Hovercraft B/C

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Hovercraft B/C

Postby Birdmusic » October 29th, 2017, 10:09 am

No hovercraft question marathon yet? I think I'll start one.

A train has an initial velocity of 4 m/s. It speeds up with constant acceleration to 8 m/s over a period of 2 seconds. What is its displacement? What equation did you use?
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Re: Hovercraft B/C

Postby heyimben » November 7th, 2017, 1:45 pm

24m? I literally just found a basic formula for displacement...

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Re: Hovercraft B/C

Postby Zioly » November 7th, 2017, 7:07 pm

I used delta(x)=v(initial)t+1/2at^2, one of the kinematic equations.

I also got 24 meters.
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Re: Hovercraft B/C

Postby Froggie » November 8th, 2017, 5:01 am

Make sure to use the “HIDE” thing next time. ;)
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Re: Hovercraft B/C

Postby Birdmusic » November 9th, 2017, 8:43 pm

heyimben wrote:24m? I literally just found a basic formula for displacement...


Correct! Since you answered first, you get to post the next question!
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2018 Results
R/S
Crime Busters: 1/2
Disease Detectives: 1/5
Hovercraft (both builds failed, rip): 11/9

2019 Results
GGSO/R/S
Disease Detectives: 6/3/10
Circuit lab:11/4/-
Sounds:3/2/1
Protein: -/-/6

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Re: Hovercraft B/C

Postby heyimben » November 10th, 2017, 12:36 pm

Birdmusic wrote:
heyimben wrote:24m? I literally just found a basic formula for displacement...


Correct! Since you answered first, you get to post the next question!

If a person were walking at the average speed of 2.5 mph from District 12 to District 13 and it takes a week to get there, how fast is the hovercraft traveling from District 12 to District 13, which makes it there in 45 min?

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Re: Hovercraft B/C

Postby WhatScience? » November 10th, 2017, 12:47 pm

heyimben wrote:
Birdmusic wrote:
heyimben wrote:24m? I literally just found a basic formula for displacement...


Correct! Since you answered first, you get to post the next question!

If a person were walking at the average speed of 2.5 mph from District 12 to District 13 and it takes a week to get there, how fast is the hovercraft traveling from District 12 to District 13, which makes it there in 45 min?


heyimben....how much of that time is gone to sleep?
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Re: Hovercraft B/C

Postby Justin72835 » November 12th, 2017, 3:29 pm

heyimben wrote:
Birdmusic wrote:
heyimben wrote:24m? I literally just found a basic formula for displacement...


Correct! Since you answered first, you get to post the next question!

If a person were walking at the average speed of 2.5 mph from District 12 to District 13 and it takes a week to get there, how fast is the hovercraft traveling from District 12 to District 13, which makes it there in 45 min?


You can assume that the person is walking nonstop (without taking rest) because he uses the term "average speed".

One week is equivalent to 168 hours. If the person walks at a speed of 2.5 mph for 168 hours, then the distance between the two points is 420 miles. To find the average speed of the hovercraft, use the formula distance/time, or in this case 420 miles/0.75 hours = 560 mph.

Next question: You have a solid cube of mass 'm' which is attached to a nearby wall using a massless, ideal spring of constant 'k'. If you launch an arrow of velocity 'v' and mass 'M' directly at the solid cube, what is the maximum compression of the spring if:

1) the arrow sticks into the solid after hitting it?

2) the arrow bounces off of the cube perfectly elastically?
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Re: Hovercraft B/C

Postby SciolyMaster » December 17th, 2017, 12:53 pm

heyimben wrote:24m? I literally just found a basic formula for displacement...

That is incorrect. 8 m/s^2 is what the train accelerates TO from the initial velocity of 4 m/s^2, not the acceleration itself, which would be (8-4)/2 = 2. Therefore, d = vi*t + 0.5*a*t^2 = 4*2 + 0.5*2*2^2 = 12, so the correct answer is 12 meters.
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Re: Hovercraft B/C

Postby UTF-8 U+6211 U+662F » January 1st, 2018, 1:50 pm

Justin72835 wrote:Next question: You have a solid cube of mass 'm' which is attached to a nearby wall using a massless, ideal spring of constant 'k'. If you launch an arrow of velocity 'v' and mass 'M' directly at the solid cube, what is the maximum compression of the spring if:

1) the arrow sticks into the solid after hitting it?

2) the arrow bounces off of the cube perfectly elastically?

Answer?
1) kMv/(M+m) 2) 2kMv/(M+m)

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Re: Hovercraft B/C

Postby Justin72835 » January 1st, 2018, 3:39 pm

UTF-8 U+6211 U+662F wrote:
Justin72835 wrote:Next question: You have a solid cube of mass 'm' which is attached to a nearby wall using a massless, ideal spring of constant 'k'. If you launch an arrow of velocity 'v' and mass 'M' directly at the solid cube, what is the maximum compression of the spring if:

1) the arrow sticks into the solid after hitting it?

2) the arrow bounces off of the cube perfectly elastically?

Answer?
1) kMv/(M+m) 2) 2kMv/(M+m)

You're definitely on the right track (I see that you already found the final velocities of the objects after the collision). Using the final velocity, you can find the kinetic energy of the object after the collision and set it equal to 1/2kx^2 and solve for x. This works because maximum compression occurs when the objects kinetic energy has been converted completely into potential energy.

