Showing posts with label mass. Show all posts
Showing posts with label mass. Show all posts

20191113

Physics quiz archive: simple harmonic motion, waves

Physics 205A Quiz 6, fall semester 2019
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855 version 1
Exam code: quiz06co6O



Sections 70854, 70855 results
0- 6 :   * [low = 3]
7-12 :   ****
13-18 :   *************
19-24 :   **************** [mean = 22.1 +/- 6.1]
25-30 :   ****************** [high = 30]

20181114

Physics quiz archive: simple harmonic motion, waves

Physics 205A Quiz 6, fall semester 2018
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855 version 1
Exam code: quiz06POr7



Sections 70854, 70855 results
0- 6 :  
7-12 :   **** [low = 9]
13-18 :   ********
19-24 :   ********************* [mean = 23.3 +/- 5.6]
25-30 :   ******************** [high = 30]

20171122

Physics quiz archive: simple harmonic motion, waves

Physics 205A Quiz 6, fall semester 2017
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855 version 1
Exam code: quiz06Ho0k


Sections 70854, 70855 results
0- 6 :   * [low = 6]
7-12 :   *
13-18 :   *********
19-24 :   ************* [mean = 23.3 +/- 5.8]
25-30 :   ********************* [high = 30]

20170929

Physics presentation: impulse and momentum

Whuuuuuuut. (Video link: "bowling strike with a ping pong ball.")

In this presentation we will introduce another new connection between forces and motion, in terms of how the net force can exert an impulse on an object in order to change its momentum. This is yet another new approach to connecting forces and motion, compared to the previous discussion in this course of using Newton's laws to relate how forces on an object result in a net force that may or many not change its motion, and analyzing how forces can do work on or against an object in order to speed up or slow down its motion.

First, defining the momentum of an object, and then expressing how the net force can exert an impulse on this object.

The introduction slide showing a ping-pong ball knocking over all ten bowling pins should seem very strange to you, as the mass of the ping-pong ball is too small to effectively bowl a strike, even if it were traveling with a supersonic speed. In order to fully account for the "knocking-over" strength of a moving object, then, we must include mass as well as its speed (and direction) to define its momentum p.

Momentum p is a vector quantity (so don't forget to draw an arrow over it) whose magnitude depends both on the mass and speed of the object, with the combined units of both mass and speed (kg·m/s).

We also need to introduce the concept of impulse J, which is the product of the net force acting on an object and the duration of time that the net force acted on this object (whether for a brief instant, or for a prolonged period). (Video link: "Teaching Tee Ball Hitting.")

Impulse has the combined units of both force and time (N·s). Here we use the somewhat obscure (but totally legit) "J" symbol for impulse, remembering to draw an arrow over it (as it is a vector quantity). (It turns out that "I" is already reserved for rotational inertia in the next chapter.)

Second, let's now explicitly make the connection between the impulse acting on an object, and the resulting change in the momentum of the object.

This "impulse-momentum theorem" emphasizes how the impulse (exerted by the net force acting over a specific duration of time) causes a corresponding initial-to-final change in the momentum of the object. And vice versa, where the initial-to-final change in the momentum of an object is caused by the impulse on the object.

Let's apply these concepts to several objects that undergo changes in momentum, with an emphasis on the directions (+/– signs) of these quantities, and how they all must be consistent with each other, starting with a golf ball initially at rest, and then has a speed of 97 m/s after being hit by a golf club. (Video link: "The Moment of Impact. An Inside Look at Titleist Golf Ball R&D.")

This golf ball is initially at rest, so its initial momentum p0 (mass times its initial velocity) is 0.

We'll define the horizontal direction to be positive to the right (and negative to the left). After it is hit by the golf club, its final momentum (mass times its final velocity) pf points to the right (and will be a positive quantity).

The initial-to-final change in momentum ∆p of the golf ball is given by:

p = pfp0,

and since get a positive quantity minus zero, then ∆p must be positive (thus pointing to the right).

Since the impulse "J" on the golf ball causes this initial-to-final change in momentum:

"J" = ∆p,

the impulse must also have the same direction as ∆p, and so it must also point to the right. (Also since the impulse "J" is the net force ΣF on the golf ball times the contact time ∆t, the net force of the golf club on the golf ball is also directed to the right.)

Now let's have you look at the directions involved in the impulse-momentum theorem for this catapult-launched F/A-18E-F Super Hornet, initially at rest, and then has a speed of 74 m/s after being it is catapulted. (Video link: "F/A-18E-F Super Hornet Catapult Launches.")

