Attempting to teach this: Momentum and Impulse

[Windrose Academy]

Homework Statement

I have taken on a task that may be beyond me. I have a homeschool co-op of 15 teens who are taking an integrated science course. Our book starts with physics. While I have an answer key to questions it does not come with explanations of how the answers were obtained. I somehow thought the instructor book would come with more details. The kids seem to grasp the concepts of Newtons laws through expieriments, but a few of the word problems they have come up with different answers and I can't seem to find where the mistakes are. Any help in being able to explain or showing the work would be greatly appreciated.

Relevant Equations

Bronco Brown wants to put Ft= ️mv to the test and bungee jumping. Bronco leaps from a high cliff and experiences free fall for 3 seconds. Then the bungee cord begins to stretch, reducing his speed to zero in 2 seconds. Fortunately, the cord stretches to its maximum length just short of the ground below. Bronco's mass is 100kg and acceleration of free fall is 10 m/s²
1. The 3 s free fall distance of Bronco just before the bungee cord begins to stretch?

2. ️ mv during the 3 second interval of free fall?

3. ️ mv during the 2 second intervals of slowing down?

4. Impulse during the 2 s interval of slowing down

5. Average force exerted by the cord during the 2 s interval of slowing down?

6. How about work and energy ? How much KE does Bronco have 3 s after his jump?
7. How much does gravitational PE decrease during this 3 s ?

For question 1.
I am stuck. I know that the equation involves time and possibly rate, should solve for distance. But not sure how to set it up with information given.

2. Ft= m ️ v
F(3)= (100kg)(30m/s)
3 s= 3000 kg m/s

Same applied to question 3.
3. F(2)= (100kg)(-30m/s)
F(2) = -3000 kg m/s

4. J=mv(f) - mv(o)
3000N•s

5. Unsure how to find the force. I believe it it requires manipulation of Work formula, but unsure how to move it around.

6. KE= ½mv²
KE=½(100)(30)²
KE= 45000 J

7. The inverse of 6. So the answer is the same 45000J

 

[Steve4Physics]

Hi @Windrose Academy. Welcome to PF.

Let’s deal specifically with Q1 only, otherwise things can get muddled - especially if you get replies from different people about different questions.

You have correctly calculated (in Q2) that the speed after 3s is 30m/s. (Though you didn’t show your working.) I’ll assume you understand.

So you know that during the first 3 seconds the speed increases, at a uniform rate, from 0 to 30m/s.

Can you answer these questions:
a) What is the average speed?
b) Therefore what distance is covered in the 3s?

Note: There are standard formulae you can (and may be expected to) use to calculate the distance (displacement really). Check earlier chapters of your textbook.

 

[PeroK]

Homework Statement: I have taken on a task that may be beyond me. I have a homeschool co-op of 15 teens who are taking an integrated science course. Our book starts with physics. While I have an answer key to questions it does not come with explanations of how the answers were obtained.

Perhaps you need to take a physics course yourself first? I don't see how you can teach physics without understanding it.

PS You might be better getting the kids to post homework questions on here. And cut out the middle-man, as it were.

 

[Mister T]

Homework Statement: I have taken on a task that may be beyond me. I have a homeschool co-op of 15 teens who are taking an integrated science course.

Hmmm.... I never understood the concept of home-schooled students taking a course at an academy or anywhere else. I thought that home-schooled students were schooled at home, not at an academy. I had a similar teaching experience and found that the only way to get the students to behave was to have one of the parents sit in on the class. The students didn't seem to want to learn or even know how to learn in a formal learning environment. But I digress. Let me try to help you.

First of all, your screen shot is not legible. Use an app like Tiny Scanner rather than just taking a photo, or better yet, use a real scanner. Also, instead of posting a series of questions followed by a series of attempted solutions, it would be easier to read, follow, and respond if you posted each question immediately followed by its attempted solution. And my final advice is for you to find a "mentor" at a local high school or college who can help you in a face-to-face situation.

Relevant Equations: Bronco Brown wants to put Ft= ️mv to the test and bungee jumping. Bronco leaps from a high cliff and experiences free fall for 3 seconds. Then the bungee cord begins to stretch, reducing his speed to zero in 2 seconds. Fortunately, the cord stretches to its maximum length just short of the ground below. Bronco's mass is 100kg and acceleration of free fall is 10 m/s²
1. The 3 s free fall distance of Bronco just before the bungee cord begins to stretch?

