Hovercraft B/C

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IvanGe
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Re: Hovercraft B/C

Post by IvanGe » May 12th, 2018, 1:39 pm

i didn't feel like posting this in the Question Marathon, so i guess i'll just post it here because it's more like a check-my-work-please.

A bungee jumper of mass 40 kg is attached to a bungee cord with a constant of 80 n/m. The unstretched length of the bungee is 15m. What is the maximum velocity of the jumper? (g = 10 m/s^2)

I did it by using the potential energy of the spring and when it is at maximum velocity, the potential energy should completely change into kinetic energy(even though it doesn't in this case). So :

Hooke's law
f = kx to find x

PE = KE + PEspring
mgh = 1/2kx^2 + 1/2mv^2

assuming h is equal to 15 + x

The answer key said 17.32 m/s and I did not get that by doing this, instead I got 18.71 m/s
Last edited by IvanGe on May 12th, 2018, 7:37 pm, edited 2 times in total.
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Re: Hovercraft B/C

Post by MattChina » May 12th, 2018, 7:12 pm

IvanGe wrote:i didn't feel like posting this in the Question Marathon, so i guess i'll just post it here because it's more like a check-my-work-please.

A bungee jumper of mass 40 kg is attached to a bungee cord with a constant of 80 n/m. The unstretched length of the bungee is 15m. What is the maximum velocity of the jumper? (g = 10 m/s^2)

I did it by using the potential energy of the spring and when it is at maximum velocity, the potential energy should completely change into kinetic energy(even though it doesn't in this case). So :

Hooke's law
f = kx to find x

PE = KE + PEspring
mgh = 1/2kx^2 + 1/2mv^2

assuming h is equal to 15 + x

The answer key said 17.32 m/s and I did not get that by doing this, instead I got 18.71 m/s^.
Idk man, Got the same answer.
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Re: Hovercraft B/C

Post by UTF-8 U+6211 U+662F » May 12th, 2018, 8:00 pm

MattChina wrote:
IvanGe wrote:i didn't feel like posting this in the Question Marathon, so i guess i'll just post it here because it's more like a check-my-work-please.

A bungee jumper of mass 40 kg is attached to a bungee cord with a constant of 80 n/m. The unstretched length of the bungee is 15m. What is the maximum velocity of the jumper? (g = 10 m/s^2)

I did it by using the potential energy of the spring and when it is at maximum velocity, the potential energy should completely change into kinetic energy(even though it doesn't in this case). So :

Hooke's law
f = kx to find x

PE = KE + PEspring
mgh = 1/2kx^2 + 1/2mv^2

assuming h is equal to 15 + x

The answer key said 17.32 m/s and I did not get that by doing this, instead I got 18.71 m/s^.
Idk man, Got the same answer.
I got the same answer.
At any particular moment at time, the Law of Conservation of Energy applies.
[math]mgh = \frac12kx^2 + \frac12mv^2[/math]
The left side is the gravitational potential energy of the bungee jumper before he jumps (h is the distance from this height to the height of the bungee jumper at this particular point in time).
The right side is the elastic potential energy of the cord (follows Hooke's law, where x is the additional stretchage of the cord past 15 m) and the kinetic energy of the bungee jumper.
To proceed, we need one more equation:
[math]h = 15 m + x[/math]
This is because at any given point, h is the length of the bungee cord.
Plug in values:
[math](40 kg)(10 \frac{m}{s^2})(15 m + x) = \frac12(80 \frac{N}{m})x^2 + \frac12(40 kg)v^2[/math]
Simplify:
[math]6000 J + (400 N)x = (40 \frac{N}{m})x^2 + (20 kg)v^2[/math]
Solve for [math]v^2[/math]:
[math]v^2 = -(2 \frac{1}{s^2})x^2 + (20 \frac{m}{s^2})x + 300 \frac{m^2}{s^2}[/math]
At this point, you can find the maximum by (1) finding the maximum of the parabola and then square rooting it or (2) plugging [math]v = \sqrt{-(2 \frac{1}{s^2})x^2 + (20 \frac{m}{s^2})x + 300 \frac{m^2}{s^2}}[/math] in your graphing calculator and letting it find the maximum for you.

Either way, you get the answer [math]5\sqrt{14} \frac{m}{s} \approx 18.7 \frac{m}{s}[/math]

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Re: Hovercraft B/C

Post by IvanGe » May 12th, 2018, 8:20 pm

UTF-8 U+6211 U+662F wrote:
MattChina wrote:
IvanGe wrote:i didn't feel like posting this in the Question Marathon, so i guess i'll just post it here because it's more like a check-my-work-please.

A bungee jumper of mass 40 kg is attached to a bungee cord with a constant of 80 n/m. The unstretched length of the bungee is 15m. What is the maximum velocity of the jumper? (g = 10 m/s^2)

I did it by using the potential energy of the spring and when it is at maximum velocity, the potential energy should completely change into kinetic energy(even though it doesn't in this case). So :

