Let's shoot some balls

Launching Angle


As for the launching angle, an angle of 45° will maximize the horizontal distance that the object will fly. In order to produce a cannon that actually shoots an object that results in a positive distance, the maximum angle at which the object could be shot is 90°. Having this knowledge, we could use the equation d = (v²/g)sin2θ. If we say that 90° is the maximum angle at which the object can be shot, the value of theta must be 45°. (Because 90/2 = 45; just in case you may be wondering why.)


Base


Because we are using only five pop cans, we would not be able to make our base super strong compared to a situation in which we could use more than five pop cans. However, the key point about the base that you should be aware of is that the base must be strong enough to withstand the force that is exerted when the cannon shoots the object. Though the object is very light, inability for the base to react to the force by exerting the same amount of force (Newton's third law right here) may cause decrease in the horizontal distance travelled by the object.

Newton's Laws - Equilibrium, Incline, Pullies, Trains

EQUILIBRIUM

- Always set positive axes so that negative numbers won't appear in the solution.
- There is no acceleration in both x and y because there is no movement in either of the component.
- There must be no friction because friction on the string or whatever is connecting the two masses can result in inequality between two values on the same axis. (The components should be in balance in equilibrium questions.)

Assumptions
- a = 0
- no friction

                                         F = ma

Fₓ = ma                                                         Fy = ma                                

T1ₓ - T2ₓ = 0                                                -Fg + T1y + T2y = 0

T1ₓ = T2ₓ                                                      T1y + T2y = Fg

                                                                      T1sinB + T2sinA = mg

                                                                      (T1sinB + T2sinA) / g = m

INCLINES

Static - the object on the inclined plane is not moving yet, hence no value for acceleration.

Assumptions

- fs = µFn

- a = 0

- +ve axes in the direction of decline
- no air resistance

µ = ?

Fy = ma
Fy = 0
Fn - Fgy = 0
Fn = Fgy
Fn = Fgcosθ

Fₓ = ma
Fₓ = 0
-f + Fgₓ = 0
Fgₓ = f
Fgsinθ = µFn

Fgsinθ = µFn

Fgsinθ = µFgcosθ

sinθ = µcosθ

sinθ / cosθ = µ

tanθ = µ

Kinetic - the object is moving, hence value for acceleration on only the x-axis (not on y-axis because the object is not jumping up and down)

Assumptions

- fk = µkFn

- aₓ ≠ 0, ay = 0
- +ve in the direction of a

- no air resistance

a = ?

Fy = ma
Fy = 0
Fn - Fgy = 0
Fn = Fgy
Fn = Fgcosθ
Fn = mgcosθ

Fₓ = maₓ      

Fgₓ - f = maₓ

Fgsinθ - µFn = maₓ        

mgsinθ - µmgcosθ = maₓ

(mgsinθ - µmgcosθ) / m = aₓ

PULLEYS

- For pulleys, you MUST remember to draw two free-body diagrams that have different positive axes.

- You can just use T for tension because the tensions present are equal.

- In pulleys, there is no force that is working horizontally, thus no Fₓ (except when the problem deals with one mass on a horizontal surface)

Assumptions

- frictionless pulleys + rope

- no air resistance

- multiple FBDs

- +ve in the direction of a

- T1 = T2

- a of the system is the same

m1

F = ma

Fₓ = ma

Fₓ = 0

(cancel out)

Fy = m1ay

m1g - T = m1a

m2

F = ma

Fₓ = ma

Fₓ = 0

(cancel out)

Fy = m2ay

T - m2g = m2ay

Find a

m1g - T = m1a    1

T - m2g = m2a    2

From 1 T = m1g - m1a    3

From 2 T = m2a + m2g   4

Set 3 = 4 m1g - m1a = m2a + m2g

collect the like terms m1g - m2g = m2a + m1a

factor out a m1g - m2g = a (m2 + m1)

(m1g - m2g) / (m2 + m1) = a

Find T

use one of the equation used in the above part

T = m1g - m1a

substitute a with the value that you get

T = (  )N

TRAINS

- Simply put, trains are like pulleys put horizontally.

- The pulling force of the foremost mass is the engine which is represented by Fₐ. The pulling force of the following masses are the according tensions.

- For trains, you MUST remember to draw 3 FBDs.

- The value of the vertical acceleration is zero because the train is not jumping up and down.

Assumptions

- 1 FBD for a

- 3 FBDs for T1 and T2

- ay = 0

- a is consistent

- no air resistance
- weightless cables

- +ve in the direction of a

After having drawn the free-body diagrams for each body of the train, you must combine everything so that the whole train becomes a little dot on your free-body diagram. This diagram you draw will be the FBD for the acceleration.

