Lab Exam Review Guide

Lab 1 Motion in 1D The speed of an object can be found from the slope of a plot of its position vs. time Lab 2 Motion in 2D One person tosses a ball to another. Define the positive y-direction as vertically upward and the positive x-direction as left to right. Which choice best describes how a plot of the ball's x- component of velocity vs. time will appear? Ignore drag. Answer: The plot will be a horizontal line indicating that the speed in the x-direction is constant over time. Lab 3 Newton’s Laws The well-known constant g is equal to 9.8 m/s2. This means that when you are standing on the surface of the Earth and holding a ball stationary in your hand, the ball has a constant acceleration of 9.8 m/s2 Answer: False Lab 4 Friction Suppose you lay a book on your desk and push horizontally on it with 5 N of force which causes it to slide across the desk at constant velocity. What is the magnitude of the friction force acting on the book? Answer: 5 N Lab 5 Uniform Circular Motion When an object moves in a circular path at constant speed, which of the following best describes the net force acting on the object? Answer: The net force acting on the object acts toward the center of the circle. It is possible for circular motion to be caused by a single force but it can also be caused by the net effect of several forces. Lab 6 Concussions The impulse is a useful metric for determining the likelihood of a concussion occurring in a particular collision. The impulse on an object in a collision can be written two ways: as the change in momentum of the object, or as the product of the force on the object and duration of the collision. Lab 7 Moment of Inertia

If they have the same radius and same mass, how will the moments of inertia of a solid disk and a bicycle wheel compare? Answer: The bicycle wheel will have a greater moment of inertia. Lab 8 Fluid Statics Archimedes’ Principle states that for any solid body immersed in a fluid, the buoyant force acting on it is equal to the weight of the displaced fluid. Lab 9 Simple Harmonic Motion A block attached to a spring undergoes simple harmonic motion. When the block passes through the equilibrium position, which of the following statements about it are true? Answer: The block's acceleration is zero. The elastic potential energy of the spring is zero. The block's kinetic energy is a maximum. Lab 10 Sound When we hear beats produced by combining two sounds, we are hearing an example of Answer: Constructive and destructive interference Lab 11 Heat Capacity and Latent Heat The amount of heat that must be added to a sample to raise its temperature depends on what factors? Answer: The material comprising the sample (e.g. copper, aluminum, glass, etc) The mass of the sample LAB REPORT QUESTIONS: Lab 1 Motion in 1D 1. How can one find the pulling speed using the dots? Briefly describe using the definition of speed. a. The pulling speed can be found by measuring the distance between each of the dots made by the timer. The timer dotted the paper 10 times per second, so the number of dots correlates to the amount of seconds the car traveled. Comparing the length between the dots and the time based on the dots, we can determine the speed by dividing the distance by the time. 2. Compare your two tapes, the one done manually vs. that done by the cart. How can you determine by looking at the spacing of the dots whether the cart was moving at a constant

