Status Update

Rough Sketch 
Figure 1.1: Rough sketch

Notes

• The right side of the rotatable piece (the side that holds in the projectile) is pushed downwards, causing the elastic to elongate (displacement > 0 )
• When the applied force is released, the elastic contracts thrusting the center piece back upwards and launching the projectile 

Materials 

• Large flat wooden platform to use as a sturdy base 
• Thin pieces of wood 
• Very thin, cylindrical wood piece 
• Elastic 
• Small hook 
• Wood glue 
• Small hand saw
• Electric drill 

Physics Concepts Covered

Hooke's law 

• Hooke's law applies to objects that can be deformed (compressed or stretched) when a force is applied on it. This includes elastics like the one that will be used in this projectile launcher. 
• The downward force applied on the rotatable arm will cause the elastic to stretch proportionally to the forced applied. 

Elastic potential energy and kinetic energy 

• These are two types of energy that will be associated with the launching of the projectile
• Elastic potential energy will be stored as the elastic is stretched and as the projectile is launched, it will possess kinetic energy 

Conservation of Energy

• This law specifics that energy cannot be created or destroyed 
• In this case, the built-up elastic potential energy will be converted into kinetic energy

Projectile motion and kinematics 

• The projectile will travel in a parabolic curve like all projectile's launched at a velocity at an angle 
• Various kinematics equations can be used to theorize the distance traveled by the projectile  

Ball Measurements 

Averages

Measurement	Average
Mass (g)	67.406
Volume of ball (cm3)	8.7
Diameter (cm)	2.63

Purpose

To design and build an effective projectile launcher that obeys Hooke's law, quantitatively theorize the distance traveled by a spherical projectile and evaluate those calculations against collected experimental data.

After a projectile launcher is home-built and data for the used projectile is collected, the theoretical distance traveled will be calculated using concepts such as Hooke's law, conservation of energy and projectile motion. The projectile launcher will then be tested and data for the actual distance traveled will be measured and compared and evaluated against the theoretical calculations using methods such as percentage error. Ideally, a small spherical mass of around 65g is to be used and travel a distance of around 10m. A basic tape measure or meter stick can be used to measure distance traveled.

Structural Drawing

Materials (updated and with measurements) 

• 1 flat wooden base (37.5cm x 13.5cm x 1.4cm) 
• 2 columns of wood (3.8cm x 3.8xm x 46.0cm)  
• 2 blocks of wood (10.4cm x 3.8cm x 5.0cm) 
• 1 piece of wood (6.8cm x 21.0cm x 3.8cm)
• 1 lightweight, flat wood piece (29.0cm x 8.0cm)  
• 1 rigid cylindrical wood piece (17.0cm in length, ~0.5cm in diameter) 
• 6 elastic bands (5.0cm in diameter) 
• 1 medium sized elastic band (exact size doesn't matter)
• 2 small sized elastic bands (exact size doesn't matter) 
• 3 3" nails (7.6cm) 
• 3 2"nails (5.1cm) 
• 4 1" nails (2.5cm) 
• 4 1" screw-eyes (2.5cm in diameter)
• 2 U-type nails 
• 3 spherical projectiles (see Status Update for specific measurements) 

Structural Drawing
Figure 2.1: Structural Drawing (angle: front)

Figure 2.2: Structural Drawing (angle: top)

Figure 2.3: Structural Drawing 3 (angle: side) 

Changes to Design

• Instead of creating a linking piece across the top of the two column's and attaching the elastic to the catapult arm from there, elastics were attached to the rotating piece from the front face of each pillar. By creating two points of attachment, double the elastic potential energy can be harnessed, and their symmetry will spread out the tension. To avoid the arm from going over 0 degrees (completely vertical), and releasing the projectile towards the ground, an elastic band will be stretched across the two columns. 

