REPORT
AIM
The aim of this experiment is to:
□ Explore the equations of uniform accelerated motion and investigate the relationship between displacement and time □ Determine the magnitude of deceleration due to friction. □ Assess the effect of mass on the car’s accelerated motion.

DESIGN
Hypothesis – A car moving in a straight line with a non-zero initial velocity will finally come to a rest as a result of friction, given that the car has no engine or external tractions. This motion can be considered as a uniform accelerated motion because: 1. The car is moving in a horizontal straight line so weight is cancelled by the normal reaction force from the ground. The only other force existed is the friction between the car and the surface therefore it will be equal to the net force 2. According to the formula Fr = μN, the amount of the friction depends on two factors: the friction coefficient and the normal reaction force, both of which are fixed. Therefore the amount of friction is constant throughout the motion 3. According to Newton’s Second Law F = ma, a constant net force will result in a uniform acceleration (deceleration). The acceleration is negative in this case as cars are slowing down to a rest.

For convenience, this decelerated motion can be inverted into an equivalent motion in which the car is acceleration from rest. It should follow the equation of x = ut + 1/2 at2,
where x = distance traveled, u = 0 (seen as the initial velocity but actually is the final velocity which is zero at rest), a = acceleration (actually deceleration) and t = time taken during that motion. This formula can be simplified as x = 1/2 at2. We will measure the variables of x and t to verify this relationship and determine the magnitude of this deceleration, which can be derived from the gradient of the regression line of x against t2.

Theoretically the mass of the car should not influence its deceleration because: Friction is the net force, Fr =μN = μmg (as...

...
E105: UNIFORM CIRCULAR MOTION
NADONG, Renzo Norien D.
OBJECTIVE
The purpose of this experiment is to quantify the centripetal force on the body when one of the parameters is held constant and to verify the effects of the varying factors involved in circular motion. Mainly, horizontal circular type of motion is considered in this activity.
Circular motion is defined as the movement of an object along the circumference of the circle or the manner of rotating along a circular path. With uniform circular motion it is assured that the object traversing a given path maintains a constant speed at all times. Centripetal force is a force that tends to deflect an object moving in a straight path and compels it to move in a circular path.
MATERIALS AND METHODS
This experiment was divided into three parts in order to further study and observe the factors that affect the centripetal force of a body. The concept of this experiment is the same on all parts, which is getting the centripetal force given with three different conditions. Every part of the experiment was executed just the same. Mass hanger plus a desired mass of weights were hanged over the clamp on pulley to determine a constant centripetal force which will act as the actual value. But on the third part of this experiment, aside from the centripetal...

...Exploration Guide: Uniform Circular Motion
Go to www.explorelearning.com and login. Please type or write your answers on a separate sheet of paper, not squished in the spaces on these pages. When relevant, data collected should be presented in a table.
Objective: To explore the acceleration and force of an object that travels a circular path at constant speed. Motion of this kind is called uniform circularmotion.
Part 1: Centripetal Acceleration
1. The Gizmotm shows both a top view and a side view of a puck constrained by a string, traveling a circular path on an air table. Be sure the Gizmo has these settings: radius 8 m, mass 5 kg, and velocity 8 m/s. Then click Play and observe the motion of the puck.
a. The puck in the Gizmo is traveling at a constant speed, but it is NOT traveling at a constant velocity. Explain why.
b. Because the velocity of the puck is changing (because its direction is changing), the puck must be experiencing an acceleration. Click BAR CHART and choose Acceleration from the dropdown menu. Check Show numerical values. The leftmost bar shows the magnitude of the acceleration, or |a|. (The other two bars show the x- and y-components of the acceleration, ax and ay.) What is the value of |a|? Jot this value down, along with radius = 8 m, so that you can refer to it later.
c. Keeping velocity set to 8 m/s, set radius to 4 m. (To quickly...

