AP Physics Lab 2

Lab #2

Katie O’Byrne, Michael O’Connell, Chris Prattos, Ryan Partain

Constant Acceleration Particle Model Lab

Objective: To determine the graphical and mathematical relationship between position and clock reading for a ball rolling down an incline.

Materials and Diagram:

-Ramp (with 10 cm increments)

-Stand with ramp attached at hole 12

-Metal Ball

-Photo Gates

-Photo Gate Timer

Procedure:

1) Set up stand and attach ramp at hole 12 from the bottom

2) Place photogate A at 10cm and photogate B at 20cm.

3) Clear all readings on the photogate timer

4)Place metal ball at top of ramp and release

5) Record time in seconds for photogate A, photogate B, and the time it took to get to photogate A to B.

6) Repeat steps 3-5 leaving photogate A at 10 cm and moving photogate b in 10 cm increments

7) On trial 8 skip 90cm and move photogate B to 100 cm for the final trial.

Data:

Table for Velocity-Time Graph

VB=(d/tB)

d=1.90cm

(For the VB and tB quantities, view the graph above)

Data Analysis:

X=(215.40cm/s^2)t^2+(80.544cm/s)t+10.637cm

X=(349.955cm/s^2)t^2+20.705m

V=(425.241cm/s^2)t+89.7cm/s

Conclusion:

1. Describe the motion studied in terms of both position and speed and explain the motion map

which represents the motion. Include qualitative motion map showing velocity and acceleration

vectors.

The position changed at faster rates per second and therefore the motion had increasing velocities. The object traveled a greater distance per second as the ball continued down the ramp. The velocity then increased as shown by the increasing length of the arrows in the motion map. The change in velocity (acceleration) however was constant.as shown by the same sized arrows in the motion map .

2. Explain the relationship between variables seen in the x-t and x-t^2 graphs. Show and explain the equations you developed and include numerical values for your constants.

As the object rolled down the ramp , it  changed positions  at a faster rate per second. This caused the graph to form a quadratic relationship or curve.  The equation for the position time graph  is x=½at^2+ V0t+ Xo.  A represents acceleration, V0 represents the velocity at time zero and X0 represents position where time is zero.  T represents time, while x represents position.  The equation for this graph was  X=(215.40cm/s^2)t^2+(80.544cm/s)t+10.637cm; however, in order to find acceleration one needs to find the slope of the velocity, which is the slope of the position vs time graph. In order to find the slope of the position vs time graph we first had to linearize the data. We did this by squaring the time values  which formed a straight line.  This gave us the equation  x= V0t^2+ X0. The equation for this graph was x= (349.955cm/s^2)t^2+20.705m.

We were able to then find the slope of this line but what it represented wasn’t clear based on the data.  We then created a velocity-time graph in order to understand the meaning of the slope.  The velocity time graph created linear line showing that the velocity was increasing steadily as the ball rolled down the ramp. The ball was accelerating at a constant rate as it rolled down the length of the ramp.  

3. Explain how you determined the velocity at a given point in order to make the v-t graph. (This

will require a bit of detail.)

We took the diameter of the ball, 1.90 cm, and divided it by the tB value. the diameter of the ball is the distance it travelled as it rolled. The time it took to travel from the start of gate b to the the end of gate b is the time elapsed)This gave us the velocity at each of the points so we could make the v-t graph.  We plotted the velocity (VB) on the y-axis and the time (tAB) on the x-axis.  The velocity was measured in cm/s, the time was measured in s, and the slope (acceleration) was measured in cm/s^2.

4. Explain the relationship between variables seen in the v-t graph and what this means in terms of the object's motion. Show and explain the equations you developed and include numerical values for your constants.

The relationship is linear (y=mx) or (V=(425.241cm/s^2)t (s))

As the object rolls down the incline the velocity of that object as it enters and exits photo gate b changes as the distance between the ball’s starting point and photogate B increases. This is because the metal ball has a constant acceleration so its velocity is constantly changing. As time goes on the velocity increases linearly as seen from the equation above.

5. Explain the significance of the slope of the v-t graph and provide an interpretation for the slope

without using the word "acceleration." Then provide a formal physics definition of acceleration.

The slope of the v-t graph shows the rate of change in the velocity per second and can theoretically be used to predict the velocity for any given second. Because the slope of the v-t graph is positive instead of zero it means that the object has an increasing change of position per elapsed time meaning that not only is the object changing positions per second but the length of this displacement is also changing.

Acceleration - The rate of change in velocity per unit of time

6. Show and describe how a general velocity equation (consisting entirely of variables) is developed from the v-t graph.

When the v-t graph has been created, an equation can be used as a mathematical model of the graph.  Just like in an x-t graph, the slope is represented by a value.  The slope in a v-t graph is the acceleration.  Instead of using the starting position (like in an x-t graph), the starting velocity is used.  The basic equation for the graph (graph of a line) is y=mx+b.  When using the units and symbols, the final equation becomes V=at+V0.

7. Show and describe how a general position equation (consisting entirely of variables) is developed from the x-t^2graph.

A position equation can be developed from the x-t^2 graph by using the slope of the x-t^2 graph to develop a v-t graph to find the area which is  ∆x which thereby must be equal to ½ of (v+v_0) t. When one makes an a-t graph, one can reason that (v+v_0)=v_0+at+v_0 therefore ∆x=½ (2v_0+at)t which can be simplified to ∆x=v_0t+½ at^2 and then x-x_0= v_0t+½ at^2 into x=½ at^2 + v_0t +x_0.

Some possible sources of error could have been that our photogates could have been imprecise and that we could have dropped the ball at slightly different areas than the exact top of the ramp due to our fingers being slightly sticky from handling our old and sticky photogates. There may also have been slight changes in the air currents in the room as the AC cut on and off but these sources of error probably gave negligible results.