Physics midterm question: distance traveled vs. displacement

Physics 205A Midterm 1, fall semester 2019
Cuesta College, San Luis Obispo, CA

The v_x(t) graph of a Physics 205A student walking along a horizontal road is shown at right. The student started at x = 0 at t = 0. Discuss whether the magnitude of the distance traveled by the student is equal to or greater than the magnitude of the displacement. Explain your reasoning using the properties of velocity, position, time, distance traveled, and/or displacement.

Solution and grading rubric:

• p:
  Correct. Demonstrates that the distance traveled is equal to the magnitude of displacement, as the student always travels in the same (positive) direction (only positive horizontal velocity values), without reversing direction (no negative horizontal velocity values).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Conceptual understanding of displacement and distance traveled, but somehow misinterprets/misapplies/ignores the given information.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x:
  Implementation of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01duCk
p: 38 students
r: 0 students
t: 5 students
v: 4 students
x: 5 students
y: 0 students
z: 0 students

A sample "p" response (from student 5808):

Online reading assignment: impulse and momentum

Physics 205A, fall semester 2019
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and previewing a presentation on impulse and momentum.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"I understand that to clearly determine momentum you must have mass to figure out the amount of force the object is projecting on to another object. Also, impulse is the change in momentum over time which is used to determine how much net force is applied to an object, or the momentum-impulse theory. Finally, impulse can change an object's direction from left-to-right (or vice versa)."

"Momentum takes into account mass and speed of an object. Impulse takes into account net force and duration of time. The impulse-momentum theorem reflects the order of effects."

"Impulse is related to average force times the change in time and linear momentum is equal to max times velocity."

"The definition of impulse seemed pretty straightforward and I feel like I understand that, but I honestly think I'm going to have trouble with this chapter. It didn't really make sense to me when I was reading it. I think I'm going to need more practice questions."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"How is the impulse-momentum theory useful? When do we need it?"

"I'm having trouble grasping the concept of the impulse-momentum theorem."

"This only has to do with when two objects are touching?"

"What I initially found confusing was the second example of the presentation preview and how the direction towards the left was negative. But understanding that the left direction was considered negative for all the examples, then it wasn't so confusing."

"I'm just having trouble visualizing how all this works for different situations."

"However, the confusing aspect of the textbook and presentation is going to be the examples using it. I feel like it might be difficult and get confusing with all the other equations."

"The setup of impulse-momentum theorem in relation to the examples given in the book, need some lecture and problems to work to make the connection to visual ideas."

"After going through the presentation preview I released impulse was slightly confusing. Once I read through it a second time I was able to understand it."

"This seems pretty straightforward."

For the child hitting the tee ball with a bat, if the bat is swung such that it exerts the same net force on the tee ball for a longer time (by giving the bat more "follow-through"), the impulse on the tee ball will be __________, and the change in momentum of the tee ball will be:

less; less.	***** [5]
less; greater.	******* [7]
greater; less.	***** [5]
greater; greater.	*************************** [27]
(Unsure/lost/guessing/help!)	* [1]

For this golf ball initially at rest, and then has a speed of 97 m/s (to the right) after being hit by a golf club, indicate the horizontal directions (+/– signs) for the various impulse-momentum theorem quantities. (Only correct responses shown.)

Golf ball's initial momentum p₀: no direction (0). [87%]
Golf ball's final momentum p_f: to the right (+). [87%]
Golf ball's initial-to-final change in momentum ∆p: to the right (+). [89%]
Golf club's impulse "J" on the golf ball: to the right (+). [76%]

For this F/A-18E-F Super Hornet initially at rest, and then has a speed of 74 m/s after being it is catapulted (to the left), indicate the horizontal directions (+/– signs) for these impulse-momentum theorem vectors. (Only correct responses shown.)

Super Hornet's initial momentum p₀: no direction (0). [87%]
Super Hornet's final momentum p_f: to the left (–). [76%]
Super Hornet's initial-to-final change in momentum ∆p: to the left (–). [71%]
Catapult's impulse "J" on the Super Hornet: to the left (–). [58%]

For this Ford Ranger, hitting a crash barrier with a speed of 11.0 m/s (to the right), and then rebounding (to the left) off the crash barrier with a speed of 2.2 m/s, indicate the directions (+/– signs) for the various impulse-momentum theorem quantities. (Only correct responses shown.)

