Cannon

A cannon works very similar to how a gun works. A charge is loaded into the cannon (such as gunpowder) and then the cannonball is loaded in on top of the charge. Wadding is placed into the top of the cannon along with the fuse. The fuse is lit which sets the wadding on fire which in turn ignites the charge. The gases from the charge will then quickly expand causing the cannon ball to fly from the end of the cannon. There are concerns about cannon. First, what's the effect of the barrel of cannons? The barrel is the place where the ignition of gun powders took place. The cannon ball which put on top of the charge receive the energy exerted by the chemical reaction of the charge. A barrel will determines whether the effect of the ignition is being sufficiently used. Normally, a longer barrel can increase the initial speed of cannonballs because it lets the cannonball receives the energy exerted by the chemical reaction longer. The second concern is the chamber pressure. Chamber pressure is the pressure exerted within the chamber of a cannon or other firearms when a cartridge is fired in it. Similar to the barrel, the higher the chamber pressure, the higher the initial speed of the cannonball being fired. However, how to increase the chamber pressure or create a cannon with a high chamber pressure? The most simple way to increase the chamber pressure within a already made cannon can be simply adding charges (i.e. gun powder)

After knowing how to increase the pressure in a cannon, how to increase the range of a cannonball that can fly after being fired? There are two factors determine that. One is explained above, by its initial velocity. The second factor is the mass and shape of the cannonball. When the mass decreases, the cannonball can be easier pushed by the force to a longer distance. When the surface area of the cannonball that directly receives the exerted force by the chemical reaction increases, the cannonball can receive more energy. This directly relates to the distance it can fly.

Equilibrium, Incline, Pullies and Trains

Equilibrium

Assumptions: 1. set positive axises

             2. no acceleration in both x and y direction becuase of the

                object is not moving

             3. no friction because friction can result in inequality.

                                

a = 0

                                     F = ma

Fx = ma                                                    Fy = ma              T1x - T2x = 0                                             -Fg + T1y + T2y = 0

T1x = T2x                                                  T1y + T2y = Fg

                                                           T1sinA + T2sinB = mg

                                                        (T1sinA + T2sinB) / g = m

Inclines - 1

Static - object have not started moving. No acceleration at all.

Assumptions:

1. fs = MFn

2. no acceleration. a = 0

3. set positive axises along the decline
4. no air resistance

No movement, acceleration is zero

Ms = ?

                               F = ma                      
Fy = ma                                              Fx = ma
Fy = 0                                               Fgx - f = 0
Fn - Fgy = 0                                         Fgsinθ- MFn = 0
Fn = Fgy                                             Fgsinθ = MFgy
Fn = Fgcosθ                                         Fgsinθ = MFgcosθ

                                                     Fgsinθ/Fgcosθ = M

                                                     tanθ = M 

Inclines - 2

Kinetic -  object is moving on x-axis thus there is acceleration.

Assumptions:

1. fk = MkFn

2. ax does not equal to zero, ay equal to zero. No movement on y-axis
3.  set positive axises along the direction of movement.

4.  no air resistance

a = ?
                               F = ma
Fy = ma                                               Fx = ma
Fy = 0                                                Fyx - fk = ma
Fn - Fgy = 0                                          Fyx - MkFn = ma
Fn = Fgy                                              Fgsinθ - Mkmgcosθ = ma
Fn = Fgcosθ                                          Fgsinθ - Mkmgcosθ/m = a
Fn = mgcosθ                                                                    

Pulley

1. two free body diagram with different positive axises.

2. Tension 1 = Tension 2

3. no horizontal force in pulley questions, however, if one mass is on a 

   horizontal surface

Assumptions:

1. no friction on rope

2. no air resistance

3. two free body diagrams

4. positive axises set along the movement

5. T1 = T2

6. acceleration is the same for either mass.

mass 1

                               F = ma

Fy = ma                                                   Fx = ma

Fg - T = ma                                             Fx = 0

m1g - T = m1a

T = m1g - m1a (1)

mass 2

                               F = ma

Fy = m2a                                                  Fx = ma

T - m2g = m2a                                         Fx = 0

T = m2g + m2a (2)

Find a

put (1) and (2) together

m1g - m1a = m2g +m2a

m1g-m2g = m1a + m2a

collect like terms:

m1g-m2g = a(m1+m2)

m1g - m2g / m1+m2 = a

Find T

use either equation (1) or (2)

(1): T = m1g - m1a

TRAINS

1. Three free body diagrams, each for different cart.

2.  no vertical movement, thus Fn = Fg

3.  only the leading cart that has the applied force. other carts' 

     force that cause movement is the tension force.

Assumptions:

1.  3 free body diagramss for T1 and T2

2.  acceleration is the same

3.  no air resistance
4.  set posivetion axises along the movement.

                               

mass 1

                                                      F = ma

Fy = m1a                                                                               Fx= m1a
Fn - m1g = 0                                                                         Fa - T1 - f = m1a
Fn = m1g                                                                               Fa - T1 - Mm1g = m1a

                                                                                                Fa - Mm1g - m1a = T1

mass 2

                                                     F = ma

Fy = ma                                                                                      Fx = ma

Fn = Fg                                                                                      T1 - f - T2 = m2a

Fn = m2g                                                                                    Fa - Mm1g - m1a - f -T2 = m2a

mass 3

                                                      F = ma

Fy = m3a                                                                                     Fx = m3a

Fy = 0                                                                                          T2 - f = m3a

Fn - m3g = 0                                                                               T2 = m3a + f

Fn = m3g                                                                                     T2 = m3a + Mm3g

 

Solving Projectile Motion Problems

There are four types of projectile motion as we learned in the physics class. They are, for example, the projectile motion one, object falls from top to the horizontal. In this diagram, the object is moving horizontally with a down force that pulls the object down to the ground. This force is know as gravity, which is 9.8m/s square.

Projectile Motion 1

Solving this type of question, usually the x value (distance travelled) is given, as well as the vx value (horizontal velocity). The common questions are: asking for time travelling the x distance; knowing the time, asking for height (dy).




Example 1:  A plane flying at the speed of 100 m/s parallel to the ground drops an object from a height of 2 km. Find (i) the time of flight (ii) velocity of the object at the time it strikes the ground and (iii) the horizontal distance traveled by the object.



This is a typical question for projectile motion 1. The commonly used equations are at = v2 - v1, and d = v1t + 1/2gt square.

i) for finding the time, we substitute the values given and used equation
d = v1t + 1/2gt square.
Since d = 2000, g (gravity, constant value one Earth) = 9.81, and v1 (start with 0 metre per second) = 0m/s

The new equation will be: 2000 = 0(t) + 1/2(9.81)(t) square; following the equation solving process, we will get a final answer of 20.20 second.
ii)We can find the velocity at the time of strike with ground by calculating component velocities at that instant in the  perpendicular directions and finding the resultant (composite) velocity as : vf = vy square + vx square. The answer will be 223.60 m/s
iii)Horizontal distance traveled by the object: time x horizontal velocity
20.20 x 100 = 2000 m

Projectile Motion 2

All questions about projectile motion 2 are given either the initial velocity and angle or vertical velocity and horizontal velocity. If initial velocity and angle are given, using cos θ multiplying initial velocity to find horizontal velocity, and using sin θ multiplying initial velocity to find vertical velocity.

Three common questions for this type of projectile motion are: asking for range, asking for height, and asking for travelling time.

Example 2: A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 30.0 degrees from the horizontal with a velocity of 40.0 m/s. How far will it travel in the air?

Using the values given, we know the initial velocity is 40.0 m/s and the angle of the motion is 30.0 degrees.  We then use cosθ to get horizontal velocity (vx), which is 36.64 m/s, and the vertical velocity (vy1), which is 20 m/s.

Since we all know distance equals time multiplying speed, and horizontal velocity is a constant value. We need to calculate the time first in order to get the distance travelled. There is a simply equation used to find time. t = 2sinθ/g
or following a more complicated method which is to calculate the time used to reach the vertical velocity from 0 m/s. In this case, I chose to use the second method. The vertical velocity is 20 m/s, and we divide it by 9.81, which is the gravity, we get 2.04 second for the object to reach the highest point and the vertical velocity. Use the time we got multiplying by 2, we will get the total time for travelling the entire distance, which is 4.08 seconds in this case. 4.08 times 34.64 (horizontal distance), we get the range, which is 141 m.

Physics behind Roller Coaster

Roller Coaster

Probably everyone have heard roller coaster and wanted to enjoy a ride. As the train cruising down from a steep hill, did anyone think about the physics behind it? What made the train crusing at a extremely high speed without a motor? What made the train loops 3 or 4 times and not falling off at the middle? They are all about physics and especially the kinetic energy and potential energy. A roller coaster is a very simple machine. The train is first carried up to the top of a lift hill and is from then powered by gravity until it reaches the end of a ride. In a roller coaster, there are two types of energy that decides the success and they are: potential and kinetic evergy. Kinetic energy is the energy of motion. When a object is moving, it has a kinetic energy on it, and when it moves faster, coordinately, a higher kinetic energy. Potential energy is harder to explain but could be simply thought as stored energy. For example, when the train is slowly moving up to the top of a hill, it gains potential energy. This energy is not used until the train start cruising down the hill. When there is more potential energy stored, the faster the train can move. And, when the train is cruising down the hill,the potential energy stored is converted to kinetic energy . The further it cruises down, the more potential energy that gets converted to kinetic energy. In other words, the train picks up speed as it falls. There is a short flash could simulate the increase and decrease of energy in a short and simple roller coaster. Flash to Show Potential and Kinetic Energy

How to add vectors

Right Triangle Figure 1

How to add vectors? We know that vectors are physical quantities that consist of a magnitude as well as a direction, for example velocity, acceleration, and displacement, as opposed to scalars, which consist of magnitude only, for example speed, distance, or energy. While scalars can be added by adding their magnitudes , vectors are  more complicated to add. When only two vectors are given, first we have to connect those vectors, for example, 4 metre west and 3 metre north. We can choose to draw them on a piece of paper and connect them together to form a right triangle when a hypothetical hypotenuse is added. The second step is to use the Pythagorean Theorem to calculate the length of the hypotenuse.

Pythagorean Theorem Figure 2

And as we calculated, the length is 5 metre. After this step, the final step is to determine the direction and angle to the y-axis. For example, in figure 1, we draw a cross at the starting point, and calculate the angle between the blue line and the y-axis. Figure 3 shows an example. the angle should be calculated is the angle between the y-axis and line A.

Use the equation in Figure 5 and use Figure 4 to determine a right triangle.

Figure 5

Figure 3

Figure 4

However, the above information are just the simplest part of adding vectors. When more and more vectors are given, how to add them all together and find the hypotenuse and angle? In such question, when 3 vectors are given and they are 5 metre 30 degree west north, 3 metre 60 degree east north and 10 metre 20 degree west south, how to solve it? This time a more complicated method would be introduced. We will add them all together, but before that we have to classify the y and x for those hypotenuses given. In figure 3, the vertical dotted line is call the y of a hypotenuse, and the horizontal dotted line is called the x of a hypotenuse. When they are all together, a right triangle could be formed. When the length of a hypotenuse and the angle are given, we use the equations introduced in the below diagram to calculate the adjacent and opposite lines.

Figure 6

When we have found all of the x and y, we will add then together and get a final x and y. We then plot them on a piece of graph paper. The shape formed by plotting the x and y can simply be the one showed in Figure 1. Since x and y are the opposite and adjacent line of the triangle, we can easily find the hypotenuse by using figure 5's equation. Lastly, we repeat the same step for finding the angle and write the answer like 20m (N 30 degree S)

How to derive equation 3 and 4 by using a graph? It is simple! imagine a trapezoid on a graph. Since the trapezoid is composed of a right triangle and a rectangle shaped figure, we can actually simply add the area of the right triangle and the rectangle for equation three:

Area for rectangle:  (v1)(t)

Area for right triangle:   (v2-v1)(t)/2

So, after combining these two results, the equation would be: 

d = v1t + ½(v2-v1)t

Because at = v2-v1 the next step would be substituting this equation into the one above. So it becomes:  d = v1t + ½at(t). As we continue, the final equation which is the third equation would be d = v1t + ½at²


Equation Four


This time, we change our steps for finding the area of the trapezoid. We previously used addition to find the area, therefore this time we will be using subtraction. We first calculate the total area of a rectangle which is the trapezoid, but we see it as a rectangle. The equation will be: (v2)(t). In order for us to get the correct area of the trapezoid, we have to subtract the excessive area we included. So the excessive area would be: (v2-v1)(t)/2

Again, since at = v2-v1, we substitute this into the result we got from above and the equation would be like: d = v2t-½(v2-v1)t, then, 

d = v2t-½at(t), continue to d = v2t-½at², and we are done!

Motion

• This is a d->t (distance and time) graph.
• When line is horizontal, it represents immobility, thus stay at one spot.
• When the line has an inclining slope, it means moving backward from the senor in this case.
• When the line has a declining slope, it means moving oppositely of backward, so in this case, move forward towards the sensor.

1. Stand 1 meter away from the origin, and stay for 1 second.

2.Walk 1.5 meter away from the origin in 2 seconds with constant speed.

3. Stand 2.5 meter away from the origin, and stay for 3 seconds.
4. Walk 0.75 meter toward the origin in 1.5 seconds with constant speed.
5. Stand 1.75 meter away from the origin, and stay for 2.5 seconds.

This is a d->t graph

         1. Stand 3 meter away from the origin, and walk 1.5 meter toward the origin in 3 seconds, with constant speed.

           2. Stand 1.5 meter away from the origin, and stay for 1 second.

           3. Walk 1 meter toward the origin in 1 second with constant speed.

           4. Stand 0.5 meter away from the origin, and stay for 2 seconds.

           5. Walk 2 meter away from the origin in 3 seconds.

1. Stand 0.8 meter away from the origin, and walk 1 meter away from the origin in 3.5 seconds with constant speed.
2. Stand 1.8 meter away from the origin, and stay for 3 seconds.

3. Walk 1.3 meter away from the origin in 3 seconds.

This is a v->t (velocity and time) graph

• Postive and constant line represents a constant velocity moving away from the sensor
• Negative and constant line represents a constant velocity moving West (toward the sensor)
• Sudden increase or decrease in the value of the line means a dramatic acceleration or deceleration.

1. Speed up for 4 seconds.

 2. Walk at a speed of 0.5m/s and move away from the origin for 2 seconds.

3. Walk at a speed of 0.4m/s and move toward the origin for 3 seconds.

4. Stay for 1 second.

Translation from Velocity to Acceleration
1. The acceleration is 0.1m/s^2, and the line is on the positive side.
2. There is no acceleration.
3. There is no acceleration.
4. There is no acceleration.

This is a velocity-time graph

1. Stay for 2 seconds.

2. Walk at a velocity of 0.5m/s away from the origin for 3 seconds.

3. Stay for 2 seconds.

4. Walk at a velocity of 0.5m/s toward the origin for 3 seconds.

Translation from Velocity to Acceleration
1. There is no acceleration.
2. There is no acceleration.
3. There is no acceleration.
4. There is no acceleration.

This is an acceleration graph

Translation from Acceleration to Velocity

1. Walk at a velocity of 0.35m/s away from the origin for 3 seconds.

2. Speed up, still away from the origin, for 0.25 second.
3. Slow down, now toward the origin, for 0.25 second.

4. Walk at a velocity of 0.35m/s toward the origin for 3 seconds.

5. Slow down, toward the origin, for 0.25 second.
6. Stay for 3 seconds.

1. There is no acceleration.
2. The line is on the negative side.
3. The line is on the negative side.
4. There is no acceleration.
5. The line is on the positive side.
6. There is no acceleration.