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Student Name: Matthew Parker
Access Code (located on the lid of your lab kit): AC-04YWHGD
Pre-Lab Questions
1. What is the acceleration of a ball that is vertically tossed up when it reached its maximum height? The ball will have a velocity of 0 m/s at its highest point and the acceleration will be 9.8 m/s down. 2. The displacement of a particle as it varies with time is given by the equation x(t) = (10.0 m/s) t + (2.50 m/s²) t². Find the particle’s instantaneous velocity and instantaneous acceleration at t = 4.00 seconds. Instantaneous velocity = 10+5t = 10+5(4)= 30 m/s Instantaneous acceleration= 0+5 = 5 m/s^ 3. What does a positive and negative slope represent for a velocity vs. time graph? If the velocity vs. time graph has a positive slope it indicates the object is speeding up/accelerating. If the graph has a negative slope it indicates that the object is slowing down/decelerating. 4. You know that a car moves with a velocity that can be modeled as v(t)= 4.0 m/s + (1.2 m/s²)t and that at t = 0 the car has a displacement of 5.00 meters from the origin. What is the position of the car at t = 4.0 seconds? 4t+1.2(t2/2)+5=position --> position=44+1.28+5=30.6 meters 5. Derive the second kinematic equation by integration of the first kinematic equation. Then derive the third kinematic equation by using algebra to combine the first and second kinematic equations. 1). v=u+at 2). s=ut+1/2at^2 t=v-u/a s=ut+1/2at^ s=u(v-u)/a+1/2a(v-u)^2/a^2=v^2-u^2=2as
Experiment 1: Distance of Free Fall
Table 1: Washer Free Fall Data
Trail Drop Height (m) Time (s)
1 1.80 m
0.55 s 2 0.64 s 3 0.49 s Average 0.56 s
Description of Auditory Observations of Equally Spaced Hex Nuts: There was about the same interval between the ‘clangs’, but as the hex nuts hit each ‘clang’ was noticeablely louder.
Description of Auditory Observations of Unequally Spaced Hex Nuts: The first three ‘clangs’ were very close to each other and the last considerably louder.
Post-Lab Questions
- Record your hypothesis from Step 1 here. Use evidence from your results to explain if your hypothesis was supported or not. If the space is in between the hex nuts is increased then the sound will be louder when the hex nuts hit the metal pan.
- What was the difference between the noise patterns for equally spaced hex nuts compared to the unequally spaced hex nuts? The sound for the equally spaced hex nuts was more uniform and after hearing the sound that the unequally spaced hex nuts you could tell that the spacing was considerably different.
- If the noise patterns were different, explain why. If they were similar, explain why. The noise patterns were different because the acceleration due to gravity had longer to act on the hex nuts that were higher giving them more momentum and this momentum was converted to sound when they hit the metal pan.
- Using the time it took a single hex nut to reach the pan, calculate the height from which it was dropped. Is this accurate compared to your known height? Explain your conclusion. The calculated height and know height were close to being the same. The only error would ccome from the air friction and being that it was a fairly quick experiment the air friction is not that much.
