Anthony+And+Sam

= = Lab: What is the acceleration of a falling body? = Objective: What is the acceleration of a falling body? =

**Materials**: Ticker Tape Timer, Timer tape, Masking tape, Mass, clamp, meter stick.

**Hypothesis:** If we drop a mass that is taped to a ticker tape going through a ticker tape timer and we graph where the dots were located at each interval of time, we will find the acceleration of a falling body due to gravity and our results should be around 9.81 m/s^2.

**Procedure:**

1. Clamp the spark timer to a cabinet 2. Feed the ticker tape through the spark timer 3. Tape a weight (200g) to the end of the ticker tape that feed through the spark timer 4. Set the spark timer to 60Hz 5. Turn the spark timer on and let go of the weight, letting the ticker tape go through the spark timer 6. Measure the distance between each dot and record This is a picture of the timer attached to the cupboard with the tape through the machine and it being taped to the mass.

This is a small snapshot of what part our ticker tape looked like after the experiment was completed.

**Excel Table:**

Time is in seconds and distance is in centimeters. Each dot (labeled 1 to 33) on this table is representative of each dot, first to last, that was drawn on the ticker tape by the timer. Each time shown is accumulative of one sixtieths of a second because each dot was drawn with intervals of one sixtieth of a second in between.

**Excel Graph:** Because our R^2 is one, we know that our data was very consistent and followed a nice parabola-shaped curve. nice results! R^2 = 1!

That table and graph can be found right here.

**Calculations:**

Y=Ax2+Bx<The equation of the line above without any specific numbers. d=(A)t2+(B)t d=Bt+At2 d=Vit+1/2at2<- Changing the variables around and replacing "x" with the time variable "t", one of the motion equations can be formulated.

A= ½ acceleration B= Initial velocity

A = 0.5a 433.97= 0.5a a=867.94 cm/s2 a=8.68 m/s2

**Discussion Questions** 1. Does the shape of your graph agree with the expected graph? Why or why not?

The shape of our graph is half of a parabolic curve starting at (0,0). This agrees with the expected graph. Since velocity = slope of a distance vs. time graph, the beginning of the curve should start having a slope close to zero. Towards the end, you can notice the slope of the curve is getting closer a closer to a vertical line, or no slope. This is due to the fact that the mass is not moving at the beginning of the experiment and then continually gets faster and faster until it reaches the ground. This is shown in our graph.

2. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.) x1 = our results x2 = average results of the class

% Difference = absolute value of ((867.94 - 848.65) / ((867.94 + 848.65) / 2)) x 100 % Difference = 2.25%

Our results are pretty close to the average of the class, only aro und 19 cm away, considering the range of the class data was about 590. Our percent difference was a low 2.25%, proving this fact even more.

3. Did the object accelerate uniformly? How do you know?

Yes, and this is due to the fact that velocity is increasing at a constant rate. Another way this can be proven because in the equation of the line, A= 1/2 acceleration and therefore acceleration is a constant and never changes.

4. What should the velocity-time graph of this object look like?

The velocity-time graph of this object should look like a straight line with a positive slope. The acceleration is positive and therefore the velocity increases each time interval.

5. Write down the expected equation of the line from this v-t graph (use specific information from your x-t graph).

y=867.94x+53.411

The slope of the velocity time graph must equal acceleration.

6. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?

It could be higher than it should be if when the mass was being dropped, an extra push by the person's hand could have increased the speed and made the object accelerate higher than usual. It could be lower too. This occurred more often in this experiment because of the forces that were going against the mass falling to the ground. The ticker tape being taped to the mass slowed it down a bit. The ticker tape was being ticked by a timer which caused a bit of decceleration. It was also slowed down by the tape sliding up against the cupboard while it was going through the timer. Since the mass slows up a bit, the acceleration is much less than it should be.

**Conclusion:**

The actual acceleration of a free falling object is 9.81m/s^2, while our experiment proved it to be 8.68m/s^2. This came out to be a percent error of 11.52%. This can be blamed on some sources of error in our experiment. The object was not truely free falling. It was attached to a tape who could have affected the speed. This tape hit the cabinet as it was going through the timer, which might have caused a sower speed. Each time the timer made a dot on the tape, the mechanism used to make the dot slowed down the tape in the tiniest way possible. Overall, our experiment went relatively well, considering there were sources of error that were impossible to eliminate if we wanted to do the experiment correctly.incomplete