Hovercraft

Hovercraft is an event where teams must design and build a hovercraft. It is a Division B and Division C event for the 2017 season. Teams will build self-propelled air-levitated vehicles, and also take a test on classical mechanics and other similar topics. The trial rules used at the 2016 National Tournament can be found at the bottom of the page in the "Links" section.

Description
Hovercraft is a dual lab, consisting of a test and a build portion. The test portion should take at least 20 minutes and competitors are allowed only 8 minutes to run their device. A 3-ring binder of any size is allowed, provided that all materials are hole punched and secure. Calculators of any type are also allowed and need not be impounded.

Construction
The hovercraft itself can weigh no more than 2000 grams. It is optimal that the hovercraft weigh as close to the maximum as possible while still being able to levitate, as this in turn will maximize score on the build portion. That being said, do not abandon functionality for the sake of attempting a perfect score. Any material is permitted for construction of the vehicle itself, but the vehicle must meet all specifications and cannot modify the track provided by the event supervisor. Brushless motors and integrated circuits are not permitted. For further information, the rules used at the the 2016 national tournament (when Hovercraft was a trial) can be found below.

Written Exam
Part 1 of the event consists of a written test which draws from the AP Physics 1, or algebra-based mechanics, curriculum. This test must contain at least five questions from each of the following topics: Newton's laws of motion, kinematics, kinetic energy, air cushioned vehicles and applications, and fluid mechanics for Div. C only. 20 – 30 minutes is suggested to complete this test. All answers must be provided in metric units along with the appropriate significant figures.

Newton's First Law
An object will remain at rest or in uniform motion unless acted on by an external force.

Newton's first law describes inertia, the tendency of an object to maintain a constant velocity in the absence of an external force.

Newton's Second Law
The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object.

[math]F = ma[/math]

Newton's second law describes the relationship between force, mass, and acceleration as represented above. Common problems for this section will give 2 of the values, and ask for the third. Make sure to practice, practice, practice if you aren't familiar with this - you should be able to instantly know and apply Newton's second law within larger, more difficult problems.

Newton's Third Law
Every action has an equal and opposite reaction

Forces always occur in equal action-reaction pairs, no matter the masses of the objects being acted upon.

For example, if a 5 (metric) ton collides with a 5 gram mosquito, both experience the same force - the mosquito just experiences a greater acceleration due to its lesser mass. Similarly, a rifle recoils, as the force that propels a bullet forward must have a opposite reaction which propels the rifle backwards.

Kinematic Equations
1. [math]v = v(0) + at[/math]

Acceleration times time equals change in velocity, or acceleration equals change in velocity over time - the definition of acceleration. Works well when no distance is given.

2. [math]\Delta x = v(0)t + 0.5at^2[/math]

Though you shouldn't need to know this for an algebra based test, change in distance is equal to the integral of velocity, which is exactly what this equation is - the integral of the above velocity equation. Another way to think of it is the average velocity times time, where final velocity is calculated using the equation 1 and average velocity is found from equation 4.

Works well when no final velocity is given.

3. [math]\Delta x = vt - 0.5at^2[/math]

The exact same thing as the previous equation, but from the opposite end. Instead of using final velocity, initial velocity is found from equation 1 and plugged into equation 4, again to get average velocity. This can be used when no initial velocity is given, but works almost the same way as equation 2.

4. [math]\Delta x = 0.5(v + v(0))t[/math]

Average velocity times time is equal to distance, pretty easy to remember. This one is great when no acceleration is given, but it only works for a constant acceleration.

5. [math]v^2 = v(0)^2 + 2a\Delta x[/math]

You'll see why this one works once you get into energy. Multiply the whole thing by 0.5m and you get this: [math]0.5mv^2 = 0.5mv(0)^2 + ma\Delta x[/math]. The last term can be converted into Fd, or work, while the first two represent final and initial kinetic energies. This is just a change in kinetic energy calculation!

Definitely put these into your notes, but if you practice enough you should end up memorizing all five of these equations. There are 5 variables in kinematics - initial velocity, final velocity, acceleration, time, and distance; you should be able to find any two quantities given the other three.

Projectile Motion
Projectile motion can be treated as 2D kinematics - 1 equation for x, and 1 equation for y. However, it adds another variable - the angle the projectile is launched at.

99% of the time, x can be treated with a simple equation: [math]x = v(0)cos(\theta )t[/math], and y uses one of the kinematics equations.