Ford Ranger's initial momentum p₀: to the right (+). [87%]
Ford Ranger's final momentum p_f: to the left (–). [78%]
Ford Ranger's initial-to-final change in momentum ∆p: to the left (–). [53%]
Crash barrier's impulse "J" on the Ford Ranger: to the left (–). [67%]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"The impulse and the change in momentum is fun."

"I think I am not understanding the relationship between momentum and impulse."

"What actually is the impulse and how is it different than momentum?"

"I still don't understand the concept of finding the momentum or how the impulse is ever different than the change in momentum because I thought that was the same thing. I think I'm just lost." (Impulse causes the change in momentum, so calculating the impulse (caused by a force acting over a period of time) allows you to find out the change in momentum, since they're mathematically equal to each other.)

"Is it possible to have a change in momentum without an impulse?" (No. Since impulses cause changes in momentum, any change in an object's momentum (which has magnitude and direction) means that there must be an impulse acting on it. Also if there is no change in an object's momentum, the there is no impulse acting on it. (If those examples sound like Newton's second law and Newton's first law, then yes, they can be applied to the impulse-momentum theorem.)

"The concept behind 'impulse' is a little confusing. The name kind of implies that it's instantaneous and only happens at one point in time." (The common meaning of "impulse" means a sudden urge to act, but the older meaning comes from "impel," or to drive forward, urge, or command.)

"Is there a way to measure impulse in a lab setting, like with a special tool?" (Since impulse is the amount of force exerted over a duration in time, then all you would need to measure impulse is the force sensor to measure how much force is exerted, and a stopwatch to time how long you would exert that amount of force. But also since you have a motion sensor (or the video analysis tool) that can track the velocities of objects, you can use that to tell you the initial velocity and the final velocity of an object, and when you multiply those velocities with the object's mass, you can calculate the initial momentum and final momentum of the object; and the change in momentum (final momentum minus initial momentum) is also equal to the impulse.)

"Something I didn't understand is how is 'J' the impulse but it also mentions that 'F⋅Δt' is also impulse. There are many equations." (It's just a definition. You exert an impulse (denoted by the vector "J") on an object by exerting a force over a period of time.)

"Will these equations be given or do we have to memorize them?" (These impulse and momentum equations are given on the quizzes and exams.)

"I am very uncertain about my answers to these examples. I thought I understood the reading but I will definitely need some clarification of these examples in class please!"

"I hope I got these questions right."

"Please go over these in class! thx"

"This case of confusion only exists if I'm mistaken in my understanding of the relationship between impulse and momentum. Since impulse is the product of time and net force, the resulting change in momentum should also increase as is the tee ball case. If not, I'm lost and need help." (You should be okay, as your reasoning sounds good.)

"I'm a bit confused on the change in momentum example where a truck has an initial velocity of 11m/s to the right and bounces off of a crash barrier with a velocity of –2.2m/s to the left. I know that change in momentum is found by subtracting (∆p_f – ∆p₀). So, for this example, I'm assuming it would be m⋅((– 2.2m/s) – (+11m/s)) to find the direction of the truck's initial-to-final change in momentum ∆p, which would be mass times –13.2m/s to the left." (This looks good. Sounds like you aren't that confused at all.)

"When it comes to the energy transfer-balance equation, I'm struggling in knowing when to drop what is not needed and a little on how to set it up after." (If there is an energy form that doesn't apply, then you drop that term (for example, if there are no springs involved, then you can drop the ∆PE_elas term. Or if there is no net initial-to-final change in that energy form, then you would also drop that term (for example, if an object is a rest on top of a vertical spring, which releases, and the object's final position is when it is (momentarily) at the highest point of its trajectory, then its initial and final translational kinetic energies are both zero, and then you would drop that term as ∆KE_trans = 0.)

"I'm enjoying this slightly cold weather."

Physics quiz question: Casio watch prototype testing

Physics 205A Quiz 2, fall semester 2019
Cuesta College, San Luis Obispo, CA

Casio engineer Kikuo Ibe tested G-Shock watch prototypes by placing them in a rubber ball dropped from a window, falling 10 m down to the ground[*]. Neglect air resistance. Choose up to be the +y direction.

"Story"
Casio America Inc.
gshock.com/technology/story

After being released from rest, the rubber ball took __________ to reach the ground.
(A) 0.49 s.
(B) 1.0 s.
(C) 1.4 s.
(D) 2.0 s.

[*] gshock.com/technology/story.

Correct answer (highlight to unhide): (C)

The following quantities are given (or assumed to be known):

(t₀ = 0 s),
(y₀ = 0 m),
y = –10.0 m (below the starting point),
v_0y = 0 m/s (no initial velocity),
a_y = –9.80 m/s².