a. Acceleration is defined as the rate of change of speed over time. So, comparing the value of the velocity versus the time gives us the acceleration at that moment in time. The slope of the graph of velocity versus time gives us the average acceleration. 10. The fit line is a way of incorporating all data into a single best estimate of the acceleration. Let’s compare this to the instantaneous acceleration calculated at each moment. Look in the acceleration column of your Capstone data table, these are the instantaneous acceleration values. How are the instantaneous acceleration values similar or different to the single acceleration value obtained from the best-fit line? a. The instantaneous accelerations vary from 0.034 m/s^2 at 0.150 s, to 0.573 m/s^2 at 0.500 s. The best fit acceleration is 0.515 m/s^2, so the instantaneous accelerations vary by approximately ∓0.4 m/s^2. So ultimately, the acceleration remains approximately constant over the recorded time interval. Lab 2 Motion in 2D 1. What is the time value when the ball in your video is at its maximum height? a. The time value for when the ball is at its maximum height is 0.833 seconds. 2. Is the time value when the ball in your video has zero y-velocity the same as the time value for when it is at maximum height? Would you expect them to be the same? Explain a. Yes, the time value when the ball has 0 y-velocity is the same as the time value for when it is at its maximum height. We do expect them to be the same. When the ball reaches its maximum height, the ball will stop going further up and then begin to start coming down. At this point, it will have a 0 y-velocity. 3. Which general kinematic equation is most like the fit equation for the x position vs time? Which kinematic equation is most like the fit equation for the y position vs time? a. The general kinematic equation that is most like the fit equation for the x position vs time is x(t) = x 0 + v 0x Δt b. The general kinematic equation that is most like the fit equation for the y position vs time is y(t) = y 0 + y 0x Δt + (½) a y Δt² 4. For the y-velocity vs time graph, how do you find the y-acceleration from the fit? What is the acceleration expected to be for such an object in free fall? a. The y-acceleration can be determined by observing the slope of the graph. The acceleration is expected to be the value of gravity, 9.8 m/s² 5. What is the x-acceleration according to your graph and fit of the x-velocity data? What would we expect it to be in this scenario according to the assumption that we have no forces in the x-direction? a. According to the x-velocity graph, the x-acceleration is 0.147 m/s², which is very close to 0, as we would expect. We would expect the acceleration to be close to zero, because the velocity is expected to be constant, since there are no forces in

the x direction. 6. One source of error in this experiment is that we ignored the effect of drag. In fact, the magnitude of the drag force is proportional to the speed of the ball. In light of this fact, is it safe to assume the magnitude of the drag force on the ball is the same at all points on its trajectory? Support your answer with your reasoning. a. No, the effect of the drag would never be uniform for all time points on the ball's trip. The drag would affect both the x and y directions of motion. Therefore, it could alter the position, velocity, and accelerated values that were analyzed and recorded. Lab 3 Newton’s Laws 1. Which of Newton’s Laws describes the hover puck’s motion in region C? a. Region C portrays Newton’s first law, which states that an object in motion will remain in motion, and an object in rest will remain at rest. The hover puck was pushed and continued to remove towards the motion sensor, which represents the first clause of Newton’s first law of motion. 2. Which of Newton’s Laws describes the hover puck’s motion in region A? a. Region A portrays Newton’s, which states that an object in motion will remain in motion, and an object in rest will remain at rest. The hover puck had not been pushed yet so it remained in rest, which represents the first clause of Newton’s first law of motion. 3. What does the value of the slope represent in our graphs of velocity vs. time? a. The value of the slope represents the acceleration of the hover puck. Since the graph is velocity (m/s) vs time (s) the slope will be m/s², which are the units for acceleration. 4. It’s very likely that the actual acceleration of the cart was less than the theoretical value you calculate. Do you think including friction in our theoretical analysis would bring our predicted value more in line with the actual value we measured or would including friction make our theoretical value even farther from what we measured? A brief qualitative answer is all that’s needed here. a. Including friction definitely would bring the 2 values closer. 5. What is the maximum acceleration of the system if you were to hang an infinitely large mass from the string? Hint: plug 1000 kg in for m2 and 1 kg in for m1. (It may seem surprising that large hanging masses don’t translate into large accelerations but remember gravity is providing the downward force!) a. The maximum acceleration of the system would be g, 9.8 m/s², because no matter the mass of the object which is falling, the acceleration of gravity is always the same. 6. As the hanging mass m2 is falling towards the ground, many people would guess that the tension in the string decreases, but this is incorrect! In fact, the tension remains constant over time. Prove that the tension remains constant over time as the mass is falling using