To run calculations faster or double check, use these three, already derived equations:

[math]t = 2v(0)sin(\theta )/g[/math]

[math]h = v(0)^2sin(\theta )^2/2g[/math] (where h is maximum height)

[math]R = v(0)^2sin(2\theta )/g[/math] (where R is range of the projectile)

When in doubt, use your basic kinematics equations!

Example Problems
Here's some examples of working through kinematics with the five equations.

1. A car moving with constant acceleration covers the distance between two points 50.0 m apart in 5.00 s. Its speed as it passes the second point is 15.0 m/s. What is its speed at the first point.

For this problem we are using equation 4.

[math]50.0 = 0.5(15.0 + v(0))*5.00[/math]

[math]50.0 = 37.5 + 2.50*v(0)[/math]

[math]v(0) = 5.00 m/s[/math]

2. For problem 1, what is the car's acceleration?

With the initial velocity already calculated, this is an easy problem, using equation 1.

[math]15.0 = 5.00 + a*5.00[/math]

[math]a = 2.00 m/s^2[/math]

However, let's say we did not have initial velocity, meaning equation 3 is more suitable.

[math]50.0 = 15.0*5.00 - 0.5*a*5.00^2[/math]

[math]50.0 = 75.0 - 12.5a[/math]

[math]a = 2.00 m/s^2[/math]

As you can see, both equations come out the same! With 3 out of the 5 values, there's always 1 direct kinematic equation to use, and 1 more indirect path using 2 kinematic equations, which helps if you happen to forget one.

3. A airplane pilot wishes to accelerate from rest with a constant acceleration of 4g to reach the speed of sound, 331 m/s. How long would the period of acceleration be?

We can again use equation 1 to quickly solve the problem.

[math]331 = 0 + 4*9.81*t[/math]

[math]t = 8.44 s[/math]

4. How far will the pilot in problem 3 travel during his acceleration?

Using the time above, we can choose to work with equation 2.

[math]\Delta x = 0 + 0.5*4*9.82*8.44^2[/math]

[math]\Delta x = 1398 m[/math]

Of course, there's another equation we could have used if we hadn't already calculated t, so let's practice with equation 5.

[math]331^2 = 0 + 2*4*9.81*\Delta x[/math]

[math]109600 = 78.48\Delta x[/math]

[math]\Delta x = 1396 m[/math]

This time the answer is slightly different due to intermediate rounding, but both answers would be accepted and would probably end up the same with proper sig fig usage.

Now let's work through some 2D kinematics problems, since we've covered all 5 basic kinematics equations.

5. A swimmer dives off a 10.00 m high diving board with a horizontal leap. What must her velocity be just before her leap if she is to clear the edge of the pool, which extends 2.00 m beyond the edge of the diving board?

Since initial velocity is perfectly horizontal, we can use equation 2 to find the time it takes her to fall without worrying about initial y velocity.

[math]10.00 = 0 + 0.5*9.81*t^2[/math]

[math]t = 1.428 s[/math]

Now it's just a basic question - what velocity is required to clear 2.00 m in 1.428 seconds? No kinematics needed.

[math]2.00 = 1.428v[/math]

[math]v = 1.401 m/s[/math]

6. A quarterback throws a football with a initial velocity of 15 m/s and angle of 30 degrees. What is the highest point of the football's trajectory?

Lets start by running a kinematic equation. We'll use equation 5 to get to the highest point, where velocity must be equal to 0 in the y direction.

[math]0 = (15*sin(30))^2 - 2*9.81*h[/math]

[math]0 = 56.25 - 19.62h[/math]

[math]h = 2.867 m[/math]

We could also use one of the projectile motion equations, though.

[math]h = 15^2*sin(30)^2/2*9.81[/math]

[math]h = 2.867 m[/math]

You can see where we got the maximum height equation from earlier.

7. How long will it take the football from problem 6 to reach a wide receiver of the same height?

We can do this equation 2.

[math]\Delta y = 15*sin(30)*t - 0.5*9.81*t^2[/math]

Solve where change in y is equal to 0 - this will yield 2 possible points, one where t is equal to 0 and the t we are looking for.

[math]0 = 7.5t - 4.905t^2[/math]

[math]t = 1.529s[/math]

We can also just use the time equation from the projectile motion equations.

[math]t = 2*15*sin(30)/9.81[/math]

[math]t = 1.529s[/math]

I originally made a dumb mistake when working through this problem, but I caught it using the second method. I cannot stress how important it is to check both ways whenever possible - don't just rely on the projectile motion equations.

