described as a one dimensional motion using only \theta. I attach a photogate to the stand, rewind the string, and display the results of \theta vs. time. We usually get a graph that looks like this for \theta vs. time:

Rotation angle vs time

<I always have trouble getting the timing of dropping the mass and starting the data collection, so my graphs are usually shifted like this one>

As we discuss the graph, the students usually notice that the graph looks similar to what we saw in Unit 2 other than the time shifted (in this example by a little less than 0.4 s). So I ask them again to mathematically describe this graph. They will usually then say that it should look like:

x = \frac{1}{2}a_o t^2 + v_o t+ x_o

When asked what the slope and curvature represent, to which they responded it should represent the “velocity” and “acceleration.” At that point I tell them that we call it “angular velocity” and give it the symbol: \omega and “angular acceleration” which gets: \alpha. I then show them the graph of \omega vs. time:

Rotation angular velocity vs time

With this, they can see that we get a fairly linear relationship where the slope is the angular acceleration:

\omega = \alpha_o t + \omega_o

We then proceed through the same derivations to get to find the area under the curve.

\Delta \theta = \frac{1}{2} \left(\omega_o + \omega\right) t

Then use that to find the equation for the \theta vs. time graph.

\theta = \frac{1}{2}\alpha_o t^2 + \omega_o t + \theta_o

After this day of work, we progress to worksheet 1, which deals with constant angular velocity problems, then worksheet 2, which deals with constant acceleration problems.

From there, we move on to the experiment for this unit. Here’s a picture of the apparatus:

U8 Apparatus

The quantities that the students manipulate are the hanging mass (external force), the radius of the mass on the “twister” and the mass on the “twister.” In this picture that mass is pretty small, later washers and an additional nut will be on each have of the “twister.” The lab groups have to determine how each of those quantities effects the angular acceleration of the system. Most groups measure the displacement of the falling mass and record that time to determine the linear acceleration, then use the radius of the “plastic wheel” (aka pulley) to determine the angular acceleration. What we eventually find at the end of the experiment is:

\alpha \propto \dfrac{r_{pulley}\times F_{ext}}{m_{system}~ r_{masses}^2}

After discussion, I reveal that the constant needed to make this an equation relates to the distribution of mass around the axis of rotation, thus we get:

\alpha = \dfrac{r_{pulley} \times F_{ext}}{k~m_{system}~r_{masses}^2}

To help relate this to Newton’s Second Law, we notice that in the numerator, is a quantity that increases rotation, and in the denominator, one that inhibits rotations. The quantity in the numerator we call torque, \tau; the quantity in the denominator we call the moment of inertia, I. Thus, we can rewrite the equation as:

\alpha = \dfrac{\tau}{I}

Before we move on, I make the point to explain that the multiplication to find torque is likely the first time my students have truly encountered multiplication with an \times. In Unit 6, we used the dot product to find work. That dot product of force and displacement meant that the two quantities had to be in the same direction, or parallel. Similarly, the cross product means that the quantities multiplied must be perpendicular.

From this point I we relate what we see here to what we saw at the beginning of Unit 3, namely to change the motion of an object we have to change the momentum of the object. Thus, we relate linear momentum, p = mv, to angular momentum, L = I\omega. We also relate translational kinetic energy, K_T = \frac{1}{2}mv^2 to rotational kinetic energy, K_R = \frac{1}{2}I\omega^2.

From there, we progress to worksheet 3, which, just like Unit 3, looks at momentum conservation of closed systems. In the worksheet, students are introduced to angular momentum diagrams. Since they have already seen the entire momentum diagram:

We jump to the complete angular momentum diagram. During this worksheet, we then have blank torque vs. time graphs.

The students use this diagram to help solve problems such as a what happens to the angular speed of a star when it collapses.

In worksheet 4, the students are introduced to Free Body Diagrams, in which they draw extended bodies instead of a small circle. I instruct the students that since we are no longer dealing with a particle, they need to draw the forces on the object where that force acts on that object. This worksheet then looks at systems in which the torques are balanced, which would constitute the traditional static torque problems.

In worksheet 5, the students now deal with conservation of momentum in open systems. Some of the problems also require the students to include kinematics relationships from worksheet 2 to solve problems.

