1. Question: (7 points, suggested time 13 minutes)
Two blocks are connected by a string of negligible mass that passes over massless pulleys that turn with negligible friction, as shown in the figure above. The mass m2 of block 2 is greater than the mass m1 of block 1. The blocks are released from rest. (a) The dots below represent the two blocks. Draw free-body diagrams showing and labeling the forces (not components) exerted on each block. Draw the relative lengths of all vectors to reflect the relative magnitudes of all the forces.
(b) Derive the magnitude of the acceleration of block 2. Express your answer in terms of m1 , m2 , and g.
Block 3 of mass m3 is added to the system, as shown below. There is no friction between block 3 and the table.
(c) Indicate whether the magnitude of the acceleration of block 2 is now larger, smaller, or the same as in the original two-block system. Explain how you arrived at your answer.
Answer/Explanation
Ans:
(a)
(b) According to Newton’s Second Law
For block 2: m2g – FT’ = m2.a
For block 1: Ft – m1g = m1.a.
According to Newton’s 3rd Law = |FT| = |FT’|
∴ m2g – m1g = m2a + m1a.
∴\(a^{2}=\frac{\left ( m_{2}-m_{1} \right )g}{m_{1}+m_{2}}\)
(c) a2 = a system = \(\frac{\left ( m_{2}-m_{1} \right )g}{m_{1}+m_{2}+m_{3}}< \frac{\left ( m_{2}-m_{1} \right )g}{m_{1}+m_{2}}=a_{2} original\)
∴ the acceleration now is smaller than that before due to the increase in the total mass of the system.
2. Question: (12 points, suggested time 25 minutes)
Some students want to know what gets used up in an incandescent lightbulb when it is in series with a resistor:
current, energy, or both. They come up with the following two questions.
(1) In one second, do fewer electrons leave the bulb than enter the bulb?
(2) Does the electric potential energy of electrons change while inside the bulb?
The students have an adjustable power source, insulated wire, lightbulbs, resistors, switches, voltmeters, ammeters, and other standard lab equipment. Assume that the power supply and voltmeters are marked in 0.1 V increments and the ammeters are marked in 0.01 A increments.
(a) Describe an experimental procedure that could be used to answer questions (1) and (2) above. In your description, state the measurements you would make and how you would use the equipment to make them. Include a neat, labeled diagram of your setup.
(b)
i. Explain how data from the experiment you described can be used to answer question (1) above.
ii. Explain how data from the experiment you described can be used to answer question (2) above.
A lightbulb is nonohmic if its resistance changes as a function of current. Your setup from part (a) is to be used or modified to determine whether the lightbulb is nonohmic.
(c)
i. How, if at all, does the setup need to be modified?
ii. What additional data, if any, would need to be collected?
(d) How would you analyze the data to determine whether the bulb is nonohmic? Include a discussion of how the uncertainties in the voltmeters and ammeters would affect your argument for concluding whether the resistor is nonohmic.
Answer/Explanation
Ans:
To answer questions 1 and 2, the students would use the voltameters and ammeters to measure the current and voltage of the circuit both before and after the lightbulb for 3 trials of 1 second. The student would record this information for calculations.
(b) i. The data from the experiment above can be used to answer question (i) by comparing the current in the circuit at the ammeter before and the ammeter after the light bulb. If a discrepancy between the two measurements in earn trial is found, fever electrons may have left the light bulb.
ii. The data from the experiment above can be used to answer question (2) by comparison the voltage measured by the voltmeters placed before and after the light bulb. If a discrepancy between the values from each voltmeter in a given trial is found, the electrical potential energy of the electrons may have changed which inside the bulb.
(c) i. The voltage of the adjustable power source must be varied for a given number of trials to create varying currents because current = \(\frac{voltage}{resistance}= I = \frac{V}{R}\)
ii. Additional data reavired would be a number of trials at given voltages from the adjustable power source. This would allow different data sets of voltage and current to be used to determinse the resistance of the bulb.