Hooke's law
f = kx to find x

PE = KE + PEspring
mgh = 1/2kx^2 + 1/2mv^2

assuming h is equal to 15 + x

The answer key said 17.32 m/s and I did not get that by doing this, instead I got 18.71 m/s^.
Idk man, Got the same answer.
I got the same answer.
At any particular moment at time, the Law of Conservation of Energy applies.
[math]mgh = \frac12kx^2 + \frac12mv^2[/math]
The left side is the gravitational potential energy of the bungee jumper before he jumps (h is the distance from this height to the height of the bungee jumper at this particular point in time).
The right side is the elastic potential energy of the cord (follows Hooke's law, where x is the additional stretchage of the cord past 15 m) and the kinetic energy of the bungee jumper.
To proceed, we need one more equation:
[math]h = 15 m + x[/math]
This is because at any given point, h is the length of the bungee cord.
Plug in values:
[math](40 kg)(10 \frac{m}{s^2})(15 m + x) = \frac12(80 \frac{N}{m})x^2 + \frac12(40 kg)v^2[/math]
Simplify:
[math]6000 J + (400 N)x = (40 \frac{N}{m})x^2 + (20 kg)v^2[/math]
Solve for [math]v^2[/math]:
[math]v^2 = -(2 \frac{1}{s^2})x^2 + (20 \frac{m}{s^2})x + 300 \frac{m^2}{s^2}[/math]
At this point, you can find the maximum by (1) finding the maximum of the parabola and then square rooting it or (2) plugging [math]v = \sqrt{-(2 \frac{1}{s^2})x^2 + (20 \frac{m}{s^2})x + 300 \frac{m^2}{s^2}}[/math] in your graphing calculator and letting it find the maximum for you.

Either way, you get the answer [math]5\sqrt{14} \frac{m}{s} \approx 18.7 \frac{m}{s}[/math]
aight guys I guess the answer key was wrong? It's from Cypress Falls Invitational Division C for those wondering.
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Re: Hovercraft B/C

Post by LittyWap » May 17th, 2018, 3:13 pm

Thought you guys would want to see this. Anyone else ready for Nats??

https://imgur.com/a/5Pw0St2
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Re: Hovercraft B/C

Post by Tesel » May 18th, 2018, 7:27 pm

LittyWap wrote:Thought you guys would want to see this. Anyone else ready for Nats??

https://imgur.com/a/5Pw0St2
An impressive score. Good luck repeating it at Nationals!
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Re: Hovercraft B/C

Post by CMS AC » May 19th, 2018, 5:04 pm

Oof... I know someone that got 14.97 seconds in division c. Good luck lol.

Also, for division b, is it just me or was there a lot of friction and the track was uphill a little bit?

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Re: Hovercraft B/C

Post by sciolycoach » May 21st, 2018, 6:44 am

Hello Everybody,

First, congratulations to everybody who competed on Saturday. I was the Division B Hovercraft supervisor at Nationals and, as always, was quite impressed with everybody's devices and tests. I was equally impressed with the politeness, kindness, hard work, and positive sportsmanship that I witnessed throughout the day, making my ballot for the Spirit Award very difficult.

The tracks were generously loaned from the Colorado State Tournament and had an MDF base with no seams. The tracks were very precisely leveled in the morning and this was checked throughout the day in several spots and were as close to level as possible. Shims were placed under the track in places to ensure levelness.

As always I was extremely impressed with the level of vehicles and even more impressed with the ability of students to make corrections on the fly as needed. It was clear that several cars were impacted by the travel to nationals and the best teams were able to make modifications as needed.

As I usually do after a National Tournament I will share some summary statistics about the event but NO specific scores or maximum/minimum scores that could be used to identify specific teams. I will not answer specific questions about the raw scores for certain places and will not release the scoring spreadsheet. Please don't inundate me with questions about your specific team's place or score as I simply cannot and will not release them.

Mass Score (MS)
Nearly 30% of all teams completed at least one run with all 16 rolls of pennies.
At least one of the top 6 teams completed both runs with less than 16 rolls of pennies and instead focused on getting strong time and exam scores.

Chart Score (CS)
Nearly 80% of all teams earned the full 10 points for the chart score. The vast majority of the teams to not earn the full points had no charts or diagrams.

Time Score (TS)
The median time score was nearly 9 points (out of 21)
The third quartile time score was nearly 17 points (out of 21)
The closest time was a 14.90 seconds.

Exam Raw Score
The exam was out of 49 total points (I threw one question out).
The median raw exam score was 16 points.
The third quartile raw exam score was 30 points.
The 90% raw exam score was 35 points.
The high raw exam score was 43 points.

Exam Score
The First Quartile was nearly 10 points (out of 41)
The Median was nearly 17 points (out of 41)
The Third Quartile was over 31 points (out of 41)
Of course the max was 41 points.

It is clear that the test was the primary separator among team scores. Every team got at least four questions right, and I worked to create a test with a mix of relatively easy, medium, and hard questions, though I am sure some would have liked a more difficult test, and others would have liked an easier test. Just remember, as the old saying goes it is impossible to please everybody but you can die trying.


Congratulations again to all of the competitors and take a well-deserved break. Enjoy your summer and I hope to see many of you next year in different tournaments around the nation.

Andy Hamm
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Re: Hovercraft B/C

Post by antoine_ego » May 21st, 2018, 7:16 pm

sciolycoach wrote:Hello Everybody,
The tracks were generously loaned from the Colorado State Tournament and had an MDF base with no seams. The tracks were very precisely leveled in the morning and this was checked throughout the day in several spots and were as close to level as possible. Shims were placed under the track in places to ensure levelness.
Out of curiosity, what were the Div C tracks made out of?
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2018 Hovercraft Nationals - 6th
2018 Mousetrap Nationals - 6th
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2019 Nationals - Team 9th Place!
[/b]
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Re: Hovercraft B/C

Post by windu34 » May 21st, 2018, 7:38 pm

antoine_ego wrote: Out of curiosity, what were the Div C tracks made out of?
Melamine. This was told to coaches during the coaches meeting on Friday night.
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