- A student ran this experiment, and instead of dropping the hex nut, he threw it. This gave the nut a velocity of v(t) = (12 m/s²)t + 5 m/s. What is the hex nut’s displacement as a function of time if its position at t = 0 seconds is 0 meters? If the hex nut dropped 1.2 meters, how long did it take for the nut to reach the ground? 6t^2+5t 6t^2+5t-1.2= 0.19sec
Experiment 2: Distance Traveled by a Projectile
Pre-Lab Questions
- In one of your experiments, you will roll a marble down a ramp to provide an initial horizontal velocity. Suppose you start the marble at rest ( v ˳= 0 m/s) and it travels a distance of, d , down the ramp. Use 1-D kinematics to predict the velocity of the ball (v ) at the bottom of the ramp. Hint: ᶠ the acceleration of the ball down the ramp is 9.81sin(θ) m/s² where θ is the angle of the ramp. Record your answer in variables ( you will calculate the velocity with magnitudes when you perform the experiment). V=(u^2+2as)^(1/2) = (0+(9.81 m/s^2)sin(theta)*s)^(1/2)
- Use the kinematic equations to derive a general equation for the time it takes a ball dropped from rest at vertical height, h , to reach the ground. Use this to write a general equation for the
1 0.205 m 2 0.205 m 3 0.210 m 4 0.205 m Average 0.206 m
Post-Lab Questions:
- Use your predictions of velocity and range from the Pre-Lab Questions and the data recorded from your experiment to complete Table 4. Table 4: Velocity and Range Data for all Ramp Distances
Ramp Distance (m)
Calculated velocity (m/s)
Predicted Range (m)
Average Actual Range (m)
Percent Error
0.235 m Click here to enter text. 0.390 m^ 0.388 m^ 0. 0.170 m Click here to enter text. 0.310 m 0.308 m 0.
0.07 m Click here to enter text. 0.210 m^ 0.206 m^ 1.
- How do your predictions compare to the observed data? Explain at least two reasons for differences. My predictions and the observed data were very close. The differences came from error in placing the marble in the exact spot every time and when releasing the marble sometimes it would stick to my finger slightly and would chage the angle of the roll.
- If you were to fire a paintball pellet horizontally and at the same time drop the same type of paintball pellet you fired from the paintball gun, which pellet would hit the ground first and why is this so? They would both hit the ground at the same time because they both have the same downward acceleration due to gravity.
- A marble slides down a wacky ramp with a velocity given by v = (0.5t²-3.0t) Î + (0.33t³-0.6t) ĵ. At t = 3 seconds the particle is shot off the ramp and behaves like a projectile. What is the magnitude of the velocity of the marble when it leaves the ramp? What is the marble’s acceleration vector while it is on the ramp? If the marble falls for 6 seconds when it is shot off the ramp at a 45° angle, what will be its displacement in the x-direction? (4.5-9)i^+(8.91-1.8)j^ = -4.5i^+7.11j^ displacement in the x-direction = -
Insert photo of your experimental setup with your name clearly visible in the background:
Experiment 3: Squeeze Rocket™ Projectiles
Table 5: Projectile Data for Rockets and Different Launch Angles
Launch Velocity (m/s)
Initial Angle Time (s)^
Average Time (s)
Predicted Range (m)
Actual Range (m)
Average Range (m)
Range % Error Click here to enter text. 90° 0.89 s
0.86 s
0 m 0.38 m
0.41 m
n/a
90° 0.88 s 0 m 0.44 m n/a
90° 0.85 s 0 m 0.42 m n/a
90° 0.82 s 0 m^ 0.40 m^ n/a
45 0.70 s 0.75 s 3.64 m 2.47 m 2.11 m 32.
45 0.77 s 4.01 m 2.04 m 49.
other examples would be a javelin thrower and a discus thrower. Both of these people have to incorporate angles and force into their games.
