Home / ap-physics-1-frq-2017-dev

1. Question: (7 points, suggested time 13 minutes)

In the three circuits shown above, the batteries are all identical, and the lightbulbs are all identical. In circuit 1 a single lightbulb is connected to the battery. In circuits 2 and 3, two lightbulbs are connected to the battery in different ways, as shown. The lightbulbs are labeled A–E.
(a) Rank the magnitudes of the potential differences across lightbulbs A, B, C, D, and E from largest to smallest.
If any lightbulbs have the same potential difference across them, state that explicitly.
Ranking:
Briefly explain how you determined your ranking.
(b) The batteries all start with an identical amount of usable energy and are all connected to the lightbulbs in the circuits at the same time.
In which circuit will the battery run out of usable energy first?
____ Circuit 1 ____ Circuit 2 ____ Circuit 3
In which circuit will the battery run out of usable energy last?
____ Circuit 1 ____ Circuit 2 ____ Circuit 3
In a clear, coherent paragraph-length response that may also contain equations and drawings, explain your reasoning.

Answer/Explanation

Ans:

(a) 

Δ V : ( A = D = E) > (B = C)

A, D, B E all have the potential difference equal to the voltage of the battery because all have direct pathways to the battery. Bulbs B & C share access to the same battery; thus, the potential difference across them is just half that of Bulbs A, D, B E. 

(b)

√  Circuit 3

√  Circuit 2

As a true  parallel circuit, the battery in circuit 3 will run out of power the soonest because it is feeding enough power to power 2 bulbs at full strength. Mathematically speaking, since \(\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + ……,\) a parallel circuit has 1/n the resistance of a single bulb circuit. Therefore, the current, I, is n times that of a single bulb circuit. In this case, n = 2, so the battery has twice the current running through it that the battery in circuit A would have. This increased current means that the battery runs out of energy faster. On the other hand, circuit 2 retains energy the longest because its resistance is higher than a single bulb circuit. Rseries = R1 + R2 + …., so Rseries = nR. Since \(I = \frac{V}{R}\), greater resistance values for equal voltage batteries produces less current. This lower current carries energy from the battery at a slower pace; thus, the battery, retains energy is keeps the lighbulbs lit longer than a single-bulb circuit.

2. Question: (12 points, suggested time 25 minutes)

A student wants to determine the coefficient of static friction between a long, flat wood board and a small wood block.
(a) Describe an experiment for determining the coefficient of static friction between the wood board and the wood block. Assume equipment usually found in a school physics laboratory is available.
i. Draw a diagram of the experimental setup of the board and block. In your diagram, indicate each quantity that would be measured and draw or state what equipment would be used to measure each quantity.
ii. Describe the overall procedure to be used, including any steps necessary to reduce experimental uncertainty. Give enough detail so that another student could replicate the experiment.
(b) Derive an equation for the coefficient of static friction in terms of quantities measured in the procedure from part (a).

A physics class consisting of six lab groups wants to test the hypothesis that the coefficient of static friction between the board and the block equals the coefficient of kinetic friction between the board and the block. Each group determines the coefficients of kinetic and static friction between the board and the block. The groups’ results are shown below, with the class averages indicated in the bottom row.

Lab
Group
Number
Coefficient
of Kinetic
Friction
Coefficient
of Static
Friction
10.450.54
20.460.52
30.420.56
40.430.55
50.740.23
60.440.54
Average0.490.49

(c) Based on these data, what conclusion should the students make about the hypothesis that the coefficients of static and kinetic friction are equal?
____ The static and kinetic coefficients are equal.
____ The static and kinetic coefficients are not equal.
Briefly justify your reasoning.

(d) A metal disk is glued to the top of the wood block. The mass of the block-disk system is twice the mass of the original block. Does the coefficient of static friction between the bottom of the block and the board increase, decrease, or remain the same when the disk is added to the block?
___ _ Increase ____ Decrease ____ Remain the same
Briefly state your reasoning.

Answer/Explanation

Ans:

(a)

i.

The force reavired to set the block in motion and the mass of the block would be measured.

ii. The student would first connect a force sensor to the logger pro interfacwe, next the scale would be used to find the mass of the small block. This block would the be placed on the Board and the student would connect a force sensor, Then using the force sensor a small increasing amount of force would be applied until the block is set in motion. This force would be recorded. 