...Uniform Circular Motion – a constant motion along a circle; the unfirom motion of a body along a circle
Frequency (f) – the number of cycles or revolutions completed by the same object in a given time; may be expressed as per second, per minute, per hour, per year, etc.; standard unit is revolutions per second (rev/s)
Period (T) – the time it takes for an object to make one complete revolution; may be expressed in seconds, minutes, hours, years, etc.; standard unit is seconds per revolution (s/rev)
Note: Period and frequency are reciprocals: T = 1/f; f = 1/T.
Sample Problems:
1. Suppose the rear wheel makes 5 revolutions in 1 minute. Find the wheel’s period and frequency.
2. As a bucket of water is tied to a string and spun in a circle, it made 85 revolutions in a minute. Find its period and frequency.
3. * An object orbits in a circular motion 12.51 times in 10.41 seconds. What is the frequency of this motion?
Tangential Speed (v or vs) – average speed; rotational speed; speed of any particle in uniform circular motion; standard unit is meters per second (m/s); v = Cf = C/T = 2πrf = 2πr/T = rω
Sample Problems:
3. What is the rotational speed of a person standing at the earth’s equator given that its radius is 6.38*106 m and that it takes 365 days for the earth to complete a revolution?
4. A ball that is whirled about on a string makes 167...

...DIFFUSION AND OSMOSIS
Chapter 3 of your textbook explains diffusion and osmosis.
Diffusion is simply the net movement of atoms or molecules from a region of higher concentration to a region of lower concentration. The force behind the movement is heat or kinetic energy (also called Brownian motion). Diffusion occurs when you spill water on the carpet floor and it spreads out, or when you open a bottle of perfume and it leaves the bottle and spreads throughout the air in the room.
Osmosis is a similar phenomenon that moves water from a region of high water concentration to a region of low water concentration.
Imagine that a cell from your body is placed in a solution of water. If the concentration of the water inside the cell is the same as the concentration of water in the solution, then we describe the water solution as being “isotonic” or having the same concentration as the water inside the cell. In this case, net movement of water will be zero and the cell will not swell or shrink. In other words, the same amount of water will move in the cells as will move out.
On the other hand, if the cell is placed in a solution of water that has a higher concentration (of water) compared to the concentration of water inside the cell, then osmosis will cause more water to move into the cell than will move out and this will cause the cell to swell. In this case, we say that the solution of water is “hypotonic”.
Now imagine that the cell is placed in a solution of...

...Lab 2
(10 points)Laboratory: Motion on an Inclined Plane.
Objective: To determine the Force, Mass, Acceleration relationship on an inclined plane with friction.
Hypothesis: Increasing the mass of the cart will decrease the acceleration.
Materials: protractor, car, calculator, meter stick, weight(s)
Procedure:
Set the inclined ramp to θ= 15.8
Measure a length of 1.0 m.
Place a 0.2 kg mass into the cart.
Release the cart and measure the time to travel the 1m.
Repeat with two additional masses.
Data:
Mass 1 0.2 Kg | ∆y cm | ∆x cm | ∆t s | Θ (degrees) |
Trial 1 | 25 | 91 | 1.4 | 15.8 |
Trial 2 | 25 | 91 | 1.2 | 15.8 |
Trial 3 | 25 | 91 | 1.2 | 15.8 |
Mass 2 0.02 Kg | ∆y cm | ∆x cm | ∆t s | Θ (degrees) |
Trial 1 | 25 | 91 | 0.9 | 15.8 |
Trial 2 | 25 | 91 | 1.1 | 15.8 |
Trial 3 | 25 | 91 | 0.9 | 15.8 |
Mass 3 0.01 Kg | ∆y cm | ∆x cm | ∆t s | Θ (degrees) |
Trial 1 | 25 | 91 | 0.8 | 15.8 |
Trial 2 | 25 | 91 | 0.8 | 15.8 |
Trial 3 | 25 | 91 | 0.9 | 15.8 |
(45)Analysis:
Find acceleration for each mass.
∆x=vi∆tavg + 1/2a∆t avg2
Mass 1= 1.08
Mass 2= 1.93
Mass 3= 2.64
Complete a full analysis for each mass.(indicate formulas used for each computation and include all units)
Kinematics:
Known: vi=0 m/s ∆x= 0.91m ∆t avg= 1= 1.3s 2=1.97s 3= 0.83s a= 1= 1.08 2= 1.93 3= 2.64 m/s squared (v/t)
Unknown: vf =1= 1.4 m/s 2= 1.9 m/s 3= 2.2 m/s (Vf= Vi + at) ρf= 1= kgm/s 2= 0.04 kgm/s 3= 0.02 kgm/s (...