So in the equations for constant acceleration motion in the vertical direction, the following quantities are unknown, or are to be explicitly solved for:

v_y = v_0y + a_y·t,

y = (1/2)·(v_y + v_0y)·t,

y = v_0y·t + (1/2)·a_y·(t)²,

v_y² = v_0y² + 2·a_y·y.

With the unknown quantity t to be solved for appearing in the third equation, with all other quantities given (or assumed to be known), then:

y = v_0y·t + (1/2)·a_y·(t)²,

(–10.0 m) = (0 m/s)·t + (1/2)·(–9.80 m/s²)·t²,

1.4285714286 s = t,

or to two significant figures, the elapsed time for the rubber ball to fall is 1.4 s.

(Response (A) is a_y/(2⋅y); response (B) is sqrt(y/a_y); response (D) is 2·y/a_y.)

Sections 70854, 70855
Exam code: quiz02Cs1o
(A) : 2 students
(B) : 17 students
(C) : 29 students
(D) : 6 students

Success level: 54%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.72

Physics quiz question: Swatch Internet Time beats

Physics 205A Quiz 1, fall semester 2019
Cuesta College, San Luis Obispo, CA

Swatch Internet Time was proposed in 1988, where 1 day is divided into 1,000 "beats."[*] A day is defined to be 24 hours, an hour is defined to be 60 minutes, and a minute is defined to be 60 seconds. The duration of 1 beat is:
(A) 3.6 s.
(B) 86.4 s.
(C) 1.5×10⁵ s.
(D) 8.64×10⁷ s.

[*] swatch.com/en_us/internet-time#main.

Correct answer (highlight to unhide): (B)

The duration of a beat in seconds can be determined by setting up conversion factors such that unwanted units cancel (beats, days, hours, minutes) while desired units remain (seconds):

1 beat = (1 beat)·((1 d)/(1,000 beats))·((24 h)/(1 d))·((60 min)/(1 h))·((60 s)/(1 min)),

1 beat = (1 beat)·((1 d)/(1,000 beats))·((24 h)/(1 d))·((60 min)/(1 h))·((60 s)/(1 min)/),

1 beat = 86.4 seconds.

(Note that all of the factors in this calculation are exact definitions or conversion factors.)

(Response (A) is ((1/1000)·60·60; response (C) is (1000)·(1/24)·60·60; response (D) is (1000)·24·60·60.)

Sections 70854, 70855
Exam code: quiz01B34t
(A) : 3 students
(B) : 46 students
(C) : 4 students
(D) : 1 student

Success level: 84%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.42

Physics midterm question: position of vertically-launched ball

Physics 205A Midterm 1, fall semester 2018
Cuesta College, San Luis Obispo, CA

A Physics 205A student slings a ball straight upwards. The v_y(t) graph of this ball is shown at right, starting from when the ball was released at y = 0 at t = 0. Neglect air resistance. Choose up to be the +y direction. At t = 6 s, discuss why the ball is at a position higher than its release point. Explain your reasoning using the properties of velocity, position, distance traveled, and/or displacement.

Solution and grading rubric:

• p:
  Correct. Supports claim that the ball is still above its release point at t = 6 s by discussing at least one of the following explanations:
  1. displacement is the bounded area between the velocity function and the time axis, and since there is a greater bounded area above the time axis (corresponding to a positive displacement for t = 0 to 4 s) than the bounded area below the time axis (corresponding to a negative displacement for t = 4 s to 6 s), the ball has traveled farther up from its starting point to its highest height, than traveling downwards from its highest height to its final position at t = 6 s; or
  2. the ball slows down from an upwards velocity of +40 m/s at t = 0 to zero velocity at its highest point at t = 4 s, and from symmetry, the ball should fall back down to its starting point with a downwards speed of –40 m/s at t = 8 s, such that the ball is still somewhere above its starting point at t = 6 s; or
  3. uses kinematic equations for constant motion to show that the final position at t = 6 s is still positive.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x:
  Implementation of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01g4iN
p: 24 students
r: 5 students
t: 21 students
v: 8 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 1414):
Another sample "p" response (from student 3691):

Online reading assignment: impulse and momentum

Physics 205A, fall semester 2018
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and previewing a presentation on impulse and momentum.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"The concept of momentum and that its magnitude depends on both the mass and speed of an object, and how impulse is the product of the net force acting on an object and the duration of time that the net force acted on this object."