Super Hornet's initial momentum p0 direction? (left (–), none (0), or right (+)?)
Super Hornet's final momentum pf direction?
Direction of Super Hornet's initial-to-final change in momentum ∆p?
Direction of catapult's impulse "J" on the Super Hornet?

Finally, consider the directions involved in the impulse-momentum theorem for this Ford Ranger, hitting a crash barrier with a speed of 11.0 m/s, and then rebounding off the crash barrier with a speed of 2.2 m/s. (Video link: "Crash Test Ford Ranger 2012....")
Ford Ranger's initial momentum p0 direction? (left (–), none (0), or right (+)?)
Ford Ranger's final momentum pf direction?
Direction of Ford Ranger's initial-to-final change in momentum ∆p?
      (Hint: watch your signs!)
Direction of crash barrier's impulse "J" on the Ford Ranger?

20161118

Physics quiz archive: simple harmonic motion

Physics 205A Quiz 6, fall semester 2016
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, 73320, version 1
Exam code: quiz06rn3T



Sections 70854, 70855, 73320 results
0- 6 :  
7-12 :   ** [low = 9]
13-18 :   *********
19-24 :   *************************** [mean = 22.9 +/- 4.6]
25-30 :   **************** [high = 30]

20151121

Physics quiz archive: simple harmonic motion

Physics 205A Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, 73320, version 1
Exam code: quiz06m45S



Sections 70854, 70855, 73320 results
0- 6 :  
7-12 :   ** [low = 12]
13-18 :   **********
19-24 :   ***************************** [mean = 24.2 +/- 4.7]
25-30 :   ******************************** [high = 30]

20141125

Physics presentation: internal energy conservation

Sigh. This equation. The ubiquitous "Q = m·c·∆T" endlessly recited in chemistry, more than likely without much understanding. And by "understanding," I mean, "physics."

Well, now that you've seen that equation, you can't unsee it, so in this presentation we'll instead emphasize the conceptual meaning of heat transfers (the "Q" on the left side of the equality), and of changes in thermal internal energy (the m·c·∆T" on the right side of the equality), and see how this equation works in the context of principles developed earlier in this course for work and mechanical energy conservation.

First, introducing internal energy systems, and especially changes in thermal internal energy.

Recall that mechanical energy systems such as translational kinetic energy and rotational kinetic energy are concerned with the macroscopic translational motion or rotational motion of an entire object. Thermal internal energy is also concerned with a motion of an object, more specifically the random microscopic motion of the individual atoms and molecules within an object. So for this just-cooked turkey coming out of the oven, it contains a large amount of thermal internal energy due to the very active microscopic motion of its atoms and molecules.

The total thermal internal energy of an object's microscopic atomic and molecular motion depends on its mass m and its temperature T (measured in kelvins), and is expressed as joules (chemists use "calories" or "Calories" for energy--be sure to ask them why they keep using those non-Système International units). A low temperature object has very little thermal internal energy, and higher temperature it has, the more thermal internal energy it has. The intensive property of different materials (due to its atomic and molecular make-up) related to its thermal internal energy is the "specific heat capacity," in units of J/(kg·K).

(Note that while your textbook explains in detail what thermal internal energy is, it does not actually define a symbol for thermal internal energy, so for our purposes we'll use "Etherm.")

Instead of finding out how much thermal internal energy Etherm an object has, often we are more concerned with its initial-to-final change ∆Etherm, which is the final amount of thermal internal energy minus the initial amount of thermal internal energy. Notice how the common factors of mass m and specific heat capacity c are pulled out, such that the amount of change in thermal internal energy is determined by the ∆T change in temperature.

Recall that mechanical energy systems such as gravitational potential energy and elastic potential energy are concerned with the storage of energy due to height of an object in a gravitational field, or due to the compression or stretching of a spring or an elastic material from its equilibrium state. Bond internal energy is concerned with the storage of energy, but stored in the spring-like bonds between the individual atoms and molecules in solids and liquids.

During vaporization (liquid turning into gas) or melting (solid turning into liquid), these bonds are broken, freeing up individual atoms or molecules. Bond internal energy increases during these processes, much like the elastic potential energy of a spring increasing as it stretches more and more, because the bonds between atoms and molecules must be stretched further apart in order to "disconnect" them from each other.

During condensation (gas turning into liquid) or freezing (liquid turning into solid) bonds begin to form between the individual atoms or molecules. Bond internal energy decreases during these processes, much like the elastic potential energy of a spring decreasing as it stretches less, because the distances between atoms and molecules must brought be closer together in order to "connect" them to each other.