In worksheet 6, the students are given first problem with a metal ball at the top of a ramp. The problem walks them through determining how long it would take the ball to go through a photogate at the bottom of the ramp without friction, and if friction causes the ball to roll without slipping. I then set up that ball on that ramp and we test their predictions. We usually get results within \pm0.0003 seconds of the time predicted for the ball rolling. We then do a similar problem for a solid disk and a hoop, with similar results.

From there, we review and take the unit test.


  1. determine and correctly use the graphical and mathematical relationships between angular position, angular velocity, and angular acceleration with time.
  2. develop mathematical models from graphs of angular acceleration vs. torque and angular acceleration vs. mass, and angular acceleration vs. distribution of mass.
  3. use rotational energy as an energy storage mode to solve problems using conservation of energy.
  4. create Angular Momentum diagrams.
  5. use conservation of angular momentum to solve problems.

AP1 Unit 7: Central Force Particle Model

We begin the 2nd semester with circular motion. Up first is the paradigm lab for uniform circular motion. Students adjust the hanging mass, the radius of orbit, and the mass of the orbiting object and determine how each effects the speed of the revolving mass.

As we analyze the results, we find that:

v^2 \propto \frac{r F_{net}}{m}

If you make a graph of v^2 vs. \frac{r F_{net}}{m}, you find that the slope if very close to unity. Thus, after rearranging the equation, we can say:

F_{net} = m \frac {v^2}{r}

From Unit 5, we know:

F_{net} = m a

We can conclude that the acceleration of an object in uniform circular motion is:

a_c = \frac{v^2}{r}

After the lab, we begin with worksheet 1, which focuses primarily on circular motion in the horizontal plane. Students are given different events and must determine which force(s) are causing the object to move in a circular path. If you would like to know a little more about this lab, you can read my post from my Modeling Workshop.

In worksheet 2, students begin to analyze quantitative problems. Circular motion in the vertical plane are also introduced. Worksheet 3 is entirely quantitative problems for both horizontal and vertical motion.

After worksheet 3, we then do a virtual experiment (PhET) to look at how gravity works beyond the Earth’s surface.

After varying the two masses, m_1, m_2, and the separation, r one at a time to determine the effect of each on the force of gravity, the students eventually come to:

F_G \propto \frac{m_1 m_2}{r^2}

We then move to the more formal equation:

F_G = G \frac{m_1 m_2}{r^2}

We then discuss the relationship between G and the field constant g from Unit 4, namely:

g = G\frac{m_{planet}}{r_{planet}^2}

From there, we move on to worksheet 4, which looks at orbital motion problems. After that is the review.


  1. determine the acceleration of an object in orbit (constant circular motion).
  2. distinguish between centripetal and centrifugal forces.
  3. correct apply Newton’s Second Law and centripetal acceleration to object moving in horizontal (including banked and unbanked) circles and vertical circles.
  4. determine the force of gravity between any two objects that have mass.
  5. determine the period of an object in orbit.

AP1 Unit 6: Energy Transfer Model

So far, we have always started this unit with a quick lab to determine the relationship between the stretch of a spring and the force the spring exerts. The result of the experiment giving us Hooke’s Law:

F_S = - k \Delta x

I’ve contemplated moving that lab to an earlier unit, likely Unit 4 after the gravity lab. So far, I have kept it at the start of this unit.

After the spring lab, we quickly proceed to the paradigm activity for this unit. You can read about that here.

The first step in deploying the model is this video that explains how to use energy diagrams. I haven’t had time to properly introduce the pie charts, but so we jump straight to energy diagrams.

We begin with worksheet 1, which has the students create the energy conservation equation from the energy diagrams for various events. For those familiar with energy diagrams, either from my previous post, Kelly O’Shea, or modeling materials, you’ll see that I’ve recently modified what my students are doing. In an effort to make them similar to momentum diagrams, I moved the identification of the system to the left side of the diagram. I then add a Force vs. position diagram between the two bar charts when the system is open, similar to the Force vs. time graph for momentum diagrams.

In worksheet two, we now add in doing calculations to the mix. The students now need to solve similar problems, but are now given numerical values.

After that, we usually are ready to celebrate Festivus!

We celebrate this holiday with the “Feats of Strength!” I begin by asking the students who is the most “awesome” at lifting a bag. Usually at least one student will ask me to clarify what I mean by most awesome. After discussion and prompting from me, we usually settle on an “awesome-ness” factor, designated A as some arrangement of:

A = \frac{m g \Delta y}{\Delta t}

At some point, the students notice that energy is in the numerator of this factor. At some point along the way. we rename this factor “power.” Before we get to that, I try to ask them what name we could give to this quantity. We usually start taking about “awesome” videos on youtube, such as Kobe Bryant jumping over a pool of snakes:

Usually, as Kobe is in the air, at least one student will say, “Whaaaattt!”