F = ma

Fₓ = maₓ

Fₐ - f = maₓ 

Fₐ - µmg = maₓ

(Fₐ - µmg) / m = aₓ

Fy = may
Fy = 0
Fn - mg = 0
Fn = mg

m1
F = ma

Fy = m1ay
Fn1 - m1g = 0
Fn1 = m1g

Fₓ = m1aₓ
Fₐ - T1 - f1 = m1aₓ
Fₐ - T1 - µm1g = m1aₓ
Fₐ - µmg - m1aₓ = T1

m3
(instead of using m2 to find T2, which is perfectly fine, it's better to use m3 because using m3 increases the chance of you getting the correct answer due to the fact that there is one less variable)

F = ma

Fy = m3ay

Fy = 0

Fn3 - m3g = 0

Fn3 = m3g

Fₓ = m3aₓ

T2 - f3 = m3aₓ

T2 = m3aₓ + f3

T2 = m3aₓ + µm3g

SUMMARY OF THE ASSUMPTIONS FOR EACH TYPE OF QUESTION

EQUILIBRIUM

- no friction

- a = 0 (for both x and y)

INCLINES (STATIC)

- fs = µsFn

- a = 0 (for both x and y)

- +ve in direction of a

- no air resistance

- µ = tanθ

INCLINES (KINETIC)

- fk = µkFn

- Fn  to surface

- aₓ ≠ 0, ay = 0

- +ve in direction of a

- no air resistance

PULLEY

- frictionless pulleys + rope

- no air resistance

- multiple FBDs

- +ve in direction of a

- T1 = T2; therefore, you can just call them 'T'

- a1 = a2

TRAIN

- 1 FBD for a (the big one)

- 3 FBDs for T1 and T2 (one per mass)

- ay = 0

- a is consistent

- no air resistance

- weightless cables

- +ve in direction of a

Common projectile motion questions (not my work)

[http://zonalandeducation.com/mstm/physics/mechanics/curvedMotion/projectileMotion/generalSolution/generalSolution.html]

                                                                 

                                                                                               ↓

Original, or initial, conditions:

The original conditions are the size of the velocity and the angle above the horizontal with which the projectile is thrown.
General:

Original size of velocity: v_o
Original angle: theta

Example:

v_o = 40.0 m/s
theta = 35 degrees

Components of original velocity:

The usual first step in this investigation is to find the x and y components for the original velocity.
General:

X component of original velocity: v_ox = v_ocos(theta)
Y component of original velocity: v_oy = v_osin(theta)

Example:

In the x direction:

v_ox = v_ocos(theta)
v_ox = (40.0 m/s)(cos(35 degrees))
v_ox = (40.0)(0.8191)
v_ox = 32.76
v_ox = 32.8 m/s

In the y direction:

v_oy = v_osin(theta)
v_oy = (40.0 m/s)(sin(35 degrees))
v_oy = (40.0)(0.5735)
v_oy = 22.94
v_oy = 22.9 m/s

How much time passes until the projectile is at the top of its trajectory?

At the top of the trajectory the y, or upward, velocity of the projectile will be 0.0 m/s. The object is still moving at this moment, but its velocity is purely horizontal. At the top it is not moving up or down, only across.

Notice that the object is still in motion at the top of the trajectory; however, its velocity is completely horizontal. It has stopped going up and is about to begin going down. Therefore, its y velocity is 0.0 m/s.
We need to find out how much time passes from the time of the throw until the time when the y velocity of the projectile becomes 0.0 m/s. This y velocity at the top of the trajectory can be thought of as the final y velocity for the projectile for the portion of its flight that starts at the throw and ends at the top of the trajectory.
We will call this amount of time 'the half time of flight', since the projectile will spend one half of its time of flight rising to the top of its trajectory. It will spend the second half of its time of flight moving downward.
General:

We can use the following kinematics equation:

v_f = v_o + at

Subscript it for y:

v_fy = v_oy + a_yt

Solve it for t:

t = (v_fy - v_oy) / a_y

Plug in 0.0 m/s for v_fy:

t = (0.0 m/s - v_oy) / a_y

If the original y velocity and the y acceleration, i. e., the acceleration due to gravity, are plugged into the above equation, it will solve for the amount of time that passes from the moment of release to the moment when the projectile is at the top of its flight.

Example:

Start with:

t = (v_fy - v_oy) / a_y

Plug in 0.0 m/s for v_fy:

t = (0.0 m/s - v_oy) / a_y

Plug in values for v_oy and a_y:

t = (0.0 m/s - 22.9 m/s) / - 9.8 m/s²
t = -22.9 / -9.8
t = 2.33
t = 2.3 s

In this example 2.3s of time passes while the projectile is rising to the top of the trajectory.

How high does the projectile rise?

Here you need to find the displacement in the y direction at the time when the projectile is at the top of its flight. We have just found the time at which the projectile is at the top of its flight. If we plug this time into a kinematics formula that will return the displacement, then we will know how high above ground the projectile is at when it is at the top of its trajectory.
General:

Here is the displacement formula:
d = v_ot + 0.5at²
We must think of this displacement in the y direction, so we will subscript this formula for y:
d_y = v_oyt + 0.5a_yt²
If now we plug in the half time of flight, which was found above, we will solve for the height of the trajectory, since the projectile is at its maximum height at this time.