(d) To determine whether the bulb is nonohmic, I would determine the resistances of the lightbulb in each trial according to the formula \(I = \frac{\Delta V}{R}\) assuming that in series no current is lost in the lightbulb (or circuit for that matter). If the resistance of the lightbulb is found to vary, it may be assumed to be nonohmic assuming the uncertainties in the voltmeters and ammeters did not affect the resorts. Because the uncertainties exist, hoever, If the change in voltage is less than or eacal to. IV, The conclusion as to whether or not the lightbulb is nonohmic cannot be verified using the given ealipment.
Question: (12 points, suggested time 25 minutes)
A block is initially at position x = 0 and in contact with an uncompressed spring of negligible mass. The block is pushed back along a frictionless surface from position x = 0 to x = -D , as shown above, compressing the spring by an amount Δx = D . The block is then released. At x = 0 the block enters a rough part of the track and eventually comes to rest at position x = 3D . The coefficient of kinetic friction between the block and the rough track is μ .
(a) On the axes below, sketch and label graphs of the following two quantities as a function of the position of the block between x = -D and x = 3D . You do not need to calculate values for the vertical axis, but the same vertical scale should be used for both quantities.
i. The kinetic energy K of the block
ii. The potential energy U of the block-spring system
The spring is now compressed twice as much, to Δx = 2D . A student is asked to predict whether the final position of the block will be twice as far at x = 6D . The student reasons that since the spring will be compressed twice as much as before, the block will have more energy when it leaves the spring, so it will slide farther along the track before stopping at position x = 6D .
(b)
i. Which aspects of the student’s reasoning, if any, are correct? Explain how you arrived at your answer.
ii. Which aspects of the student’s reasoning, if any, are incorrect? Explain how you arrived at your answer.
(c) Use quantitative reasoning, including equations as needed, to develop an expression for the new final position of the block. Express your answer in terms of D.
(d) Explain how any correct aspects of the student’s reasoning identified in part (b) are expressed by your mathematical relationships in part (c). Explain how your relationships in part (c) correct any incorrect aspects of the student’s reasoning identified in part (b). Refer to the relationships you wrote in part (c), not just the final answer you obtained by manipulating those relationships.
Answer/Explanation
Ans:
(a)
(b)
i. The student is correct to say that any eassing nor spring will result more a longer kinetic energy value for the block when it leave the spring. Springs store more energy when compressed further and that greater amount of energy will be transferred to the block.
ii. The student B me or not to suppose that the block will go twice as far. The amount of energy shared in a spring B proportioned to the amount of is compressed squared, Because it is compressed there as far, it will have firmes the potential energy which will be transferred to the block, which will also have foretimes as much energy and will go four times as far.
(c) d= final position
EK = EU EU= Ff . d = mgμ. d \(d_{1}=\frac{k}{mg\mu }.D^{2} = 3D\)
\(E_{k}=\frac{1}{2}kx^{2}\) \(\sum Fy = F_{N}-mg=0\)
\(F_{f}=mg\mu \) \(d_{2}=\frac{k(2D)^{2}}{mg\mu }=4\left ( \frac{KD^{2}}{mg\mu } \right )\)
= 4 (3D)
\(\frac{1}{2}kx^{2}= mg\mu .d\) \(D = \frac{kx^{2}}{mg\mu }\) = 12 D
Four times the energy will be stored in the spring, four times the energy will go into the block, and fraction will take 4 times the distance to disperse the energy.
(d) More potential energy will be stored on the spring when it is compressed further as the student stated and as illustrated by the relationship \(E_{u}=\frac{1}{2}k\Delta x^{2}.\) This relationship also shows that when the spring is compressed twice as far (Δx B doubled) four times no energy is stored in the spring. The force of fraction is the same in both instances, so the same amount of energy is dissipated by friction per unit of length translated by E=Ff.d) so 4 times the energy will take 4 times the distance to be dissipated, not twice the distance as the student who assumed that potential energy is proportional to ΔX supposed.