For question 1.
I am stuck. I know that the equation involves time and possibly rate, should solve for distance. But not sure how to set it up with information given.

The speed changes uniformly from 0 to 30 m/s, so the average speed is 15 m/s. For 3 s. Thus the distance is 45 m. But this is not relevant to the question asked.

2. ️ mv during the 3 second interval of free fall?

2. Ft= m ️ v
F(3)= (100kg)(30m/s)
3 s= 3000 kg m/s

The last line should be (F)(3 s) =(100 kg)(30 m/s), which results in F = 1000 N. This is the same result you'd get using F = mg. Again, this is not relevant to the solution of the problem.

3. ️ mv during the 2 second intervals of slowing down?

3. F(2)= (100kg)(-30m/s)

F(2) = -3000 kg m/s

So F = -1500 N.

4. Impulse during the 2 s interval of slowing down.

4. J=mv(f) - mv(o)

3000N•s

That's right.

5. Average force exerted by the cord during the 2 s interval of slowing down?

See above.

6. How about work and energy ? How much KE does Bronco have 3 s after his jump?
7. How much does gravitational PE decrease during this 3 s ?

These are not relevant. The problem asks you to use impulse-momentum, not work-energy.

 

[Chestermiller]

With regard to question 1, the gravitational force acting of the man F is constant at F= mg = (100)(10)= 1000 Newtons. So, for 3 seconds, the impulse of the force on the man is 3000 Newton.secs. He starts out with zero velocity and ends up with a velocity of v. So his change in momentum is 100(v-0)=1000v. So we have $$Ft=m\Delta v=3000=100v$$

 

[haruspex]

100(v-0)=1000v.

Typo

 

[Windrose Academy]

Thank you all for your help. A few hours sleep made a big difference as well. While my screen name does say Academy please know I do not have an actual Academy- I am a mother who happens to have 2 teens who homeschool. They wanted to take a physics class.

As many of you know science is more fun and easier to understand when done hands on. And of course studing with friends and working in groups is a big part of science, so I reached out to our local homeschool community and asked if other teens would be interested in learning as well. This is how I ended up with 15 teens.

When I chose the curriculum I thought the instructor manual would come the solutions to the problems. Not just the answers. I did take physics in high school. But that was 30+ years ago and not my current field of work. This particular course covers physics, chemistry, biology, and earth science. As a working Emergency Room RN I am confident in teaching chemistry and biology.

Again thanks for the help.

 

[Mister T]

When I chose the curriculum I thought the instructor manual would come the solutions to the problems. Not just the answers. I did take physics in high school.

To teach physics in high school the minimum education required is a bachelor's degree. Unfortunately, many of the teachers have no more experience in physics than you do. It's the students who get short-changed in this process.

I would suggest that you find another book to use for your students, something that comes with a detailed instructor's guide, or at least an instructor's manual that has the solutions worked out.

 

[Windrose Academy]

Yes for #1
We used d=½gt² rounding up to 10 for g.

For #2.
The answer is looking for the delta mv for the 3 s free fall. I used
Ft= ️ mv
F(3)= ️ (100 kg) (30 m/s)
F(3)= 3000 kg • m/s

For #5. Looking for the force. I used
F=m( vi-vf)/t
F=100kg(30m/s-0)/2s
F=100kg(30m/s)/2s
F=1500N

I really think I was just very tired after working 12 hour shift and trying to reason math I hadn't done in 30+ plus years in the middle of the night.

These kids have had so much fun exploring laws of motion with balloon races, building perpetual motion devices with bottle caps and staws, pulling table cloths and seeing effects of mass in a homemade water-bottle sling shots. It was time to put all the math on behind the laws on paper.

Truly thank you so much for answering the questions. Offering support in a forum that is meant to help each other. Class today was a huge success.

 

[berkeman]

I really think I was just very tired after working 12 hour shift and trying to reason math I hadn't done in 30+ plus years in the middle of the night.

I had wondered why your post was from 1:30AM...

building perpetual motion devices with bottle caps and staws

Oops, probably best if you read this for background: https://www.physicsforums.com/insights/why-we-dont-discuss-perpetual-motion-machines-pmm/

Glad that the class is going better.

 

[kuruman]

To @Windrose Academy : It took me a few seconds to decipher this as Δ(mv).

️ mv

For future reference there is a menu with special characters that you might wish to use. See screenshot below