Procedure

Building the Projectile Launcher

1. Nail the 6.8cm x 21.0cm x 3.8cm wood piece into the front length of the base board, so it is centered. Use 2" nails and nail from the bottom 
2. Nail the 10.4cm x 3.8cm x 5.0cm wood block into the center back of the support piece previously added using a 3" nail. 
3. Nail each wood column to the front support piece, using a 3" nail, so that they lie on each side of the block 
4. Slip another 10.4cm x 3.8cm x 5.0cm wood piece between the pillars. This swill act as a support piece to keep the two columns straight 
5. Nail a screw-eye on each side of the pole at 3cm and 18cm from the top
6. Tie three elastics (5.0cm in diameter each) to each of the top pair of screw-eyes
7. Attach the thin cylindrical wood piece to the remaining piece of wood (29.0cm x 8.0cm) by stapling it with U-type nails 5cm from one of its ends 
8. On the other end, lightly hammer in two 1" nails 0.75cm from each side of the center, 2.0cm from the edge. These will act as a placeholder for the spherical projectile 
9. Lightly hammer in two 1" nails to the sides of the wood piece so they are 7cm from the other side. Make sure most of the nail is sticking out as the elastic bands will be hooked onto here
10. Slip the ends of the cylindrical wood piece through the screw-eyes on the bottom and wrap a small elastic band around each end as a stopper 
11. Slip the ends of the bundle of elastics onto the nails sticking out from the sides of the attached wood piece 
12. To finalize, put the remaining elastic band over top of the two pillars so it rests on top, above the screw-eyes. 

Figure 4.1: Finalized projectile launcher

Using the Projectile Launcher and Collecting Data 

1. Pick a flat, even surface that has nothing that would potentially interfere with the path of the projectile 
2. Place the projectile launcher on the ground so that there is around 10 meters of open space for the projectile's path  
3. Place the ball (projectile)  between the two nails sticking out of the rotating arm 
4. Put weight on the base board, for example by stepping on it or getting someone to step on it, so it does not lift up from the force applied to the arm 
5. Apply force to the arm, pulling it back until it is at 90 degrees to the vertical (completely horizontal). You may need to lightly keep the spherical ball in place as it may begin to roll backwards 
6. Let go of the arm. The arm will thrust upwards, putting the projectile in motion 
7. Mark where the projectile lands. The distance it rolls after landing does not count in its distance traveled 
8. Measure the distance from the projectile launcher to the landing spot and record the data 
9. Repeat steps 3-8 twice, each time with a "different" ball (same ball with same measurements and maybe some negligible differences) 

--------------------

Further Changes to the Structural Design and Procedure 

Problem 

After sketching a final structural drawing and designing a procedure for the building and testing the projectile launcher, modifications were made to increase the efficiency of the design. 

The flaw to the structural design/procedure was the two 1" nails that were used as a holder for the projectile as rotating arm was retracted While this method was effective in keeping the projectile in place with minimal manual support (i.e. lightly supporting it with your hand), it was found to interfere with the motion of the projectile, specifically its release. It was observed that with this design, the projectile would travel extremely minimal distances, despite the force behind the launcher, and on one occasion case have a backward displacement. This is due to the two nails providing a backwards force against the direction of motion of the projectile.

As the arm begins to rotate from a position horizontal to the ground to the starting position of 20 degrees, the projectile begins to leave its resting position. While theoretically the ball leaves rest at an angle perpendicular to the surface it lies on, meaning that the nails nailed straight into the wood should have little contact against the ball, experimentally, they push against the projectile, pushing the projectile upwards. On top of that, friction may play a role and the heads of the nails may even exert a downward force onto the projectile, forcing it into its resting position. 

Modification 

The solution to the problem would be a feature that would be able to keep in-place the projectile without interfering with its flight. The two nails were uniquely replaced by part of a contact lens package that has a spherical shape to it. With a diameter just under that of the projectile, the projectile is able to firmly sit on it without any parts coming in major contact with it at the sides; the ball is kept in place from the bottom and not the sides this time. 