3. Was the value of μS you obtained reasonable? To support your answer, refer to the reported values here for a similar pair of materials noting that the block is wood. a. After referring to the reported values, the wood on wood coefficient of friction range is between 0.4N - 0.6N, our coefficient of friction was 0.3N. tan(16.5)= . 296N. Therefore, the value of μS obtained is a little lower than reasonable because it lies outside the reported range of values by 0.1N. 4. Why is the angle to get the block to start sliding larger than the angle to get the block to stop sliding? Referring to the graph in the figure below may help you. a. The discrepancy in angles is due to the various properties of static and kinetic friction. In general, static friction is stronger and requires more energy to overcome, whereas kinetic friction is weaker and allows things to glide more easily. As a result, a larger angle is required to overcome the higher static friction and move the object, whereas a smaller angle is necessary to counterbalance the lower kinetic friction and stop the object. 5. Observe the figure below showing the force of friction on the y-axis, and the applied force on the x-axis. Indicate on the graph the following points A. Some amount of force is applied but the box remains stationary. Increasing the force slightly does not make the box move. B. No force is applied. The box is stationary. C. Enough force is being applied that the box is moving. D. Some amount of force is applied but the box remains stationary. Increasing the force slightly causes the box to start moving. 6. When trying to stop a car on icy pavement in as short a distance as possible, is it better to slam on the brakes and skid to a stop or apply the brakes more gently and roll to a stop? (Hint: What type of friction is acting between your tires and the roadway if your tires are sliding along the

road? What if your tires are rolling along the road?) Briefly explain. a. Kinetic friction is lower than static friction, therefore when the car is sliding on the road, the grip of the tires is less. Since the kinetic friction is low, it is preferable to apply the brakes slowly and roll the car to a stop. This results in a shorter stopping distance and more control over the car's motion. Lab 5 Uniform Circular Motion 1. What force or forces are acting on the cube as it rotates in a circle at constant speed? Draw a free body diagram for the cube as seen from the side. On your diagram, indicate the direction of the force(s) acting on the cube. 2. How can the hypothesis for this lab be tested experimentally? In a few sentences, describe at least one way to test the hypothesis experimentally by using the rough and smooth surfaces available on the turntable. a. This hypothesis can be tested by placing a cube on a flat surface and seeing how the cube progressively moves outward as the turntable spins. Then, place a cube on the rough/friction surface and notice how the cube comes to a stop as a result of friction. Friction prevents the block from moving, hence the direction of friction must be inward toward the center of the turntable. 3. When the cube falls off the edge of the turntable, which path will it take according to Newton’s 1st Law? Use the figure at right showing the top-down view of the turntable to select a path A through E. Consider the horizontal plane only (not vertical motion). a. It will take path B. 4. Angle in radians is often an unfamiliar unit, so let’s gain more familiarity by looking at our graph of radians vs. time. About how many radians did your cube move through from start to finish of your tracking? Check for internal consistency: How many radians are there in a circle? Is the number you are reporting reasonable? a. According to Graph #1, our cube moved .9 radians from start to finish of the tracking. In the video, we tracked the cube for only a quarter of a rotation so we used the angle θ vs time graph and used the difference in y. There are 2π radians F N F g F cp

6. Figure 2 is a graph of the empirical cumulative distribution function (CDF) vs the linear acceleration for two different data sets (HITS and NFL). The CDF is the fraction of observations that are below the specified value on the x axis. For example, about 90 percent of the sub-concussive impacts in the HITS data (solid black line) occurred at accelerations below 50 g. According to the NFL data set, what was the highest acceleration experienced that did not cause a concussion? What was the minimum acceleration that did cause a concussion? a. According to the NFL graph, the data for sub-concussive is represented by the black dotted line and the concussive data is represented by the red curve. b. According to the graph, the highest acceleration that was experienced that did not cause a concussion was 100g. The minimum acceleration that did cause a concussion was 50g. Lab 7 Moment of Inertia 1. Both barbells have the same total mass, so what is it about Barbell 2 that makes it difficult to move quickly? a. The weights in barbell 1 are placed closer to the center of the rod than compared to barbell 2. This makes it harder to move barbell 2 quickly because the moment of inertia is greater than barbell 1. 2. Recall that for translational motion it takes more force to accelerate a larger mass, which is explained by Newton’s 2nd Law: Net = . This law also holds in rotation form: Net = . In the rotational version of the law, force is replaced by torque , and acceleration is replaced by angular acceleration ; what takes the place of mass? a. Moment of inertia takes the place of mass 3. Another group has hypothesized that adding 100 g extra mass to the center of the barbell where the hand is placed would have no effect on the ability of the person to rotate the barbell back and forth. Do you agree or disagree? Why? a. I agree with this hypothesis because if we were to add 100g of extra mass to the center of the rod, the moment of inertia of the additional mass added would equal 0 because r=0 (distance from center of mass). I=Fr, so r=0 means the moment of inertia would also be 0. 4. If you were to double the mass, but halve the radius of one of the barbells, how would the moment of inertia compare to the original value? a. The moment of inertia would be half of the original value. 5. If you were a tightrope walker, which barbell would you rather be carrying? a. I would rather be carrying the barbell with the weights on either end because the moment of inertia is greater. Also, it is difficult to move quickly with the barbell with weights on either end, which is helpful for a tight walker because you are supposed to move slow and steady in order to keep your balance. 6. How did the angular acceleration change with the new moment of inertia? Was your prediction correct?