Usually followed by, “That can’t be real!” If a student doesn’t say it, I’ll lead them to it, but once they say, “Whaaatt!” they’re ready. I let them know that the unit of power is the watt.

I now reveal the two challenges for the Feats of Strength, the Bag Lift, and the Ascension of a Flight of Stairs. The students each must compete, but they don’t have to try too hard if they don’t want. In the end, they must calculate their power output in watts. I do this, so that the students get a physics sense of the SI unit of power. We usually wrap up the activity by asking how many of the students could have powered a 100 W light bulb or a 1200 W microwave oven.

From there, we progress to worksheet 3 which now includes calculations of power along with the energy diagrams. After that it’s time to review and take the test. After this unit test, we review for our midterm, take that midterm, and go off for Christmas Break.

Energy Transfer Model Paradigm Lab

The setup for this lab is a loop spring launching a cart up an inclined ramp. The spring is attached to an old pulley bracket. We just purchased the new Go Direct Sensor Cart, so it can do all the measurements for force, position, and velocity needed. The data is so clean! If you have the means to get some, I highly recommend it. If you don’t, I talked a little about how I used to do this experiment here.

The students collect data for force vs. compression to determine the spring constant of the loop spring.

Then they collect the amount of compression (\Delta x) and the maximum displacement up the ramp (\Delta r) as the minimum and maximum values from the position vs. time graph for each launch.

They also record the maximum velocity (v) from the maximum of the velocity vs. time graph.

(I’m not going to show the graphs for the rest of the analysis in case other teachers are using this for an actual lab report.)

The analysis of this lab is usually tough for my students, but we first try to determine the relationship between the compression of the spring (\Delta x) and the maximum vertical displacement of the cart (\Delta y), found after measuring the angle of the ramp and finding the component of the displacement along the ramp. After several trials and the subsequent plotting of our two variables, we get a curved graph. Although by this point the students know to square the displacement to try to linearize the graph, after discussion, they agree that the slope, with units (\frac{1}{meters}) has no significant meaning.

From there, I eventually steer the discussion to finds something else that quadruples when the compression is doubled. We eventually settle on relating the area under the graph for F_S vs. \Delta x_{spring} to the vertical displacement. Before we can make that graph, we need to get an equation for that area as a function of only the compression of the spring. Since the line can be represented by the equation:

F_S = - k \Delta x

and we can find the area by finding the area of the triangle from:

Area = \frac{1}{2} F_S \left( \Delta x \right)

We can combine these two equations and get and equation for the magnitude of the area as:

Area = \frac{1}{2} k \left( \Delta x \right)^2

If we now make a plot of Area as a function of vertical displacement, we again get a proportional graph. With analysis, we find that the slope has units of the inverse of the weight of the cart.

\Delta y = \frac{1}{mg}Area

Thus we can rearrange to get:

mg \Delta y = \frac{1}{2}k \left(\Delta x \right )^2

From here, we can now begin to analyze the relationship between the maximum velocity (v) versus compression. We find that we again have a proportional relationship, with another meaningless slope. Since area worked last time, we try this again and plot the maximum velocity versus Area. In this case, we see a curved graph. To linearize the data, we plot the square of the maximum velocity versus Area and find that the slope is approximately half of the mass of the cart.

v^2 = \frac{1}{\frac{1}{2}m} Area

After rearranging, we get:

\frac{1}{2}mv^2 = \frac{1}{2}k\left( \Delta x\right)^2

Thus, we now have a relationship between the amount the spring was compressed, the maximum vertical displacement, and the maximum velocity. We discuss how, similar to momentum in Unit 3, there is something that is being conserved, which we call Energy. From there we take the traditional definition of Energy they remember from middle school science, “the ability to do work” and try to make sense of what that means. We first realize the work is often defined as the use of Energy, so they realize that this circular definition doesn’t work. I eventually lead them to Kelly O’Shea’s definition of energy as the “ability to cause pain.” We then begin to discuss the various flavors of energy through this flow chart:

Energy Flow Chart

Usually at this point in the year, we won’t discuss Translational vs. Rotational Energy, we add those later when we get to Unit 8. Later in the year, we will rename the “Chemical Energy” as Electric Energy, but since all of my students have had chemistry, that flavor of energy always comes up. We also briefly discuss some of the flavors of dissipated energy, but note that for AP1, we usually will just refer to it as “dissapated.” We finish the post lab discussion by beginning to fill in the cover sheet as seen here:

The section at the top is for kinetic energy. The second section is for all the flavors of potential energy. The third section is for dissipated energy, and the last section is for means of transferring energy between the system and it’s surroundings.