Added to the materials list would be 

• Cut out contact lens package (approx. 2.0cm - 2.5cm in diameter) 
• Hot glue gun 

Step 8. of building the projectile launcher would be replaced with placing the contact lens package piece, at the same position 2.0cm from the edge in the center of the rotating piece. 

Figure 4.2: Modification to the structural design 

Theoretical Calculations: Hooke's Law

Hooke's Law 
Hooke's law is a concept that can be applied to objects that can deform, when acted on by a force, and return to their original state when alleviated of that force. This includes objects such as springs, which can be both compressed or stretched, as well as elastics. But what exactly does Hooke's law tell us about these objects? Hooke's law can be defined with the following formula: 

The formula states that the force required to compress or stretch an object that is elastic in its properties is equal to, and directly proportional to the product of its spring constant (k), and displacement of length (x) from an initial neutral position known as equilibrium. The relationship is simple: to compress or stretch an elastic objects greater distances requires more force, and also requires more force the greater the spring constant is.
 

Finding the Spring Constant 

The spring constant, k, is a unique value, measured in newtons per meter (N/m), that is different for different types of springs/elastics. Since it is proportional to the force applied to displace a spring/elastic, we can deduce that rigid objects such as large compression springs used in industrial machinery will have a large spring constant while an easily stretchable household elastic will have a low spring constant. 

In order to find the spring constant, we must isolate k. We can do this by simulating a situation in which the spring force is proportional to another force. In order for this to occur, the net force must equal 0, meaning that acceleration must equal 0. 

Method 

By hanging a mass from an elastic band, we can create a situation in which spring force is equal to another type of force - in this case, gravitational force. By attaching an elastic band to a peg in the wall and attaching different masses to it, we can find the spring constant k by measuring the displacement of the elastic that is a result of the gravitational force exerted by each of the masses hung on the elastic. This is shown in the below equation: 

Figure 3.1: Method used to find the displacement of an elastic under different gravitational forces. 

Masses that the elastic could sustain such as this water bottle weighing 515g were hung from the elastic in a bucket (mass = 102g).

Raw Data 

Figure 3.2: Mass and Length of Elastic Band 

Mass (g)	Length (cm)
0	6.5
102	7.7
249	8.5
457	9.5
604	10.1
764	12.5
972	15.6

Processed Data

We first must convert mass and length into standard units before calculating gravitational force and displacement. Our initial length for displacement is the length of the elastic band when it is holding 0 g. 

Sample Calculations 

Fg = mg 

     = (0.457kg)(9.8m/s)

     = 4.48 N 

d = Final Length - Initial Length

   = 0.085m - 0.0065m 

   = 0.030 m 

Figure 3.3: Gravitational Force of Mass and Displacement of Elastic Band 

Gravitational Force (N)	Displacement (m)
0	0
1.00	0.012
2.44	0.020
4.48	0.030
5.92	0.036
7.49	0.060
9.53	0.091

Finding the Slope
While it is possible to algebraically find the spring constant, k, by dividing the gravitational force by the displacement for each case and finding the average k value, we can also do this graphically by finding the slope using a linear regression analysis. Since k = mg/x, the resulting slope should equal the value of k the same way the slope of a distance vs time graph will give velocity.
*Note that Fg must be on the Y axis and d must be on the X axis in order for the value of the slope to reflect that of the spring constant.

Figure 3.4: Linear Regression to Find the Spring Constant 

According to the regression analysis, the collected data shows a strong correlation in which the R-squared value is 0.9357. The resulting slope of the best fit line is 109 and therefore, we can say that the spring constant for the elastic band is 109 N/m.

We can confirm this value manually by picking two points on the best fit line and and finding the divisor of change in gravitational force and change in displacement. In this case, we pick the lowest and highest points on the best fit line.