a. The angular acceleration decreases with the new moment of inertia. Before the new moment of inertia, the angular acceleration was 5.65 m/s^2 and after the new moment of inertia was added the angular acceleration was 2.65 m/s^2. We predicted that it would decrease so yes, our prediction was correct. Lab 8 Fluid Statics 1. Use the expressions for W(in air) and W(in water) from the figure above to show that the buoyant force is equal to the difference between the weight in air and the weight in water. Show your work. a. W(in air) = mg i. 1.75N (weight in air) b. W(in water) = mg - F B i. 1.75N (weight in air) - 0.21N (buoyant force) = 1.54N (weight in water) c. These expressions show that the weight in air - the buoyant force = the weight in water 2. Show that one can find the density of the unknown object by dividing its weight by the buoyant force W object /FB = P object /P fluid . Show your work. a. P object = P F (W object / F B ) b. 1(1.75N/ 0.21N) = 8.33 g/cm³ (Density) 3. Do you think the metal is gold? To answer this, use a reliable online resource to look up the density of gold and other common metals. If you don’t think your metal is gold, what is it likely to be? Use your data to support your answer. a. No, the metal is not gold. The density of gold is 19.3 g/cm^3 and the calculated density of the metal is 8.33 g/cm³. 4. The total buoyant force acting on both objects is simply the sum of the buoyant force on each individual object: FB, total = FB,wood + FB, slug. Explain why this is true using the fact that the buoyant force is due to the volume of the fluid displaced. a. The buoyant force of the wood, 0.32 N, is much higher than the buoyant force of the brass, 0.21 N. When the two objects are connected to each other, the net force acting on them is going down, and since the buoyant force of the wood is higher than that of the brass, the buoyant force of the two objects combined must equal their sums to account for both. 5. Is the value of the density of the wood you found greater than or less than that of water? When fully submerged, how does the buoyant force acting on the wood compare to the gravitational force acting on the wood? Does this explain why it floats to the surface when not weighted down? a. The value of the density of the wood is less than the density of water. The buoyant force acting on the wood is pushing it to the surface of the water when fully submerged unlike gravity which is pulling the wood down towards the center of gravity. The buoyant force is greater than the gravitational force acting on the wood which explains why it is floating and not being weighed down.