"That momentum is the mass of an object times the speed of that object and it is a vector because of the direction of that object's speed. Impulse is dependent on the average net force applied on an object and how long that force is applied."

"According to the impulse-momentum theorem, impulse causes a change in momentum."

"Both momentum and impulse are vector quantities. Momentum is the product of mass and velocity and change in momentum could be caused by impulse which is equal to the product of force and time."

"How to find the change in momentum. The change in momentum is the final momentum minus the initial momentum."

"It seemed like I was applying momentum and impulse the correct way during the examples but now going back over it I don't think I have a clear understanding of what either of them are."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"I found the impulse-momentum theorem a bit confusing."

"I didn't understand the concept of impulse, 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."

"I'm struggling to understand what exactly impulse and momentum are. More like what they do, and how to describe them in a problem."

"I don't understand why time plays a role in impulse and the change in momentum, but I think I understand now."

"Figuring out the signs (+ or –) when finding the change in momentum. I think it depends on what direction the object is going, but on some of the examples I was not sure what the signs would be."

"I honestly didn't find much in this chapter confusing. The subjects of momentum and impulse were not very difficult to me."

For the child hitting the tee ball with a bat, if the bat is swung such that it exerts the same net force on the tee ball for a longer time (by giving the bat more "follow-through"), the impulse on the tee ball will be __________, and the change in momentum of the tee ball will be:

less; less.	[0]
less; greater.	******* [7]
greater; less.	***** [5]
greater; greater.	***************************** [29]
(Unsure/lost/guessing/help!)	*** [3]

For this golf ball initially at rest, and then has a speed of 97 m/s (to the right) after being hit by a golf club, indicate the horizontal directions (+/– signs) for the various impulse-momentum theorem quantities. (Only correct responses shown.)

Golf ball's initial momentum p₀: no direction (0). [86%]
Golf ball's final momentum p_f: to the right (+). [86%]
Golf ball's initial-to-final change in momentum ∆p: to the right (+). [81%]
Golf club's impulse "J" on the golf ball: to the right (+). [70%]

For this F/A-18E-F Super Hornet initially at rest, and then has a speed of 74 m/s after being it is catapulted (to the left), indicate the horizontal directions (+/– signs) for these impulse-momentum theorem vectors. (Only correct responses shown.)

Super Hornet's initial momentum p₀: no direction (0). [86%]
Super Hornet's final momentum p_f: to the left (–). [39%]
Super Hornet's initial-to-final change in momentum ∆p: to the left (–). [36%]
Catapult's impulse "J" on the Super Hornet: to the left (–). [45%]

For this Ford Ranger, hitting a crash barrier with a speed of 11.0 m/s (to the right), and then rebounding (to the left) off the crash barrier with a speed of 2.2 m/s, indicate the directions (+/– signs) for the various impulse-momentum theorem quantities. (Only correct responses shown.)

Ford Ranger's initial momentum p₀: to the right (+). [86%]
Ford Ranger's final momentum p_f: to the left (–). [39%]
Ford Ranger's initial-to-final change in momentum ∆p: to the left (–). [36%]
Crash barrier's impulse "J" on the Ford Ranger: to the left (–). [45%]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"Can you go over the impulse-momentum theorem?"

"Is it possible to go over more conceptual aspects of this rather than examples? I'm having trouble figuring out from the reading what each of these concepts mean and I feel that I should know better what they are before I do example practice problems."

"Can you do more problems in class?"

"No questions on today's assignment, I think I got this so far!"

"I think I understood this stuff...after answering these questions, let's hope!"

"What?"

"Which is a scarier thought? That the human race is the most advanced form of life in the universe, or that we are mere amoebae compared to other life forms?"

"PANICKING!! WORRIED ABOUT MIDTERM!! Practice midterm NOT helpful. Sure we have the long weekend to study, but it will be difficult to ask you for help face-to-face (which is more effective than email)." (I will specifically go over what you need to know for the midterm, and target the Wednesday's review session such that ideally you can spend the long weekend how to apply the tools that you're given for the midterm, rather than trying to determine what tools are needed for the midterm.)

Physics quiz question: vertical golf ball throw

Physics 205A Quiz 2, fall semester 2018
Cuesta College, San Luis Obispo, CA

"1700 GOLF BALLS VS. TRAMPOLINE from 45m!"
How Ridiculous
youtu.be/gz7nk2PylPM

Scott Gaunson for the "How Ridiculous" YouTube channel threw a golf ball up towards the top platform of the Gravity Discovery Centre and Observatory's Leaning Tower of Gingin, near Perth, Australia[*']. Video analysis shows that from the moment it was released from his hand (with an initial upwards speed), the golf ball took 1.8 s to travel 45 m upwards to the top platform, where it still had an upwards speed of 15 m/s. Neglect air resistance. Choose up to be the +y direction. At the moment it left his hand, the golf ball had an initial speed of:
(A) 25 m/s.
(B) 27 m/s.
(C) 33 m/s.
(D) 43 m/s.