AP1 Unit 5: Unbalanced Forces Particle Model

The fifth unit, although not an introduction to the term impulse, is an in-depth study of momentum swaps between a system and its surrounding. The unit takes about three weeks. It begins with a Modified Atwood Machine activity.

The angle of the ramp is adjusted until the cart rolls down the ramp with zero acceleration. After a brief discussion of what is happening, what can be measured, and what can be manipulated, we settle into looking at how mass of the system and the magnitude of the unbalanced force are each related to the acceleration of the system. By moving mass from on the cart (m_1) to hanging from the string (m_2), the force can be adjusted without changing the mass of the cart-string-hanging mass system. Adding mass to the cart without changing the hanging mass, changes the mass of the system. By having a motion detector, the acceleration of the cart can be calculated for each run.

The data that is produced shows a proportional relationship between the net force and acceleration, and an inversely proportional relationship between the mass of the system and the acceleration. As you combine these two proportionalities you get:

a \propto \dfrac {F_{Net}}{m}

To write it as an equation, we must add a constant:

a = k \dfrac{F_{Net}}{m}

We then briefly discuss how for the SI System, this value of k defines the derived unit newton as the force required to accelerate 1 kilogram at a rate of 1 \dfrac{m}{s^2}. While the value of k for the English/American System is chosen such that the magnitude of the force is the same as the magnitude of the mass, and both are given the unit: pound.

We wrap up the post-lab discussion by looking at how the results from this activity change out momentum diagram. By including the impulse into or out of a system, we now have the complete momentum diagram we will use for the course. Here is an example for the diagram for a canon shooting a canon ball, treating the ball alone as the system.

Any object within the system goes inside the circle, any object in the surrounding that is involved in the event is written outside the circle. Every object inside the system is then displayed in the first and last bar graph. The width of the bar represents the mass of that object. The height corresponds to its velocity. The area of each bar represents the momentum of that object. The middle bar graph represents the impulse into or out of the system. The width corresponds to the elapsed time for the impulse. The height is the force. Thus the area of this graph is the impulse. For this level class, the shape tends to be a rectangle or a triangle. In a calculus based class, this could be any shape or function. Most of my students are familiar with Newton’s Second Law from their middle school science classes, but most haven’t seen the momentum diagram as a way to represent that equation.

From there, we progress to the model deployment phase of the unit. In worksheet 1, we look at elevator problems and other impulsive events. One of them features a momentum diagram with a trapezoidal impulse graph to challenge the students. In worksheet 2, we then combine the momentum diagram and kinematics from Unit 2. In worksheet 3, we begin to introduce 2-dimensional problems. The motion of the object is still in 1-dimension, but at least one force in the problem is in 2-dimensions. Students must find the components of that force in solving the problem.

After worksheet 3, we do the lab for this unit, which is the friction lab. Students must find the graphical and mathematical relationship between the normal force and the maximum static and constant velocity friction forces. Worksheet 4 then extends the problems seen in worksheet 3 to now include the resulting friction equations. We conclude the unit in worksheet 5. In this worksheet we work up to actual 2-dimensional projectile motion problems.

At the end of this unit, students should be able to:

  1. create an interaction diagram including identification of the system and the types of interaction
  2. use the relationship between normal force and friction correctly
  3. create momentum diagrams that include impulses
  4. solve problems using Newton’s 2nd Law correctly
  5. apply Newton’s 2nd Law to predict the path of objects moving in two dimensions

AP1 Unit 4: Balanced Forces Particle Model

The fourth unit introduces Forces. It takes about two and one-half weeks. It begins with a series of Bowling Ball Activities; e.g. make the ball speed up, slow down, turn 90^{\circ}, move in a circular arc. In the previous unit, a simplified “Interaction Diagram,” what many call a “System Schema.” As we begin to use this diagram, we see that multiple interactions are occurring.

