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### Re: Hovercraft B/C

Posted: March 30th, 2018, 3:14 pm
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?
32?

### Re: Hovercraft B/C

Posted: March 30th, 2018, 3:19 pm
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

### Re: Hovercraft B/C

Posted: March 30th, 2018, 5:38 pm
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!

### Re: Hovercraft B/C

Posted: March 31st, 2018, 6:16 pm
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.

### Re: Hovercraft B/C

Posted: April 1st, 2018, 10:44 am
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.
First, find the individual components of the 2D velocity vector:

$v_x = vcos\theta$
$v_y = vsin\theta$

The amount of time the ball will be in the air can be found using

$\Delta y = v_{y}t + \frac{at^2}{2}$

Since $\Delta y = 0$ and $a = -9.8$:

$0 = t(vsin\theta - 4.9t)$

$t = 0, \frac{vsin\theta}{4.9}$

To then get the total distance traveled, we simply use $v_{x}t = \Delta x$.

$vcos\theta*\frac{vsin\theta}{4.9} = \Delta x$

$\frac{v^2sin(2\theta)}{9.8} = \Delta x$ is the final equation.

### Re: Hovercraft B/C

Posted: April 1st, 2018, 11:18 am
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.
First, find the individual components of the 2D velocity vector:

$v_x = vcos\theta$
$v_y = vsin\theta$

The amount of time the ball will be in the air can be found using

$\Delta y = v_{y}t + \frac{at^2}{2}$

Since $\Delta y = 0$ and $a = -9.8$:

$0 = t(vsin\theta - 4.9t)$

$t = 0, \frac{vsin\theta}{4.9}$

To then get the total distance traveled, we simply use $v_{x}t = \Delta x$.

$vcos\theta*\frac{vsin\theta}{4.9} = \Delta x$

$\frac{v^2sin(2\theta)}{9.8} = \Delta x$ is the final equation.

### Re: Hovercraft B/C

Posted: April 2nd, 2018, 8:24 am
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?

### Re: Hovercraft B/C

Posted: April 2nd, 2018, 6:30 pm
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 $17.3 \frac{m}{s} \textrm{ south}$

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

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

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

### Re: Hovercraft B/C

Posted: April 2nd, 2018, 8:31 pm
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 $17.3 \frac{m}{s} \textrm{ south}$

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

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

$19 \frac{m}{s} \textrm{ at } 1.14 \textrm{ south of west}$
That’s correct. You turn!

### Re: Hovercraft B/C

Posted: April 9th, 2018, 5:20 pm
1. A uniform disk is rolled down a hill of height 11m. What is the speed of the disk when it reaches the bottom of the hill?
2. As soon as the disk reaches the bottom of the hill, it hits a horizontal surface of frictional coefficient 0.13. What is the acceleration of the disk?
3. A ball of mass 2kg is attached to a string of length 0.75m and rotated vertically with an angular velocity of 7rad/s. What is the ratio of the string tension at the top of the loop to the string tension at the bottom?
Ignore significant figures.