(While there is a "latent heat" equation to calculate the change in the bond internal energy during phase changes (when bonds are broken or made between individual atoms and molecules), we will focus primarily on changes in thermal internal energies, so only consider situations where there are changes in temperature, but no changes in phase.)

Next, heat--the "Q" in the "Q = m·c·∆T" equation.

Recall that work is the transfer of mechanical energy on the macroscopic level, whether put in by an external agent, or taken away because of friction/drag. Heat, then, is the transfer of internal energy on a microscopic level.

If there is no energy transferred into or out of the thermal internal energy of a system (as with the contents an extremely well-insulated Thermos® bottle), then it is effectively thermally isolated from the environment, and the heat exchanged between the system and the external environment is zero.

If the thermal internal energy of a system increases, its temperature increases, and thus external heat from the environment is positive, being added into the system (as for this blowtorch on a marshmallow).

If the thermal internal energy of a system decreases, its temperature decreases, and thus external heat from the environment is negative, being removed from the system (as for these sous-vide cooked portions of duck meat being chilled in an ice-water bath).

The direction of heat flow is determined by the relative temperatures of two objects interacting with each other--the object at a higher temperature will "heat up" (transfer energy to) an object at a lower temperature. The two objects will stop transferring energy between each other when they attain the same final temperature (such that there is no longer a direction for heat to flow), thus reaching thermal equilibrium.

(Don't ever expect heat to spontaneously flow from a lower temperature object to a higher temperature object--unless you "do work" (expend mechanical energy) to make this happen, which is why it will cost you to run a refrigerator or air conditioner to remove heat from the low temperature contents, and dump it to the warmer environment outside. We'll focus on the "natural" direction of heat flow that occurs between two different temperature objects that thermally interact with each other.)

Third, accounting for the transfers between different thermal internal energy systems, and also for the contribution to or taking from these energies by an external source of heat.

This is expressed in a format in order to emphasize the parallels with the total mechanical energy conservation equation developed earlier. The transfer/balance equation here shows how the transfer of energy (as heat exchanged by external agents and/or the environment) causes corresponding changes in the thermal internal energies of all objects in our system. Any or all of these thermal energy forms on the right-hand side of the equation can increase or decrease (due to an increase or decrease in temperature), but together all of their changes must add up to the corresponding heat exchange term on the left-hand side of this equation.

When you substitute the individual ∆Etherm terms on the right-hand side of the equation with the equivalent m·c·∆T expressions, then this results in the oh-so-familiar-but-maybe-not-quite-so-meaningless-anymore "Q = m·c·∆T" equation.

Note that in the idealized case that the system is thermally isolated from both external agents and the environment, then the left-hand size of this equation would be zero. Then the individual energy terms on the right-hand side of this equation can then trade and balance amongst themselves, instead of with the outside world.

So now let's see how this transfer/balance equation can be applied to idealized situations where heat gain/loss exchanges with the outside world are negligible compared to the exchanges within a system.

Raw seafood is placed on a block of salt that has already been heated up in an oven. The energy contained in the high-temperature block of salt is then transferred to the seafood, cooking it. While it is being cooked, does the internal thermal energy of the seafood increase, decrease, or not change? Does the thermal internal energy of the salt block increase, decrease, or not change?

Assuming that the seafood and salt block system is thermally isolated from the environment (such that Qext = 0), which thermal internal energy experienced a greater amount of change: the seafood, or the block?

As the seafood cooks, its internal thermal energy increases, as the temperature of the food increases. (If you calculated the change in internal energy of the seafood, you would get a positive value, which is consistent with an increase.)

As the salt block cooks the seafood, its internal thermal energy decreases, as the temperature of the salt block decreases. (If you calculated the change in internal thermal energy of the salt block, you would get a negative value, which is consistent with a decrease.)

Since we are assuming idealized situations where heat gain/loss exchanges with the outside world are negligible compared to the exchanges within a system, then there is no heat given off or taken in from the environment, so the left-hand side of the transfer/balance equation is zero:

Qext = ∆Eseafood + ∆Esalt block,

0 = ∆Eseafood + ∆Esalt block,

then we are only left with the changes in the thermal energies of the seafood and the salt block:

0 = ∆Eseafood + ∆Esalt block,

0 = (+) + (–).