Evaluation of the Method 
The R-squared value of the collected data does not show an entirely perfect correlation; while it is not weak, it's not the strongest either. While fault in the employment of the procedure may have been the resistance of the wall. Masses were hung on the elastic by placing them in a bucket and its large size caused it to have contact with the wall. On top of this, the elastic band was hooked from a peg in the wall that was located in an inside corner, increasing the contact with the wall. Often, the elastic had to be manually stretched before allowed to naturally hang so the friction of the wall wouldn't hold up the stretching. This may explain  inconsistencies in the data such as the 4th and 5th data points. 

Theoretical Calculations: Conservation of Energy

Elastic Potential Energy 

Stretch an elastic and then stop at a position. A quick release of your finger can cause the elastic to snap back to its original state but keeping your finger on it will leave it motionless; you're in full control of what the elastic does. Whether you wait a few seconds, a whole minute or longer, the elastic will always snap back. A storage of potential energy is what is being experienced by an elastic and is a similar situation to a ball in your hand that will always fall to the ground, whenever you decide to let go of it. Whether it's the ball or elastic, both objects have the potential to be displaced and consequently do work. 

The formula for elastic potential energy is as follows: 

The formula contains the same variables as those in Hooke's laws; k is the spring constant, measured in Newtons/meter, and x is the displacement, measured in meters. 

We can predict the elastic potential energy gathered by stretching the arm of the projectile launcher. We must use the previously found spring constant of the bundle of elastic bands (109 N/m) and we also must find the displacement of the elastic bands when stretched as the arm is brought to 90 degrees. We also must calculate the cosine of each product since each elastic lies on an angle inward from 90 degrees. 

When measured, the elastic slanting 34 degrees inward stretched to 21.0cm while the elastic slanting 10 degrees inward stretched to 20.5cm. 

Displacement of elastic 1 (34 degrees) = 21.0cm - 6.0cm = 15.0cm 

Displacement of elastic 2 (10 degrees) = 20.5cm - 5.5cm = 15.0cm 

Conservation of Energy 

The concept of conservation of energy is easier to understand in examples such as this one where potential energy is involved. Like we defined earlier, potential energy is a stored energy that can be released, in this case by letting go of the catapult arm. This trigger creates a thrusting force in which we see movement; when potential energy is released, it doesn't just disappear somewhere but it causes a consequent event. This consequent event involves a different type of energy that is converted from the previous stored potential energy. This scenario can be generalized with the law of conservation of energy. 

Like the law of conservation of mass, this law specifies that energy cannot be created or destroyed but can be transferred between different types of energies. In this case involving elastic potential energy, we see that as the potential energy is released, the projectile is given movement and therefore, we can say that there's been a conversion of energy into kinetic energy. This law can be applied to different situations such as a breaking car in which kinetic energy is converted to heat energy which is dissipated through as friction occurs. 

The transformation of energy doesn't happen all at once either. When we are still holding the elastic, we observe no kinetic energy because the projectile is not in motion yet and therefore can say only potential energy is present. As the elastic is released from its stretch displacement, it begins to revert back into its equilibrium position. The closer it gets to equilibrium, the more potential energy is converted into kinetic energy. By the time the elastic reaches equilibrium, all of the potential energy will have been converted into kinetic energy. 

Using this law of conservation of energy, we can algebraically find the velocity of the projectile since kinetic energy = (1/2)mv^2 and is equal to the previously found value of elastic potential energy. 

Theoretical Calculations: Projectile Motion

Projectile Motion 
As elastic potential energy is transferred into kinetic energy and the projectile leaves the arm of the catapult at an initial velocity, it begins to travel on its parabolic path, like any projectile launched at an angle; as the negative acceleration of gravity brings it to a peak point, it proceeds to pull the projectile back down to earth.

Horizontal and Vertical Components 
In order to examine the motion of the projectile, we must individually look at its components and therefore, we must break down the initial velocity into horizontal and vertical components.

Let's now compile all of our known values for each component.