[*] youtu.be/gz7nk2PylPM.

Correct answer (highlight to unhide): (C)

The following quantities are given (or assumed to be known):

(t₀ = 0 s),
(y₀ = 0 m),
y = +45 m (level of the top platform above the throw release),
v_y = +15 m/s (moving upwards at the level of top platform),
t = 1.8 s (time when at the level of top platform),
a_y = –9.80 m/s².

So in the equations for constant acceleration motion in the vertical direction, there are no quantities that are unknown, and only one to be explicitly solved for:

v_y = v_0y + a_y·t,

y = (1/2)·(v_y + v_0y)·t,

y = v_0y·t + (1/2)·a_y·(t)²,

v_y² = v_0y² + 2·a_y·y.

Note that as the quantity v_0y to be solved for appears as the only unknown in every one of the equations above, with all other quantities given (or assumed to be known), then any one of these equations could be used to solve for the initial velocity. Here we only demonstrate how the first equation is used:

v_y = v_0y + a_y·t,

+15 m/s = v_0y + (–9.80 m/s²)·(1.8 s),

v_0y = +15 m/s –(–9.80 m/s²)·(1.8 s) = +15 m/s + 17.64 m/s = +32.64 m/s,

or to two significant figures, the initial speed of the golf ball was 33 m/s (in the upwards direction).

(Response (A) is y/t; response (B) is v_f·t; response (D) is v_f·t – (1/2)·a_y·(t²).)

Sections 70854, 70855
Exam code: quiz02HwRd
(A) : 5 students
(B) : 5 students
(C) : 41 students
(D) : 1 student

Success level: 79%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.31

Physics midterm question: comparing distances traveled

Physics 205A Midterm 1, fall semester 2017
Cuesta College, San Luis Obispo, CA

The x(t) graph of a Physics 205A student walking along a straight line is shown at right. The student started at x = 0 at t = 0. Discuss why the student traveled a farther distance from t = 0 to t = 7 s than the distance traveled from t = 7 s to t = 10 s. Explain your reasoning using the properties of position, distance traveled, and displacement.

Solution and grading rubric:

• p:
  Correct. Supports claim that student traveled a farther distance from t = 0 to t = 7 s than from t = 7 s to t = 10 s by discussing:
  1. distance traveled counts both forwards and backwards motion, such that from t = 0 to t = 7 s the student traveled a total distance of 3 m (2 m in the forwards direction, then 1 m in the backwards direction); compared to
  2. the distance traveled by the student from t = 7 s to t = 10 s is 2 m (always in the forward direction).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. At least discussion demonstrates general understanding of distinction between distance traveled and displacement, but does not compare differences in distances traveled for the two time intervals.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. May compare areas, slopes, elapsed times, average speeds and/or average velocities.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855
Exam code: midterm01mOoL
p: 25 students
r: 3 students
t: 10 students
v: 15 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 2030):

Online reading assignment: impulse and momentum

Physics 205A, fall semester 2017
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and previewing a presentation on impulse and momentum.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"This is another new approach to connect forces with changes in motion. This new approach does not involve Newton's laws but instead uses impulse and momentum (both are vectors)."

"Momentum is clicking very well for me. I understand that momentum is mass times velocity. It is also a vector quantity."

"An object's momentum is its mass and velocity combined. Impulse is the change in momentum, which is the average net force over an elapsed time."

"Impulse causes a change in momentum, and impulse is defined by the average net force exerted on an object multiplied by the contact time, so how long that force was exerted on the object. Momentum is mass times velocity."

"Impulse is equal to the change in momentum. Impulse is also a vector quantity and has the same direction as the average force, and is force times ∆t."

"Impulse is related to momentum because impulse is equal to the change of momentum."

"The impulse must have the same direction as the change in momentum, and so it must also point to the right if ∆p is pointing or going to the right."

"If a force is applied on an object for a given amount of time, that object feels an impulse."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"After reading this section, I think I still need additional help understanding the whole impulse concept."

"Most of this is confusing. Mainly how to calculate impulse of an object. I'm not really sure still what impulse is or when to use it."