Now we can see that amount that the seafood's internal thermal energy increases (where ∆Eseafood is positive) is directly related to the amount that the salt block's internal thermal energy decreases (where ∆Esalt block is negative), in order to equal the zero on the left-hand side of the equation. So the salt block's internal thermal energy "feeds" (or is "transferred to") the seafood's internal energy during this process.

Frozen meat is placed in a water bath, in order to defrost it. At the very start of this defrosting process (where the frozen meat just begins to warm up from its below-freezing temperature, and the ice crystals inside have not yet reached the melting point), does the internal thermal energy of the meat increase, decrease, or not change? Does the thermal internal energy of the water increase, decrease, or not change?

Assuming that the meat and water system is thermally isolated from the environment (such that Qext = 0), which thermal internal energy experienced a greater amount of change: the meat, or the seafood?

A shot of whiskey is mixed with a pint of beer to make a boilermaker. Assuming that the whiskey and beer have approximately the same temperature before they are mixed together, does the internal thermal energy of the shot of whiskey increase, decrease, or not change? Does the thermal internal energy of the pint of beer increase, decrease, or not change?

Assuming that the whiskey and beer system is thermally isolated from the environment (such that Qext = 0), which thermal internal energy experienced a greater amount of change: the whiskey, or the beer?

(If you haven't noticed the type of vocabulary used in this presentation, we are deliberately avoiding the confusion between "hot" (in terms of high temperature, high thermal internal energy objects) and "heat" (thermal energy transferred between objects).)

20141120

Physics quiz archive: simple harmonic motion, waves

Physics 205A Quiz 6, fall semester 2014
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, 73320, version 1
Exam code: quiz06eAg7



Sections 70854, 70855, 73320 results
0- 6 :  
7-12 :   ***** [low = 12]
13-18 :   **********
19-24 :   ********************** [mean = 23.3 +/- 5.3]
25-30 :   *************************** [high = 30]

20141108

Physics presentation: waves

Watch as the very end of this whip cracks and breaks the sound barrier. As for me, when I look at this whip cracking...I'm thinking, dinosaur butt. (Video link: "WorldWideWhips introducing the Whip-Cam.")

This was the infamous Tacoma Narrows Bridge, where high winds set up standing waves that, unchecked, eventually destroyed it. They don't build bridges like that these days...or do they? (Video link: "Tacoma_Narrows_Bridge_destruction.ogg.")

We'll introduce different wave phenomena, parameters that specifically describe periodic waves, and consider "standing waves."

First, and overview of different types of one-dimensional waves.

For a transverse wave (here, a pulse) the disturbance is sideways to the direction of the wave motion. (Video link: "Transverse wave travel along a bungee cord.")

For a longitudinal wave (here, also a pulse) the disturbance is along the direction of the wave motion. (Video link: "Longitudinal waves in a spring in slow motion.")

If the disturbance (here, transverse) repeats itself periodically, then a periodic wave is set up along this rope. (Video link: "Transverse Waves.")

If the disturbance (here, longitudinal) repeats itself periodically, then a periodic wave is set up along this spring. (Video link: "Transverse & Longitudinal Waves.")

Next, we'll focus on waves on strings (and ropes, cables, and other similar media), and specifically on periodic waves along strings.

The speed of a transverse pulse or transverse periodic wave along a strong depends on the square root of the string tension (here denoted by F rather than T, which we'll reserve for period, as well as for temperature near the end of this course) divided by the linear mass density, essentially the "thickness" (mass per unit length) of the string.

Which leads us to whips and dinosaur tails. A whip must be constructed that it tapers with decreasing thickness, such that as its mass per unit length (m/L) decreases, then the wave speed increases (assuming that tension remains approximately constant). If the wave speed at the end of the whip is faster than the speed of sound (nominally 343 m/s in air), then that part of the whip will move fast enough to break the sound barrier and create a sonic boom. Computer models of certain dinosaurs indicate that their tails may have been used as whips for defense and/or signaling.

The speed of periodic waves along strings is set by the string tension and thickness, but the frequency of the wave is set by the source. The resulting spatial repeat interval is the wavelength.

Note the hierarchy of these wave parameters. Since the wave speed is determined by properties of the string (independent of the source), and the frequency is determined by the source (independent of the string), these are said to be independent wave parameters. In contrast, the wavelength of the wave is dependent on both the independent speed and frequency parameters. Algebraically there is nothing wrong with expressing this relation as v = λf and f = v/λ, as long as you recognize that the dependency of λ doesn't change.