the spring. 6. Our results support the idea that an object’s natural vibration frequency depends on properties intrinsic to the system itself (like k and m) but is independent of the external force that creates the vibration in the first place. Based on this fact, how does the natural frequency of vibration of a 1 m 3 block of lead compare to that for 1 m 3 block of aluminum? Assume the atoms are a lattice connected by springs (shown at right), and that the stiffness of the bonds between neighboring atoms is the same for aluminum as for lead. Hint: look at the relative masses of these two elements on the periodic table. a. I looked up the values of the atomic masses of aluminum: 26.98 amu, and lead: 207.2 amu. Because mass is in the denominator in the equation used to calculate frequency, a larger mass will have a smaller frequency and a larger period. So in this case, lead would have the lower frequency and higher period, and aluminum would have the higher frequency and lower period. Lab 10 Sound 1. How could you prove to a skeptic that the beats are an interference effect that requires both sound sources? Test out your method to convince yourself that the beats require both sound sources and it is not a trick caused by one source alone. a. To prove to a skeptic that beats are an interference effect that requires both sound sources play a pre-recorded sound and it will not have beats. Then, take two sound sources which have different frequencies. When both of the sources are switched on we hear the sound which increases and decreases at equal intervals of time. This is what we call the beats. 2. How many times per second should you hear a beat? Does the beat frequency you hear seem to agree with this calculated value? a. There are five beats in 1 second. The beat frequency we hear is 5 beats per second because 90 Hz - 85 Hz = 5 Hz. 3. In the experiment diagrammed above, how far does the pulse travel between the time it leaves the fingers and returns to the microphone? Answer in terms of the distance L. a. L = 0.79m b. ΔT = 0.0046 c. V sound = 2(0.79)/(0.0046) = 343 m/s 4. How close is your value to the accepted value for the speed of sound in air? Refer to the percent difference. a. ((343 - 340)/ 340) x 100 = .8% b. Our percent difference is very low which proves that our value is very close to the accepted value. 5. Why do we need to use a quick sound like a finger snap for this experiment? For example, bats make a very short duration clicking sound for their ultrasonic sensing. Why does this work better than perhaps a slower sound like a squeak? To answer, refer to the duration of the pulse peak and echo peak shown in your graph.

a. We use a quick sound like a finger snap for this experiment because high frequencies tend to produce stronger echoes than low frequencies. Bats use a short duration clicking sound because a sound of that nature and frequency is sharper and more accurate than a slower sound with a low frequency. Like shown in the graph, the click sounds give a higher frequency which can be heard more accurately than a low frequency sound. 6. How would your calculation and for this experiment change if the pulse and echo traveled through human tissue instead of air? To answer this, assume the distance stays the same and look up how the speed of sound in tissue such as bone or muscle compares to that in air. a. The speed of sound in human tissue is higher than in air, with the speed being approximately 343 m/s in air, and between 1500-1600 m/s in human tissue. Because of this, the time it would take for a sound pulse to travel through tissue would be shorter than if it was traveling through air. Additionally, the acoustic properties of tissue, such as absorption and attenuation, would need to be considered for accurate interpretation of results. Lab 11 Heat Capacity and Latent Heat 1. What does the resulting slope imply about the relation between the heat lost by the warm water and the heat gained by the cold water? a. Considering the values should be equal in theory due to the first law of thermodynamics, the resulting slope suggests that there is a direct relationship between the heat gained and the heat lost. 2. Assuming that the system is isolated from the surroundings, derive an expression for the latent heat of fusion of water in terms of measurable quantities. To do this start with the fact that all the heat that leaves the hot water will go into melting the ice and further heating the resulting water: leaving hot water * = melting ice + into cold water. To * * * * * * * * * * * * * * * simplify our analysis, assume that the mass of the melted ice is the same as the mass of the cold water (i.e., once enough heat flows into the ice to melt it, some heat will also go into warming the 0 °C water to its final temperature of about 5 °C). Let: M HW = Mass of hot water (0.2 Kg) M i = Mass of ice ( VmL x 10 △ -6 ) = 190 mL x 10 -6 x 1000 = .19Kg T HW = Temperature of hot water = 80.5 ºC T CW = Temperature of cold water (ice) = 9.3 ºC C W = Specific heat capacity of water (4181 J/KgK) L = Latent heat of fusion of water Expression: L = -M HW C W T △ HW - M i C W T △ CW L =[-(0.2Kg)(4181 J/KgK)(9.3 - 80.5 ) - (.19Kg)(4181 J/KgK)(9.3 - 0 )] / .19Kg ℃ ℃ ℃ ℃ L = 274,471.6474 J/Kg