...Uniform Circular Motion
PES 115 Report
Objective
The purpose of this experiment is to determine the relationships between radiuses, mass, velocity and centripetal force of a spinning body. We used logger pro to accurately measure the orbital period of the spinning mass and used these measurements to determine the interrelated interactions of the specified properties and viewed the results graphically.
Data and Calculations
The black markings on the string are about 10 cm apart in length, measured from the center of the spinning mass.
Part A: Factors that influence Circular Motion
Velocity versuse Centripetal Force
Fill out the table holding the Spinning mass (M) and the radius (R) constant.
Figure 1: Experimental setup for the lab
Which Spinning Mass did you select _hook with foam wrapping_ (Tennis ball, etc..)
What is the mass of the Spinning mass _0.0283_ kg.
What Radius did you select _0.30_ m (around 20 cm is a good choice).
Fill out the tables for five different hanging mass values.
Hanging Mass (m) [kg]
0.1001 kg
0.1992 kg
0.2992 kg
0.4000 kg
0.4997 kg
Revolution Number and Time per Revolution (T) [sec]
1
0.61337 s
0.413210 s
0.367288 s
0.316510 s
0.271455 s
2
0.613087 s
0.403737 s
0.370600 s
0.310189 s
0.274200 s
3
0.613727 s
0.393689 s
0.374100 s
0.316308 s
0.273700 s
4
0.611319 s
0.39364 s
0.368047 s
0.309619 s
0.279400 s
5
0.618954 s
0.388600 s
0.365853 s
0.300742 s
0.282000 s...

...I. Objectives:
• To verify Newton’s Second Law of Motion with the use of state-of-the-art devices
II. Materials and Equipment:
• 2.2 m Track- 1 pc
• Plunger Cart- 1pc
• Super pulley with clamp – 1pc
• .500gram mass- 1pc
• Stopwatch1- 1pc
• Block ( to act as bumper)- 1pc
• Beam Balance- 1 unit
• String – 2m long
• Set of Weights-1 set
III. Data and Results
Cart Mass Hanging Mass Trial1 Trial2 Trial3 Trial4 Trial5 Average Time
512g 13g 2.16s 2.15s 2.06s 2.0s 2.1s 2.09s
1016.5 27g 1.98s 2.11s 2.19s 2.09s 2.01s 2.08s
Cart Mass Acceleration (m1+m2)a FNET=m2g %Error
512g 0.18m/s2 0.09N 0.12N 25%
1013g 0.18m/s2 0.19N 0.26N 27%
IV. Observations
• The more force we apply on an object, the faster an object goes.
• The more mass an object has, the more difficult it is to change its state of motion.
• Though the second attempt has more mass, it still has the same acceleration as the first one.
• Air resistance from ceiling fan also affects the force to an object.
V. Conclusion
We therefore conclude that when the force applied to the object increased, the acceleration also increased, and that the mass of an object is inversely proportional to acceleration.
We also conclude that the mass in the cart affects the force within it as it moves towards one of the road block. Also the friction between the cart and the track affects the force and acceleration. When the cart has a large amount of mass, then the...

...The lighting was kept constant during the complete test. No additional light sources were added throughout the experiment, nor were any light sources removed throughout the experiment. This minimized the errors involved with trying to read and record the indicated measurements on the ticker tape, as well as the ability to analyze the experiment.
2. The same person measured and recorded the height and length of the incline plane and level horizontal track to ensure consistency in the recording of data. By having the same person read all of the measurements minimized the degree of uncertainty in position and angle of reading values.
3. The same person recorded all measuring values on the ticker tape to ensure consistency in the recording of data. By having the same person read all of the measurements minimized the degree of uncertainty in position and angle of reading values.
4. The spark timer was set to a frequency of 10Hz (10 dots/s). This eliminated the possible random error that could have been associated with the recording of time if a person were to record the time, due to delays in reaction time. By having the consistency of the spark timer record the time values the total accuracy and precision for the experiment was increased.
5. There was no wind or abnormal air movements during the complete test. No doors were opened or closed during the experiment, nor were any windows opened or...