Example:

Starting with:
d_y = v_oyt + 0.5a_yt²
Then plugging in known values:
d_y = (22.9 m/s)(2.33 s) + (0.5)(-9.8 m/s²)(2.33 s)²
d_y = 53.35 - 26.60
d_y = 26.75
d_y = 27 m

How much time passes until the projectile strikes the ground?

General:

With no air resistance, the projectile will spend an equal amount of time rising to the top of its projectile as it spends falling from the top to the ground. Since we have already found the half time of flight, we need only to double that value to get the total time of flight.

Example:

t = 2(2.33 s)
t = 4.66
t = 4.7 s
This is the total time of flight.

How far away does the projectile land from its starting point?

The distance from the starting point on the ground to the landing point on the ground is called the range of the trajectory. This range is a displacement in the x direction. It is governed by the x velocity of the projectile. This x velocity does not change during the flight of the projectile. That is, whatever is the value of the x velocity at the start of the trajectory will be the value of the x velocity throughout the flight of the projectile. The x velocity remains constant because there are no accelerations in the x direction. The only acceleration is in the y direction, and this is due to the vertical pull of gravity. Gravity does not pull horizontally. Therefore, the calculation for the range is simplified.
General:

Let us start with the general displacement formula:
d = v_ot + 0.5at²
Since we are working in the x direction, we should subscript this equation for x:
d_x = v_oxt + 0.5a_xt²
Now, since the acceleration in the x direction is 0.0 m/s², the second term in the above equation drops out, and we are left with:
d_x = v_oxt
The velocity in the x direction does not change. The projectile maintains its original x velocity throughout its entire flight. So, the original x velocity is the only x velocity the projectile will have. We could, therefore, think of the last equation as:
d_x = v_xt
If we plug in the original x velocity for v_x and the total time of flight for t, we will solve for the horizontal displacement, or range, of the trajectory. 

Projectile motion/Type 1 and 2

Projectile motion is the motion of an object whose path is affected by the force of gravity. We are all affected by gravity, but it profoundly alters the motion of objects that are thrown or shot upward. The arching of a thrown ball is caused by gravity, as well as its falling motion in general.

When faced with a projectile motion problem, there are things you must do and remember.

Do
1. determine the horizontal and vertical components of the initial velocity.


For an object projected horizontally, the vertical component of initial velocity is zero.
For an object thrown at 20 m/s 53º above the horizontal, the vertical component of the initial velocity is 20sin53º = 16.0 m/s [up], and the horizontal component is 20cos53º = 12.0 m/s.

2. break down the problem into two problems: horizontal and vertical.
3. assign - and + signs appropriately

Remember:
1. acceleration is zero in the horizontal component and g in the vertical.
2. at the top of its rise, the projectile has a vertical velocity = zero.

(For the two types of projectile motion below, I had to defy Mr. Chung's use of certain letters for certain values because the diagrams that I found used different letters.)

TYPE 1

x-components

1. Component of initial velocity along x-axis.
u_x=u

2. Acceleration along x-axis
a_x=0
(Because no force is acting along the horizontal direction)

3. Component of velocity along the x-axis at any instant t.
v_x=u_x + a_xt
=u + 0

» v_x=u
This means that the horizontal component of velocity does not change throughout the projectile motion.
4. The displacement along x-axis at any instant t
x=u_xt + (1/2) a_xt²
» x=u_xt + 0
» x=u t

y-components

1. Component of initial velocity along y-axis.
u_y=0
2.Acceleration along y-axis
g=9.8m/s²
3. Component of velocity along the y-axis at any instant t.
v_y=u_y + a_yt
=0 + gt
» v_y=gt
4. The displacement along y-axis at any instant t
y= u_yt + (1/2) a_yt²
» y= 0 + (1/2) a_yt²
» y=1/2gt<sup>2

TYPE 2</sup>

                                                                     x-components

1. Component of initial velocity along x-axis.
u_x=u cosØ  [Ø = theta]

2. Acceleration along x-axis
a_x=0
(Because no force is acting along the horizontal direction)

3. Component of velocity along the x-axis at any instant t.
v_x=u_x + a_xt
=ucosØ + 0

= ucosØ

» v_x=ucosØ
This means that the horizontal component of velocity does not change throughout the projectile motion.
4. The displacement along x-axis at any instant t
x=u_xt + (1/2) a_xt²
» x=(ucosØ)t

y-components

1. Component of initial velocity along y-axis.
u_y=u sinØ
2.Acceleration along y-axis
g= -9.8m/s²
3. Component of velocity along the y-axis at any instant t.
v_y=u_y + gt
v_y=usinØ + gt
 4. The displacement along y-axis at any instant t
y= u_yt + (1/2) gt²
» y= (usinØ)t + (1/2)gt²