The hand on the left oscillates up-and-down as the source of a wave that travels left-to-right along this apparatus. The top case is where the hand oscillates up-and-down with a certain frequency and a small amplitude, while the bottom case is where the hand oscillates up-and-down with the same frequency and larger amplitude. Which wave travels with the faster speed? Which wave has the longer wavelength? (Video link: "141108diffA.")

Let's take a look at the frequency, which was stated to be the same for both these waves, but check if that really is the case. You'll need a friend to watch this animation with you. For the top wave, every time you see the hand at the left moves up to make a crest (or "hump"), say "now." "Now. Now. Now..." Keep doing that. For the bottom wave, convince your friend to say "now" every time the hand at the left moves up to make a crest there as well." "Now. Now. Now..." If the two of you do this correctly, the rate that you say "now" should more or less be the same rate that your friend says "now." This means that these waves should have (approximately) the same frequency.

To see that the speed is the same for both these waves let's time how long each wave takes to travel across the screen from left-to-right. Watch when a crest (or "hump") starts at the left for both waves, and then say "go." You'll watch the top wave crest move all the way across to the right end of the apparatus, while your friend will watch the bottom wave crest move all the way across to the left end of the apparatus. When your wave crest reaches the right end of the apparatus, stay "stop." There should be a tie (more or less) for the time it takes for these wave crests to travel from left to right, and since they travel the same distance in the same amount of time, then they must have (approximately) the same speed.

For the wavelength, note the horizontal distance from crest-to-crest for the top wave, and mark it with two of your fingers. Compare it to the horizontal distance from crest-to-crest for the bottom wave, which your friend can also mark with two fingers. This horizontal crest-to-crest distance should be more or less the same. This means that these waves have (approximately) the same wavelength.

In this experiment, amplitude of the two waves was different, while the frequency of the two was the same. The speed of the two waves was not affected by the difference in amplitude; and the wavelength of the two waves was also not affected by the difference in amplitude. Thus both wave speed and the wavelength do not depend on changes in amplitude, and wave speed and wavelength are both independent of the amplitude of a wave, whatever it is.

Here the hand oscillates up-and-down with the same amplitude, but the top case has a lower frequency and the bottom case has a higher frequency. Which wave travels with the faster speed? Which wave has the longer wavelength? (Video link: "141108difff.")

In this setup, the hand oscillates up-and-down with a given frequency and amplitude. As the wave travels from left-to-right, it is then transferred to another section that allows the wave to travel at a much greater speed. Along which section does the wave have a higher frequency? Along which section does the wave have a longer wavelength? (Video link: "Wavelength & Frequency: Different Media.")

Third: "standing waves," which we'll discuss in terms of resonance, rather than the more involved approach of wave superposition and reflections.

A string of finite length will naturally oscillate at its fundamental frequency f1 if it is plucked. (Video link: "111104-1270795.")

If this string is periodically disturbed at the same frequency as its fundamental frequency, then the string will undergo resonance, as the timing of the periodic disturbances matches the natural vibration of the string. (Video link: "111104-1270800.")

You can also set the string into another resonance by oscillating it at exactly twice the fundamental frequency, this results in an interesting pattern where there is a node at the center, where the string is always fixed. This is where the "standing" in standing waves comes from. (Video link: "111104-1270801.")

And same goes for oscillating the string at exactly three times the fundamental frequency (resulting in two equally spaced nodes), and so on. (Video link: "111104-1270803.")

It can be shown from a variety of proofs that the fundamental frequency of a string depends on the wave speed v (which depends on its tension and thickness), and length L. The frequencies that this string will resonate at are then merely integer multiples of the fundamental frequency.

In order to play different notes on a guitar, each string, while the same length, has different thicknesses (assuming that tensions are approximately equal), such that each string will vibrate at different fundamental frequencies when plucked. Would the thinnest or thickest strings have the slowest wave speed? Would the thinnest or thickest strings have the lowest fundamental frequency? (Video link: "Slow Motion GUITAR Strings -2000/4000% slower.")

In order to play a larger variety of different notes, a finger will hold down a string at a certain point between its ends, effectively shortening its length, and thus changing its fundamental frequency. Would decreasing the length of a string increase or decrease its wave speed? Would decreasing the length of a string increase or decrease its fundamental frequency? (Video link: "Music in slow motion - Guitar, Bass, Drum Kit, Piano and Violin.")

Which leads us to this bridge in Volgograd, Russia. High winds buffeting this bridge have apparently excited a very high resonant frequency along it--observe the pedestrian: is he on a node or antinode? (Video link: "A bridge across the Volga river in Volgograd, Russia.")