8. How far away does the wide receiver need to be to catch the ball?

Thankfully, we already have t, which makes this a bit easier. We can now just plug this to a basic x equation.

[math]\Delta x = 15*cos(30)*1.529[/math]

[math]\Delta x = 19.862 m[/math]

Given how long this calculation has been, it's a good idea to check with the range equation.

[math]R = 15^2*sin(30*2)/9.81[/math]

[math]R = 19.862 m[/math]

As you can see, it's always possible to rely on those 5 kinematic equations, so if you ever get confused, just go back to those.

Calculations
Even though the section says kinetic energy, tests are likely to cover multiple energy calculations.

Kinetic energy: [math]K = 0.5mv^2[/math]

Gravitational potential energy: [math]U(g) = mgh[/math]

Elastic potential energy: [math]U(s) = 0.5kx^2[/math]

Work: [math]W = Fd[/math]

Just like F = ma, these equations should be second nature to you by the time you're done practicing. Momentum problems will be tough and could require instant application of multiple of these equations.

Momentum
Momentum is an important part of all collisions.

Momentum: [math]p = mv[/math]

Impulse: [math]J = \Delta p = F\Delta t[/math]

In any collision, momentum is conserved, called the Law of Conservation of Momentum.

[math]\Sigma p(initial) = \Sigma p(final)[/math]

Inelastic
In a perfectly inelastic collision, the two objects combine to form essentially one mass. Kinetic energy is not conserved in such collisions, and instead can be lost to factors like sound or heat.

[math]m(1)*v(i) = (m(1) + m(2))*v(f)[/math]

Inelastic collision problems are relatively easy as they only require solving 1 problem.

Elastic
In a perfectly inelastic collision, both kinetic energy and momentum are conserved as both objects simply bounce off of each other.

Thus two equations can be used to solve for the final velocities of both objects:

[math]m(1)∗v(1,i) + m(2)*v(2,i) = m(1)*v(1,f) + m(2)*v(2,f)[/math]

[math]m(1)∗v(1,i)^2 + m(2)*v(2,i)^2 = m(1)*v(1,f)^2 + m(2)*v(2,f)^2[/math]

Not that all of the 0.5 terms cancel out, but are present in each of the kinetic energies used in the second equation. Both equations should yield a quadratic equation with two solutions - the initial conditions and the more useful final conditions.

Example Problems
1. A 70 kg astronaut is installing a new sensor onto the outside of a space station. If he were to throw the 10 kg sensor at a speed of 2.8 m/s, how quickly would he begin to move, and in which direction?

We will use conservation of momentum, assuming an initial momentum of 0.

[math]0 = 70*v + 10*2.8[/math]

[math]v = -0.4 m/s[/math]

The negative sign indicates the astronaut is moving in the opposite direction of the sensor.

2. A 1000 kg car moving at 15.0 m/s collides with a stopped 6000 kg truck. What is the final velocity of the two vehicles, assuming they lock together after colliding?

Since this is an inelastic collision, we only use conservation of momentum.

[math]1000*15.0 = (1000+6000)*v[/math]

[math]15000 = 7000v[/math]

[math]v = 2.142 m/s[/math]

Pretty easy, right?

3. A 1.00 kg electric vehicle veers off course and collides with a 0.20 kg can. If the initial velocity of the vehicle is 5.00 m/s and the can is at rest, what are the final velocities of both objects assuming a perfectly elastic collision?

Let's start with conservation of momentum:

[math]1.00*5.00 = 1.00*v(1) + 0.20*v(2)[/math]

[math]v(1) = 5.00 - 0.20v(2)[/math]

And now for conservation of kinetic energy, since this is perfectly elastic:

[math]1.00*5.00^2 = 1.00*v(1)^2 + 0.20*v(2)^2[/math]

[math]25.0 = (5.00 - 0.20v(2))^2 + 0.20*v(2)^2[/math]

[math]25.0 = 25.0 - 2.00v(2) + 0.24v(2)^2[/math]

[math]0.24v(2) = 2.00[/math]

[math]v(2) = 8.33 m/s[/math]

[math]v(1) = 5.00 - 0.20(8.33)[/math]

[math]v(1) = 3.33 m/s[/math]

And we're done! Feel free to plug the values in for both equations to verify your results, that's the best way to check this type of problem.

Links
Trial Events rules