4. Question: (7 points, suggested time 13 minutes)
Two identical spheres are released from a device at time t = 0 from the same height H, as shown above. Sphere A has no initial velocity and falls straight down. Sphere B is given an initial horizontal velocity of magnitude v0 and travels a horizontal distance D before it reaches the ground. The spheres reach the ground at the same time tf , even though sphere B has more distance to cover before landing. Air resistance is negligible.
(a) The dots below represent spheres A and B. Draw a free-body diagram showing and labeling the forces (not components) exerted on each sphere at time \(\frac{t_{f}}{2}.\)
(b) On the axes below, sketch and label a graph of the horizontal component of the velocity of sphere A and of sphere B as a function of time.
(c) In a clear, coherent, paragraph-length response, explain why the spheres reach the ground at the same time even though they travel different distances. Include references to your answers to parts (a) and (b).
Answer/Explanation
Ans:
(a)
(b)
(c)
Reaching the ground from the table concerns only vertical distance. Thus, the only component of velocity that impacts the time to reach the ground is the y component of velocity. The X-component makes a ball travel farther horizontally during the same time, but it won’t make it fall at a faster speed toward the ground. This can also be shown using the following Kinematic equation:
\(y = y_{0}+y_{g0}t.\frac{1}{2}a_{g}t^{2}\)
y final and y0 are the same for both spheres because they start and end at the same heights. The initial y component of velocity is zero for both spheres. Acceleration is also equal, for gravely is the only think acting on the spheres in the vertical direction. Thus, the time if takes for the spheres to reach the ground must be equal.
5. Question: (7 points, suggested time 13 minutes)
The figure above shows a string with one end attached to an oscillator and the other end attached to a block. The string passes over a massless pulley that turns with negligible friction. Four such strings, A, B, C, and D, are set up side by side, as shown in the diagram below. Each oscillator is adjusted to vibrate the string at its fundamental frequency f. The distance between each oscillator and pulley L is the same, and the mass M of each block is the same. However, the fundamental frequency of each string is different.
The equation for the velocity v of a wave on a string is \(v = \sqrt{\frac{F_{T}}{m/L}},\) where FT is the tension of the string and m/L is the mass per unit length (linear mass density) of the string.
(a) What is different about the four strings shown above that would result in their having different fundamental frequencies? Explain how you arrived at your answer.
(b) A student graphs frequency as a function of the inverse of the linear mass density. Will the graph be linear? Explain how you arrived at your answer.
(c) The frequency of the oscillator connected to string D is changed so that the string vibrates in its second harmonic. On the side view of string D below, mark and label the points on the string that have the greatest average vertical speed.
Answer/Explanation
Ans:
\(\frac{\lambda }{2} = L\)
\(\lambda =2L\)
\(\lambda f =v\)
\(f = \frac{v}{\lambda }\)
\(f = \frac{\sqrt{f_{T }}}{\sqrt{m/L}}.\frac{1}{2L}\)
By manipulating the equation given and fλ = v, it can be found that \(f = \frac{\sqrt{f_{T }}}{\sqrt{m/L}}.\frac{1}{2L}\). Since FT and L are the same for all four strings, this means that the frequency changed based on the linear mass density of each storing. So all four string have different linear mass densities.
(b) As was found in part a), \(f\alpha \frac{1}{\sqrt{m/L}}\) , so \(f^{2}\alpha \frac{1}{\sqrt{m/L}}\) since the tension in the string out the length of the string are constant. Therefore, to graph f as a function of \( \frac{1}{\sqrt{m/L}}\) (as the student has done) would not yield a linear graph. Since \(f\alpha \sqrt{m/L}\), the graph would appear to be a sideways parabola, concave down for all values of (m/L) – so definitely not linear . \(f\alpha \frac{1}{m/L}\)