"I have a hard time visualizing how impulse works. I guess until now I was just thinking about impulse and momentum as the same thing?"

"Wrapping my head around the impulse-momentum theorem."

"I am confused by impulse but I am okay with momentum."

"The units of impulse because I find them nonintuitive. I think it's more of a personal conceptual issue but sometimes units are nonintuitive."

"Nothing was necessarily confusing, I just need practice with the new round of formulas."

"I found this pretty simple."

For the child hitting the tee ball with a bat, if the bat is swung such that it exerts the same net force on the tee ball for a longer time (by giving the bat more "follow-through"), the impulse on the tee ball will be __________, and the change in momentum of the tee ball will be:

less; less.	[0]
less; greater.	*** [3]
greater; less.	******* [7]
greater; greater.	************************** [26]
(Unsure/lost/guessing/help!)	* [1]

For a golf ball initially at rest, and then has a speed of 97 m/s after being hit by a golf club, indicate the horizontal directions (+/– signs) for the various impulse-momentum theorem quantities. (Only correct responses shown.)

Golf ball's initial momentum p₀: no direction (0). [89%]
Golf ball's final momentum p_f: to the right (+). [95%]
Golf ball's initial-to-final change in momentum ∆p: to the right (+). [86%]
Golf club's impulse "J" on the golf ball: to the right (+). [76%]

For this catapult-launched F/A-18E-F Super Hornet taken from rest to a final speed of 74 m/s, indicate the directions (+/– signs) for the various impulse-momentum theorem quantities. (Only correct responses shown.)

Super Hornet's initial momentum p₀: no direction (0). [%]
Super Hornet's final momentum p_f: to the left (–). [32%]
Super Hornet's initial-to-final change in momentum ∆p: to the left (–). [19%]
Catapult's impulse "J" on the Super Hornet: to the left (–). [22%]

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, indicate the directions (+/– signs) for the various impulse-momentum theorem quantities. (Only correct responses shown.)

Ford Ranger's initial momentum p₀: to the right (+). [73%]
Ford Ranger's final momentum p_f: to the left (–). [78%]
Ford Ranger's initial-to-final change in momentum ∆p: to the left (–). [49%]
Crash barrier's impulse "J" on the Ford Ranger: to the left (–). [51%]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"Could you please go over the above examples in class?"

"Please review the Ford Ranger problem!"

"Going over the above examples would be extremely helpful. Thank you for the extra examples and explanations in class lately!"

"If two objects moving in opposite directions hit each other, how do we measure the impulse?" (You just need to know the mass of each car, and each car's initial and final velocities. Accident reconstruction experts can look up the mass of a vehicle, and then deduce the initial and final velocities of each car by looking at various clues such as length of tire skid marks, vehicle sensor recordings, etc.)

"Great job so far! I'm really enjoying your class."

"Sorry, I have been loaded with research and studying lately. I don't mean to skip your assignments, but I appreciate that you take into account busy schedules and let that slide. Thank you." (You still get credit for completing as much as you can on these reading assignments, even if it's just touching bases with me.)

"Will there be questions on the midterm with similar format as the homework?" (Yes, since many of these questions are taken from previous semesters' midterms. We'll start reviewing for the midterm today.)

"I think having the answers to these reading questions would help with understanding." (We go over these in class; also part of each reading assignment is to read through the answers to the previous reading assignment; follow the links to the blog. If you find that just the answers (without explanations) aren't enough for you, then come and seem me during office hours, or ask questions via e-mail. I'm not going to make answers available while the reading assignment is open, though, as I really want to see how the class will do without influencing their responses.)

"I'm confused about the first slide--so did the ping-pong ball actually knock over the bowling pins? I don't understand the relevance." (Since momentum is mass times velocity, in order for a ping-pong ball to have the same momentum as a bowling ball (which can knock down pins), the ping-pong ball would need to have an absurdly fast velocity to make up for its tiny mass.)

I'm having a little bit of trouble distinguishing impulse versus momentum. I'm okay with momentum more than impulse since I've worked more with it, but if you could go over them in relation to each other I'd appreciate it."

"How is the impulse-momentum theorem really that different from Newton's laws about net forces, etc.?" (They basically say the same thing (impulse causes change in momentum; net force causes change in motion); so it's just using different terminology and concepts to explain the same thing. But conservation of momentum is where things gets interesting, when we apply it to collisions between two objects, after the midterm.)

"No comment." (Eh, you just did.)

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) p_f 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 = p_f – p₀,

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 p_f 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 p_f 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?