Horizontal Component		Vertical Component
Initial velocity	7.63 m/s	Initial velocity	2.78m/s
Acceleration	0 m/s2	Acceleration	–9.8  m/s2
Time	?	Time	?
Displacement	?	Displacement	–0.44 m

• The displacement in the vertical component is negative since the projectile leaves a point above the ground. 

Time In the Air 

Next, we can solve for time in the vertical component. 

Since we are dealing with a quadratic, we are given two variable values as an answer. Since one of the values is negative, we can deem it incorrect since time cannot have a negative value. We can also sketch the parabolic path of the parabola to understand where the second answer (-0.129s) comes from.

If we were to continue to sketch the path of the projectile's motion before it is launched from its height of 0.44m, we would eventually reach an imaginary point at which the displacement is -0.44m. This is the imaginary point that the projectile would reach in -0.129s.

Distance Traveled 
Upon finding the time the projectile will spend in the air, we can use this same time value to find the displacement in the horizontal direction (distance traveled).

Data

Calculating the Average 
mean = sum of trials / number of trials
          = (4.10m + 4.30m + 4.30m) / 3
          = 4.23 m

Error Calculation 
The difference in the theoretical distance and actual distance traveled by the projectile can be calculated in a percentage.
% error = [1.00 - (actual result / theoretical result)] x 100
             = [1.00 - (4.23m / 5.31m)] x 100
             = (1.00 - 0.80) x 100
             = 20%

Conclusion

The purpose of this experiment was to design a projectile launcher that utilized spring power (obeys Hooke's law), to calculate the theoretical distance that would be traveled when a 65g spherical projectile is launched and compare it to experimental data. The theoretical calculations for the built launcher stated that the 65g projectile would travel a horizontal distance of 5.31m. This theoretical value was moderately met with 20.0% discrepancy at a mean distance of 4.23m over a series of three trials. The consistency over the three trials, represented with a standard deviation value of 0.115, proves a precision in the launcher's function.

Evaluation

The experimental data's low standard deviation value paired with the 20.0% error calculation when compared to the theoretical result shows that there's a factor in the projectile launcher that consistently limits its experimental performance.

A possible explanation for part of this difference could be the ball place-holder (contact lens package) that was used. While this place-holder doesn't exert an obvious negative force on the projectile, like the nails used in the initial design, a negative frictional force between the surface of the projectile and contact lens package could contribute to a smaller actual value than the experimental value that negates any form of resistant force.

A more likely reason for this 20.0% error though is the point of release of the projectile. If the ball was released at a point before reaching 20 degrees, then its distance traveled would have decreased. This is not due to the greater angle of projection, since this angle doesn't effect horizontal displacement, but less kinetic energy achieved at the point of release. The estimate of distance traveled was calculated with the presumption that the elastic would entirely return to equilibrium. If the projectile was released before this point, then kinetic energy would be less because there is still some potential elastic energy left in the elastic that has yet to be converted. Consequently, the displacement of the elastic is also less. A reason for an early release of the projectile could simply be gravity, since an angle of 20 degrees to the vertical is quite steep and it is very possible that the projectile could've left its place-holder on the rotating piece at an earlier incline/position.

Eliminating the difference between theoretical and experimental results could be by basing your calculations and device around an angle of projection of less vertical incline. For example, instead of making the point of release 20 degrees to the vertical, which could be unrealistic, an equilibrium that lies at an angle of 45 degrees may be better. In order to make up or the decrease in displacement of the elastic, there are several factors that could be taken into consideration. These may also increase the overall effectiveness of the projectile launcher and make the projectile go further.

Using a material with a larger spring constant will prove very effective since the elastics used were of ordinary household material. Even incorporating a spring that is not too rigid may prove drastically different in result. Also, using a longer rotating arm-piece will increase the displacement of the elastic, creating a larger build-up in elastic potential energy. Developing a way to attach the elastics to the rotating piece at a more parallel placement will also have some help on a longer horizontal displacement. The inward slant (in this case, 10 degrees and 34 degrees) takes away from the elastic potential energy collected since a resultant value must be considered.