## Hovercraft B/C

Justin72835
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### Re: Hovercraft B/C

Ohh, I see. $\frac12(\frac13m(0.4 m)^2)(3.5 \frac{m}{s} * \frac{1 rad}{0.4 m})^2 = mg(\frac12 * 0.4 m * cos \theta + \frac12 * 0.4 m)$
$\theta = 1.53 \textrm{radians}$ Not really sure since I still haven't quite wrapped my head around rotational stuff.

EDIT: Over-complicated things the first attempt
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### Re: Hovercraft B/C

Cool (finally!)

A 1 lb point mass is held up by two ideal ropes. One rope is vertical and 55 cm long. The other is at an angle from the vertical rope. It is attached to the same "ceiling" as the other rope, so it looks something like this:

Code: Select all

--------------------- | / | / | / |/ O 
Imagine you suddenly decrease the length of the rope to the right by half of the length of the rope to the left. Find the tension in each rope (magnitude and direction) after the system reaches static equilibrium if the angle between them was originally
a) as close to 0 degrees as possible
b) 30 degrees
c) 45 degrees
d) 60 degrees
e) as close to 90 degrees as possible

EDIT: Unfortunately, I made an error when I first did this problem, and the algebra is more complicated than I thought (meaning very time-consuming). Thus, I will ask another problem :/

Three masses are held up to a ceiling by three ropes, like so:

Code: Select all

___ | O | O | O 
The ball on top has a mass of 5 kg, the ball in the middle 3 kg, the ball on the bottom 1 kg. Assuming the ropes are ideal, find the tension in each rope.

Justin72835
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### Re: Hovercraft B/C

Cool (finally!)

A 1 lb point mass is held up by two ideal ropes. One rope is vertical and 55 cm long. The other is at an angle from the vertical rope. It is attached to the same "ceiling" as the other rope, so it looks something like this:

Code: Select all

--------------------- | / | / | / |/ O 
Imagine you suddenly decrease the length of the rope to the right by half of the length of the rope to the left. Find the tension in each rope (magnitude and direction) after the system reaches static equilibrium if the angle between them was originally
a) as close to 0 degrees as possible
b) 30 degrees
c) 45 degrees
d) 60 degrees
e) as close to 90 degrees as possible

EDIT: Unfortunately, I made an error when I first did this problem, and the algebra is more complicated than I thought (meaning very time-consuming). Thus, I will ask another problem :/

Three masses are held up to a ceiling by three ropes, like so:

Code: Select all

___ | O | O | O 
The ball on top has a mass of 5 kg, the ball in the middle 3 kg, the ball on the bottom 1 kg. Assuming the ropes are ideal, find the tension in each rope.
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

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### Re: Hovercraft B/C

Justin72835
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### Re: Hovercraft B/C

Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.

How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?
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MattChina
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### Re: Hovercraft B/C

Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.

How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?
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### Re: Hovercraft B/C

Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.

How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?
Assuming there is no change in lift (which affects friction) and the hovercraft gets to max speed instantly:
One penny roll is around the mass of 50 (new) pennies, 50 * 3.11 g or 0.156 kg.
Required speed is 1.8 m / 15 s = 0.12 m/s.
(2.25 kg)(0.2 m/s) = (2.25 kg + n * 0.156 kg)(0.12 m/s)
n = 9.6

10

Justin72835
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### Re: Hovercraft B/C

Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.

How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?
Assuming there is no change in lift (which affects friction) and the hovercraft gets to max speed instantly:
One penny roll is around the mass of 50 (new) pennies, 50 * 3.11 g or 0.156 kg.
Required speed is 1.8 m / 15 s = 0.12 m/s.
(2.25 kg)(0.2 m/s) = (2.25 kg + n * 0.156 kg)(0.12 m/s)
n = 9.6

10
Oh sorry, I forgot to mention that you should assume that each roll of pennies has a mass of 125 grams (it's mentioned in the rules manual). However, your answer is technically correct, so you can go ahead with the next one!
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### Re: Hovercraft B/C

Derive the formula for the distance a projectile travels given the angle of elevation of its launch, theta, and the initial speed, v. Assume the projectile encounters no air resistance and starts at ground level.

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### Re: Hovercraft B/C

Derive the formula for the distance a projectile travels given the angle of elevation of its launch, theta, and the initial speed, v. Assume the projectile encounters no air resistance and starts at ground level.
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### Re: Hovercraft B/C

Derive the formula for the distance a projectile travels given the angle of elevation of its launch, theta, and the initial speed, v. Assume the projectile encounters no air resistance and starts at ground level.

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### Re: Hovercraft B/C

There is a 20kg ball heading east at 26m/s. A 36kg ball is traveling southwest at 38m/s. If both balls undergo an inelastic collision, what is the speed and the direction of the balls after the collision?
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### Re: Hovercraft B/C

There is a 20kg ball heading east at 26m/s. A 36kg ball is traveling southwest at 38m/s. If both balls undergo an inelastic collision, what is the speed and the direction of the balls after the collision?
It doesn't seem like there's enough information given to solve the problem, but given that the balls "stick together", then
$(20 kg)(26 \frac{m}{s}) - (36 kg)(38 \frac{m}{s} * \frac{\sqrt{2}}{2})) = (56 kg)(\vec{v}_x)$ where east is positive.

$\vec{v}_x = -7.99 \frac{m}{s}$ or $7.99 \frac{m}{s} \textrm{ west}$

$(36 kg)(38 \frac{m}{s} * \frac{\sqrt{2}}{2} = (56 kg)(\vec{v}_y)$ where south is positive.

$\vec{v}_y = 17.3 \frac{m}{s}$ or $17.3 \frac{m}{s} \textrm{ south}$

$||\vec{v}|| = \sqrt{\vec{v}^{\,2}_x + \vec{v}^{\,2}_y} = 19 \frac{m}{s}$

$\arctan{\frac{17.3}{7.99}} = 1.14$

$19 \frac{m}{s} \textrm{ at } 1.14 \textrm{ south of west}$

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### Re: Hovercraft B/C

There is a 20kg ball heading east at 26m/s. A 36kg ball is traveling southwest at 38m/s. If both balls undergo an inelastic collision, what is the speed and the direction of the balls after the collision?
It doesn't seem like there's enough information given to solve the problem, but given that the balls "stick together", then
$(20 kg)(26 \frac{m}{s}) - (36 kg)(38 \frac{m}{s} * \frac{\sqrt{2}}{2})) = (56 kg)(\vec{v}_x)$ where east is positive.

$\vec{v}_x = -7.99 \frac{m}{s}$ or $7.99 \frac{m}{s} \textrm{ west}$

$(36 kg)(38 \frac{m}{s} * \frac{\sqrt{2}}{2} = (56 kg)(\vec{v}_y)$ where south is positive.

$\vec{v}_y = 17.3 \frac{m}{s}$ or $17.3 \frac{m}{s} \textrm{ south}$

$||\vec{v}|| = \sqrt{\vec{v}^{\,2}_x + \vec{v}^{\,2}_y} = 19 \frac{m}{s}$

$\arctan{\frac{17.3}{7.99}} = 1.14$

$19 \frac{m}{s} \textrm{ at } 1.14 \textrm{ south of west}$
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### Re: Hovercraft B/C

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