Saturday, January 8, 2011

Constructive vs. Destructive Wave Interference

When two waves moving through a medium meet with each other at a particular point in space at the same time, they interact with each other by combining and forming a net wave. The amplitude of this new net wave is equal to the vector sum of all the wave amplitudes that come together at that specific point. This phenomenon is called interference and it is governed by the principle of superposition (the concept of the sum of all amplitudes).

There are two types of wave interference: constructive and destructive. Constructive wave interference occurs when two waves meet and cause an amplification of the amplitude in the resultant. Destructive interference is the opposite, occurring when two waves meet and cause a decrease in the amplitude in the resultant. (see diagram below)

Constructive vs. destructive wave interference.

Tuesday, December 7, 2010

Different Forms of Energy

Energy is an essential part of our daily lives. Yet, it has puzzled and fascinated scientists for many centuries. Energy is a very abstract concept that is often defined as "the ability a physical system has to produce changes on another physical system". Therefore, energy can take on many different forms. Some of these forms include:

  • Thermal energy
  • Chemical energy
  • Electrical energy
  • Radiant energy (energy of electromagnetic waves)
  • Nuclear energy
  • Magnetic energy
  • Elastic energy
  • Sound energy
  • Mechanical energy

Wednesday, December 1, 2010

How To Maximize a Cannon's Horizontal Range

In order to maximize a cannon's horizontal range, the angle of the barrel must be set to 45 degrees above the horizontal (assuming all other things like the size of the projectile and the power of the propellant are equal, level ground, and no air resistance). This optimal angle can be proven using the formula for the range of a projectile:

R = V²sin(2θ)/g


Using this formula, we can determine the value of θ that will maximize the value of sin(2θ), which will maximize the range of the projectile since v and g do not change. The maximum value of sin(2θ) is 1, which occurs at sin(90). Therefore the value of θ must be 45 degrees above the horizontal. 

Tuesday, November 30, 2010

The Best Cannons Ever

A cannon is a piece of artillery that uses an explosive-based propellant to launch a projectile at the enemy. It is mostly used to destroy the enemy's defensive structures and vehicles. The simplest design of a cannon consists of a barrel with one end sealed shut. A quantity of propellant and a projectile are put into the barrel so that they collect at the shut end of the barrel. The propellant is then ignited and the projectile is propelled out of the barrel. Throughout history there have been many examples of totally gnarly cannons. Here are some examples:

1. The Tsar Cannon
















The aptly named Tsar Cannon is the largest old-fashioned cannon ever made. It was built by the Russians in 1586, but it was never used in combat.

2. The Paris Gun


















The Paris Gun was a long-range siege cannon that the Germans used to bomb Paris during WWI. This gun had such a great range that the French thought they were being bombed by a new type of high-altitude zeppelin rather than a cannon.

3. V-3 Cannon




















The Nazis had planned to use this super-cannon to bomb London from a secret bunker across the English Channel. However, it was damaged beyond repair before it was completed.

4. 2B1 Oka
















This self-propelled cannon made by the Soviets was so powerful that its recoil caused damage to the vehicle itself.

5. Mallet's Mortar
















This cannon was built by the British in 1857, but it was never used in combat. Just look at the size of those projectiles!!

