Conservation+Energy

Trey Carpenter Mr. Richardson AP Physics B

The purpose of this project is to prove the conservation of energy from a projectile, between when a ball is launched and when it strikes a timing pad. My goal in this project is too see if the total energy of the steel ball bearing when initially launched is equal to that of the ball bearing when it lands on the timing pad. To prove the conservation of energy, I will need very precise measurements of the launching mechanism, as well as for the time of flight and the initial and final speeds of the ball. I will need these measurements to plug into the potential and kinetic energy equations.

Procedure:

Supplies :

Projectile Launcher



Air Pump

Steel Ball

Labquest 2

Time of Flight Pad



I used the vernier projectile launcher to launch a 21.8g ball through the air onto the timing pad

These products were connected to a Labquest 2 to retrieve and display the data gathered by the launcher and timing pad

To meet the stated purpose above I used the gathered information from the 10 tested runs and applied the results to the Energy equations: (KE_i)+

(PE_i)=(KE_f)+(PE_f)

Before I could start my test trials I had to site in the launcher so that I would know where to place the time of flight pad

I decided to set the launch degree to 60˚ and the air pressure to 80psi

For each run I followed the same procedure

I placed the ball into an air tight cylinder in the launcher

I connected the air pump to a valve, that was located in the back of the launcher

I pumped the pump until the pressure gauge read 80psi

I pressed the play button on the labquest 2 so that it would begin taking data

Last of all I pressed the “Arm” and “Fire” buttons on the front face of the launcher to launch the ball onto the time of flight pad

I then proceeded to press the stop button on the Labquest 2 device and saved the data for that run

After this preliminary data was collected I then used the conservation of energy equations (while assuming that the air resistance on the ball was

negligible so that the horizontal velocity of the ball would be constant) to find the initial and final energies of the ball for each of the ten runs.

I then compared the two energies to find that they were extremely close to being equal if not exactly the same


 * Analysis: **

Using the tables below, along with the constants of the height from which the ball was launched (146mm), the height off of the ground that the ball landed (30mm), and the mass of the ball (21.8g) i used the Kinetic and Potential Energy equations to find the total energy of the ball when it was first launched and when it hit the timing pad.

To reach these numerical values I found the initial total energy by using the equation:
 * E=mgh+(1/2)mv^2**

I then used the velocity vector in the "y" direction to find the maximum potential energy the ball would have during its flight and added it to its kinetic energy in the "x" direction by using the "x" component of the ball's velocity in order to find the final total energy of the ball. The equation i used for this was:
 * E=mg[({sin(60)v}^2)****/2g+(0.146)]+{(21.8/2)[cos(60)v]^2}**

I decided on using this equation to find the final energy because when an object is at the peak of its flight, it has it's maximum potential energy and minimum kinetic energy, and since potential and kinetic energies are relative to each other, you then know that although the object will be moving much faster when it lands, the total energy will remain the same.


 * Run 1 ||
 * Time || Gate State || Speed ||
 * 0.00000 || Blocked ||  ||
 * 0.00365 || Unblocked ||  ||
 * 0.01112 || Blocked || 4.495 ||
 * 0.01490 || Unblocked ||  ||
 * 0.81318 || Blocked ||  ||


 * Run 2 ||
 * Time || Gate State || Speed ||
 * 0.00000 || Blocked ||  ||
 * 0.00366 || Unblocked ||  ||
 * 0.01118 || Blocked || 4.474 ||
 * 0.01497 || Unblocked ||  ||
 * 0.80950 || Blocked ||  ||


 * Run 3 ||
 * Time || Gate State || Speed ||
 * 0.00000 || Blocked ||  ||
 * 0.00366 || Unblocked ||  ||
 * 0.01116 || Blocked || 4.478 ||
 * 0.01496 || Unblocked ||  ||
 * 0.81026 || Blocked ||  ||


 * Run 4 ||
 * Time || Gate State || Speed ||
 * 0.00000 || Blocked ||  ||
 * 0.00366 || Unblocked ||  ||
 * 0.01117 || Blocked || 4.477 ||
 * 0.01496 || Unblocked ||  ||
 * 0.80949 || Blocked ||  ||


 * Run 5 ||
 * Time || Gate State || Speed ||
 * 0.00000 || Blocked ||  ||
 * 0.00365 || Unblocked ||  ||
 * 0.01114 || Blocked || 4.489 ||
 * 0.01492 || Unblocked ||  ||
 * 0.81168 || Blocked ||  ||


 * Run 6 ||
 * Time || Gate State || Speed ||
 * 0.00000 || Blocked ||  ||
 * 0.00364 || Unblocked ||  ||
 * 0.01110 || Blocked || 4.503 ||
 * 0.01487 || Unblocked ||  ||
 * 0.81472 || Blocked ||  ||


 * Run 7 ||
 * Time || Gate State || Speed ||
 * 0.00000 || Blocked ||  ||
 * 0.00365 || Unblocked ||  ||
 * 0.01113 || Blocked || 4.493 ||
 * 0.01491 || Unblocked ||  ||
 * 0.81249 || Blocked ||  ||


 * Run 8 ||
 * Time || Gate State || Speed ||
 * 0.00000 || Blocked ||  ||
 * 0.00365 || Unblocked ||  ||
 * 0.01113 || Blocked || 4.494 ||
 * 0.01491 || Unblocked ||  ||
 * 0.81261 || Blocked ||  ||


 * Run 9 ||
 * Time || Gate State || Speed ||
 * 0.00000 || Blocked ||  ||
 * 0.00365 || Unblocked ||  ||
 * 0.01112 || Blocked || 4.497 ||
 * 0.01490 || Unblocked ||  ||
 * 0.81251 || Blocked ||  ||


 * Run 10 ||
 * Time || Gate State || Speed ||
 * 0.00000 || Blocked ||  ||
 * 0.00364 || Unblocked ||  ||
 * 0.01110 || Blocked || 4.503 ||
 * 0.01487 || Unblocked ||  ||
 * 0.81408 || Blocked ||  ||

Results:

The results of this project are fairly straight forward. It is clear that energy is conserved when air resistance is considered negligible.

(KE i )+(PE i )=(KE f )+(PE f )


 * || Run 1 || Run 2 || Run 3 || Run 4 || Run 5 || Run 6 || Run 7 || Run 8 || Run 9 || Run 10 ||
 * Initial Energy (J) || 251.43 || 249.37 || 249.76 || 249.67 || 250.84 || 252.21 || 251.23 || 251.33 || 251.62 || 252.21 ||
 * Final Energy (J) || 251.43 || 249.37 || 249.76 || 249.67 || 250.84 || 252.21 || 251.23 || 251.33 || 251.62 || 252.21 ||