- A student buys a high powered toy rocket gun and tries to do this experiment. He decides to shoot the first rocket with an initial angle of 0 degrees. The student knows that the rocket has an initial velocity of 5 m/s when he shoots it off of the 3 meter high table. The toy gun is able to give the rocket a horizontal acceleration of (2.1 m/s³)t in the same direction as the initial velocity. The vertical acceleration, directed downwards, is g. Assume air resistance can be ignored. What is the horizontal displacement of the student’s toy rocket? t=(2h/g)^(1/2)=(6/9.8)^(1/2)=0.78 s d=ut+0.5at^2=50.78+0.52.1*0.78^2=4.55 m
Experiment 4: Balancing Centripetal Force
Pre-Lab Questions:
1. In this lab, you will be rotating a mass on one side of a string that is
balanced by a second mass on the other end of the string (Figure 5). Apply Newton’s Second Law of Motion to mass 1, m ₁, and mass 2, m ₂, to solve for the period of mass 1. Hint: assume m ₁ = 4 m ₂. How is the centripetal force on m ₁ related to the force of gravity on m ₂. T=m2g T=m1rw^2 w=2(3.1415)/T m2g=m1r(2(3.1415)/T)^ T=((4(3.1415)^2)m1r/m2g)^(1/2)
2. Draw a free body diagram and solve for the
centripetal acceleration in terms of θ and g for one person riding on the amusement ride in Figure 3. Centripetal acceleration=g*tan(theta)
3. The wheel of fortune is 2.6 meters in diameter. A
contestant gives the wheel an initial velocity of 2 m/s. After rotating 540 degrees, the wheel comes to a stop. What is the angular acceleration of the wheel? r=2.6/2=1.3 m Vi=2/1.3=1.538 rads/s 540 degrees= 9.62 rads angular accel.= (1.538)^2/2*9.62=0. rad/s^
4. The angle that a spoke on a bicycle wheel has rotated behaves according to function θ(t) = at² +
bt where a = 0.6 rad/s² and b = 0.3 rad/s. Find the angular velocity of the spoke as a function of time and the angular acceleration as a function of time. Then find the instantaneous angular velocity and the instantaneous angular acceleration at t = 3 seconds. 2at+b=IAV 6*0.6+0.3=3.9 rad/s^2 2a=IAA 2(0.6)= 1.2 rad/s^
Data:
Table 1: Rotational Data
Radiu s (m)
Time per 15 revolutions (s) Period (s)^ Expected Value^
Percent Error (%) 0.25 8.15 s 0.54 s 0.61 -11. 0.40 8.83 s 0.59 s 0.70 -15. 0.15 6.63 s 0.44 s 0.52 -15.
Post-Lab Questions:
5. Compare your measured data to your predicted values with a percent error calculation. Explain
any difference with an error analysis. There is a lot of human error in this experiment. The speed has to remain constant, the angle of rotation needs to be kept perpendicular to the tube, counting the revolutions, and stopping the stopwatch watch at the exact moment of 15 revolutions.
6. Draw a circle to represent the path taken by your rotating mass. Place a dot on the circle to
represent your rotating washer. Add a straight line from the dot to the center of the circle, representing the radius of rotation (the string). Now label the direction of the tangential velocity and the centripetal force. Insert photo of the circle with your name clearly visible in the background:
7. Use your data to calculate the average velocity, angular velocity, and centripetal acceleration for
the mass in each radius.
Average velovcity 0.15m=2.14 m/s 0.25m=2.9 m/s 0.40m=4.26 m/s
Angular velocity 0.15m= 14.28 rad/s 0.25m=11. rad/s 0.40m=10.65 rad/s
Centripetal acceleration 0.15m= 30.53 m/s^ 0.25m= 33.64 m/s^2 0.40m=45.37 m/s^
8. Refer to the picture in Figure 3 again (picture in Pre-
Lab question #2). Before the apparatus begins to spin, the wires connecting the swings to the top of the structure will be completely vertical. Once the apparatus begins to spin the swings move outward radially, but also upwards vertically. From where does the force causing this vertical acceleration come? The tension in the wires provides the vertical acceleration
9. Refer to the picture in Figure 3. Imagine that the swings are rotating around the center with a constant
speed, and the wire connecting the swings to the center pole is at a 45 degree angle. The angular velocity of the center pole is then doubled. Does this mean that the chairs’ velocities will increase by a factor of two, less than two, or more than two? Explain your reasoning. The velocity will increase by more than a factor of 2.
10. A uniform disc is rotating around a frictionless, vertical axle that passes through its center has
as radius of R=0.300m and a mass of 25.0 kg. The disc rotates according to θ(t) = (2.20 rad/s²) t²
- (5.63 rad/s) t. When the wheel has rotated 0.200 rev, what is the resultant linear acceleration of any point on the disc? w=(4.4t+5.63) rad/s a=4.4 rad/s^2 linear acceleration= aR=4.40.3=1.32 m/s^