Friday, November 26, 2010

Solving Newton's Second Law Problems

Our physics class has been spending the past few days perfecting the art of solving problems that apply Newton's second law of motion. These problems use the formula: F=ma, or the net force acting on an object is equal to its mass multiplied by its acceleration. This formula can be used to solve problems involving Newton's second law.
Follow these simple steps to solve F=ma problems:
  1. Draw a free body diagram (FBD) with the information given to you in the question. Depending on the type of question, you might have to draw more than one FBD in order to solve the question. For example, pulley and train type questions require multiple FBDs.
  2. Create your list of assumptions. Is there any friction? Is there air resistance? How many FBDs do I use? Is there acceleration? Which way is positive? Are the x and y axes tilted? These are some of the questions that must be answered before solving the problem. Drafting a list of assumptions will answer these questions and allow you to solve the problem without it becoming unnecessarily difficult to solve. The following is a list of assumptions for each of the four types of second law problems:
    • Equilibrium problems:
      • no friction
      • a = 0
    • Incline problems (static):
      • fs = µsFn
      • a = 0
      • x and y axes are tilted to line up with the ramp's surface
      • (+) in direction of a
      • no air resistance
      • µ = tan θ
      • Fn is perpendicular to surface of incline
    • Incline problems (kinetic):
      • fk = µkFn
      • ax ≠ 0, ay = 0
      • (+) in direction of a
      • no air resistance
      • Fn is perpendicular to surface of incline
    • Pulley problems:
      • frictionless pulleys + rope
      • no air resistance
      • multiple FBDs
      • (+) in direction of a
      • T1 = T2
      • a of system is the same
    • Train problems:
      • 1 FBD for a
      • 3 FBDs for T1 and T2
      • ay = 0
      • a is consistent
      • no air resistance
      • weightless cables
      • (+) in direction of a
3. Once you have your list of assumptions, you are ready to solve the problem. Split each of the FBDs into the x and y axis. Then solve for Fx = max and Fy = may for each of the FBDs. You will likely get two separate equations that you can then sub into each other to solve for the missing variable.


Congratulations! You have just solved a problem using Newton's second law of motion.


Some pictures:
An example of a free-body diagram.

Another example of a free-body diagram.

      Tuesday, November 2, 2010

      Projectile motion

      Now that we have learned how about vector components and how to add vectors, the next logical step would be to apply our newly learned knowledge to solve projectile motions.

      We started studying vector components by doing an experiment with a marble and a small ramp. The goal of this experiment was to prove gravity; that is, prove that gravity has an acceleration of 9.8 m/s^2. The experiment involved rolling the marble off the small ramp when it was placed on a table. Then we had to record the time it took for the marble to hit the ground after it left the ramp. With a piece of chart paper and a sheet of carbon paper, we also had to find out the horizontal distance from the table to where the marble landed. We then had to use this information to calculate the acceleration of gravity and the final velocity of the marble.

      So, in order to solve problems involving projectile motion, one first has to separate the x and y components of the motion. The x component represents the horizontal motion of the object. For the problems that we are doing right now, the velocities in the x direction are always constant. Also, the effects of air resistance are ignored. The y component  represents the vertical motion of the object. For the problems that we are solving currently, the vertical motion is almost always downwards. The initial velocity is 0 m/s and the acceleration is 9.8m/s^2 downwards. Like the x component, the effects of air resistance are ignored.

      One uses both of these x and y vector components to calculate the actual motion of object. The components are simply added to each other to find the resultant motion.

      Here is a picture to illustrate the motion of a projectile dropped from an airplane:

      Saturday, October 30, 2010

      My favourite roller coaster

      My favorite roller coaster would have to be Behemoth at Canada's Wonderland. It is the tallest and fastest roller coaster in all of Canada, reaching a maximum height of 70 m and a maximum speed of 124 km/h. These two factor combine to make Behemoth, in my opinion, the coolest roller coaster in all of Canada.

      As soon as you get on the roller coaster, you know you are going to have a blast, because the restraint system only presses down on your lap, leaving the rest of your body free to move around.

      As soon as you start ascend up the lift hill, you are captivated by a magnificent view of the park as well as the iconic CN tower if you glance over to your left. As soon as you reach the apex of the lift hill, you start to plummet down a near-vertical hill, reaching death-defying speeds. You then travel up and down the remaining hills, your body alternating between a feeling of weightlessness and a feeling of being pushed into your seat. This lasts for a total of 3 minutes and 10 seconds, 3 minutes and 10 seconds of pure exhilaration and fun.

      These are the reasons why Behemoth is my favorite roller coaster.

      Here are some pictures of Behemoth: