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.