For those who want it


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Re: Hovercraft B/C

Postby Justin72835 » January 10th, 2018, 5:56 pm

Since nobody is going:

You have a very long ramp with an inclination of 35 degrees. You give a hollow sphere a translational velocity of 23 m/s toward the base of the ramp. The sphere has a mass of 4 kg and a radius of 0.25 m.

a) What is the hollow sphere's total kinetic energy before rolling up the ramp?

b) To what height above the ground will the sphere roll up the ramp before rolling back down? Assume that it rolls without slipping.
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Re: Hovercraft B/C

Postby Riptide » January 10th, 2018, 6:59 pm

Justin72835 wrote:Since nobody is going:

You have a very long ramp with an inclination of 35 degrees. You give a hollow sphere a translational velocity of 23 m/s toward the base of the ramp. The sphere has a mass of 4 kg and a radius of 0.25 m.

a) What is the hollow sphere's total kinetic energy before rolling up the ramp?

b) To what height above the ground will the sphere roll up the ramp before rolling back down? Assume that it rolls without slipping.


Answer
There are 2 different kinetic energies to consider when solving for the total kinetic energy: Translational KE and Rotational KE.

To find the initial total KE, simply add the 2 energies.


The Moment of Inertia ( ) for a hollow sphere is equal to

Given a mass of 4 kg and a radius of 0.25 m, = 0.167

Now to solve for :



Plugging back into the equation results in:


The answer to part A is 1763 J.

The law of conversation of energy states that the total energy of a system remains constant. There are 3 different energies being used in this scenario: Translational Kinetic Energy, Rotational Kinetic Energy, and Gravitational Potential Energy. This gives us the equation:



The initial potential energy is 0, and the final kinetic energies are also 0. This simplifies the equation down to:



From the previous part, we already have the total initial KE. Substituting into our equation gives us:



With basic algebra, we find that the answer to part B is 44.98 m.
Last edited by Riptide on January 10th, 2018, 7:18 pm, edited 1 time in total.
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Re: Hovercraft B/C

Postby Justin72835 » January 10th, 2018, 7:03 pm

Riptide wrote:
Justin72835 wrote:Since nobody is going:

You have a very long ramp with an inclination of 35 degrees. You give a hollow sphere a translational velocity of 23 m/s toward the base of the ramp. The sphere has a mass of 4 kg and a radius of 0.25 m.

a) What is the hollow sphere's total kinetic energy before rolling up the ramp?

b) To what height above the ground will the sphere roll up the ramp before rolling back down? Assume that it rolls without slipping.


Answer
There are 2 different kinetic energies to consider when solving for the total kinetic energy: Translational KE and Rotational KE.

To find the initial total KE, simply add the 2 energies.


The Moment of Inertia ( ) for a hollow sphere is equal to

Given a mass of 4 kg and a radius of 0.25 m, = 0.167

Now to solve for :



Plugging back into the equation results in:


The answer to part A is 1763 J.

The law of conversation of energy states that the total energy of a system remains constant. There are 3 different energies being used in this scenario: Translational Kinetic Energy, Rotational Kinetic Energy, and Gravitational Potential Energy. This gives us the equation:



The initial potential energy is 0, and the final kinetic energies are also 0. This simplifies the equation down to:



From the previous part, we already have the total initial KE. Substituting into our equation gives us:



With basic algebra, we find that the answer to part B is 44.98 m.

Hey, great job! Now its your turn.
Last edited by Justin72835 on January 10th, 2018, 7:37 pm, edited 1 time in total.
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But in ourselves, that we are underlings."


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Re: Hovercraft B/C

Postby Justin72835 » January 10th, 2018, 7:33 pm

Riptide wrote:
Justin72835 wrote:Since nobody is going:

You have a very long ramp with an inclination of 35 degrees. You give a hollow sphere a translational velocity of 23 m/s toward the base of the ramp. The sphere has a mass of 4 kg and a radius of 0.25 m.

a) What is the hollow sphere's total kinetic energy before rolling up the ramp?

b) To what height above the ground will the sphere roll up the ramp before rolling back down? Assume that it rolls without slipping.


Answer
There are 2 different kinetic energies to consider when solving for the total kinetic energy: Translational KE and Rotational KE.

To find the initial total KE, simply add the 2 energies.


The Moment of Inertia ( ) for a hollow sphere is equal to

Given a mass of 4 kg and a radius of 0.25 m, = 0.167

Now to solve for :



Plugging back into the equation results in:


The answer to part A is 1763 J.

The law of conversation of energy states that the total energy of a system remains constant. There are 3 different energies being used in this scenario: Translational Kinetic Energy, Rotational Kinetic Energy, and Gravitational Potential Energy. This gives us the equation:



The initial potential energy is 0, and the final kinetic energies are also 0. This simplifies the equation down to:



From the previous part, we already have the total initial KE. Substituting into our equation gives us:



With basic algebra, we find that the answer to part B is 44.98 m.

Also, I want to add a little bit to your explanation.

Cool shortcut?
The cool thing about these types of problems is that you can always cancel stuff out and/or combine like terms.



Looking at the second term:



...where k is the coefficient in front of the respective moment of inertia equation.

This gives:



From this simple expression, you can do a multitude of different operations, including the problem from part B. :D
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."


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