Students are asked to make a quick sketch of momentum vs. time for the bowling ball starting from rest, being pushed by the broom, and then rolling at “constant” velocity. Most agree that it should be a horizontal line at p = 0, then a diagonal line as it is speeding up, and then a horizontal line at the final constant velocity.

In previous units, students saw the slopes are often important quantities, so they are asked to try to figure out what this slope of momentum vs. time represents. Usually, with little guidance, they can figure out that this slope represents the rate at which momentum is swapping, which we define as a “Force.” Thus, if an interaction is the swapping of momentum, it’s derivative is a Force.

From there we introduce the different types of interactions present during our activities. We identify three contact forces: push (Normal), pull (Tension), and slide (Friction). We also name one non-contact force, Gravity. From there, these types of interactions are added to the interaction diagram. Then show how this interaction diagram, can be used to make Force Diagrams.

We conclude the day by noticing that multiple interactions can balance out, thus multiple forces on a single particle or systems can balance out. We go on to notice that the system will only accelerate when the forces are unbalanced, which I mention is Newton’s 1st Law (when in an inertial frame of reference).

So the next step in the sequence is Worksheet 1 in which they practice making the Interaction Diagram with forces included and Force Diagrams. They are aided by a reading that help show in greater detail how to create Force Diagrams including equality marks.

From there, we do a lab to try to begin to understand the non-contact force of gravity by hanging various masses on a spring scale. Thus finding the relationship between mass and the force of gravity.

Once that’s complete we add this calculation to begin to add numerical values into Force Diagrams in worksheet 2. Along with a second reading, we end the unit with worksheet 3 as we add component forces into the mix.

The student goals for this unit are:

  1. can draw a properly labeled free body diagram of all the forces acting on an object including equality marks
  2. given one interaction between two objects, determine the direction of force exerted on each object
  3. determine the direction of acceleration of an object from a free body diagram
  4. determine whether or not the forces are balanced given information about the motion of the object
  5. determine the force of gravity on an object within a gravitational field

AP Physics 2 Syllabus

I was recently able to get my syllabus approved by College Board. The approval number is # 1485589v1 Authorized

You can download a copy of the .docx version here.

AP Physics 1 Syllabus

I recently was able to get my syllabus approved by College Board. The approval number is # 1485588v1 Authorized

You can download the .docx copy here

Energy Diagrams in Thermodynamic processes

I’d like to show how Energy Diagrams can be extended to help solve problems for thermodynamic processes. As an example, imagine the problem tells us that a 1 cubic meter sample of gas at a pressure of 101 kPa, is compressed isothermally at a temperature of 300 kelvin to half its original volume.

Sample super lol2

The first thing we can do is write the given information, which is done in red ink. Next we can try to figure out the rest of the variables for the initial and final states. From the Ideal Gas Equation

PV = nRT

we can calculate the number of moles at the initial state (blue ink). If we assume it’s a sealed container, then the moles would be the same in the final state (pink number). We were told the process is isothermal, so the final temperature must be the same as the initial temp (orange ink). We can determine the pressure at the second state by again using the Ideal Gas Equation.

The next step is to draw the approximate size of each container below the state variables. Since the initial volume is twice that of the final, we make the container on the left about two times bigger than the one on the right (grey ink).

Next up is the energy present at each state. To find these values, we can use the other equation developed during the computer simulation/paradigm lab:

U = \frac{3}{2}nRT

Since this only depends on the number of moles (a constant) and temperature (not constant) the relative size of the energy bars will be proportional only the temperatures  of the two states (purple ink).

The next step is drawing the PV diagram (brown ink). This shouldn’t be anything new for people familiar with thermodynamics or from most textbooks. For the sake of brevity, I’m not going to explain that. After we make the PV graph we can now determine the working that occurred between the system and the surrounding by finding the area under the curve (brown shading). To find the area, we can use calculus or the equation given in most textbooks for the work of an isothermal compression:

W = -nRT ln \frac{V_1}{V_2}

The last step is to complete the energy flow diagram (the “O” of the “LOL Diagram” at the bottom). By having the grey pistons drawn, we have a clear indication that the gas was compressed, which would mean that energy flowed into the system from an outside force moving the piston inwards (black ink). To determine the amount of heating, we can use the First Law of Thermodynamics equation:

\Delta U = Q + W

We know the initial and final energy (purple ink), we know the work done (black ink), so we just have to calculate the heating by subtraction (dark green ink).