(b)       \(\sum = ma\)                     Ff ≤ NNf

                                                                                                                         Nf

                                                                    Nf = m9

                                                                   F5   = F                                                    Nfu  = F

                                                                                                                                    M9u  = F

                                                                                           \(\frac{M9}{F} = u\)

(c) 
____ The static and kinetic coefficients are not equal.

The data points in group 5 are an outliner and should be retested If these data points are not used the averages are Fk = .44 and Fs = .54 

(d)

____ Remain the same

The frictional force can be defined as μsN = Fs     When the disk is added the Normal force increases but the coefficient of friction stays the same. coefficients of friction deal with the types of surfaces rubbing together not the  wieght.

3. Question: (12 points, suggested time 25 minutes)

The left end of a rod of length d and rotational inertia I is attached to a frictionless horizontal surface by a frictionless pivot, as shown above. Point C marks the center (midpoint) of the rod. The rod is initially motionless but is free to rotate around the pivot. A student will slide a disk of mass mdisk toward the rod with velocity v0 perpendicular to the rod, and the disk will stick to the rod a distance x from the pivot. The student wants the rod-disk system to end up with as much angular speed as possible.
(a) Suppose the rod is much more massive than the disk. To give the rod as much angular speed as possible, should the student make the disk hit the rod to the left of point C, at point C, or to the right of point C ?
____ To the left of C ____ At C ____ To the right of C
Briefly explain your reasoning without manipulating equations.

(b) On the Internet, a student finds the following equation for the postcollision angular speed w of the rod in this situation: ω = \(\frac{m_{disk}xv_{0}}{I}\) . Regardless of whether this equation for angular speed is correct, does it agree with your qualitative reasoning in part (a) ? In other words, does this equation for ω have the expected dependence as reasoned in part (a) ?
____ Yes ____ No
Briefly explain your reasoning without deriving an equation for ω.

(c) Another student deriving an equation for the postcollision angular speed w of the rod makes a mistake and comes up with w =  \(\frac{Ixv_{0}}{m_{disk}d^{4}}\)  . Without deriving the correct equation, how can you tell that this equation is not plausible—in other words, that it does not make physical sense? Briefly explain your reasoning.

For parts (d) and (e), do NOT assume that the rod is much more massive than the disk.
(d) Immediately before colliding with the rod, the disk’s rotational inertia about the pivot is mdisk x2 and its angular momentum with respect to the pivot is mdisk v0x . Derive an equation for the postcollision angular speed ω of the rod. Express your answer in terms of d, mdisk , I, x, v0 , and physical constants, as appropriate.

(e) Consider the collision for which your equation in part (d) was derived, except now suppose the disk bounces backward off the rod instead of sticking to the rod. Is the postcollision angular speed of the rod when the disk bounces off it greater than, less than, or equal to the postcollision angular speed of the rod when the disk sticks to it?
____ Greater than ____ Less than ____ Equal to
Briefly explain your reasoning.

Answer/Explanation

Ans: 

(a)

____ To the right of C

Due to the rod being much more Massine than the disk, the collision can be approximated as an impulse defined to the rod. To maximize angular speed and thus angular momentum, the largest possible angular impulse should be delivered, so the lever ann should be maximized which happens  when the collision point is to the right of C.

(b)

Yes

In this equation, increasing x, the lever arm, means increasing ω. This agrees with part (a) because the lever arm is longer when the disk hits to the right of C.

(c)

In this equation, increasing mdisk would decrease ω, all the remaining constant.

This is impossible because a longer disk delivers a longer impulse to the rod due to it having more momentum to begun with, thus measuring ω.

(d)

by conservation at angular momentum:

mdisk v0x = (I + mdisk x2) ω

ω = \(\frac{m_{disk}v_{0}x}{I + m_{disk}x^{2}}\)

(e)

√  Greater than

By conservation at angular momentum, total angular momentum is constant before and after collision. When the disk sticks, the disk’s final angular momentum is positive while it’s negative if it faunas back. Thus for the sum of the disk and rod’s angular momentum to be equal, the angular momentum and thus the angular speed of the rod must be greater if the disk faunas back.

Question: (7 points, suggested time 13 minutes)

A physics class is asked to design a low-friction slide that will launch a block horizontally from the top of a lab table. Teams 1 and 2 assemble the slides shown above and use identical blocks 1 and 2, respectively. Both slides start at the same height d above the tabletop. However, team 2’s table is lower than team 1’s table. To compensate for the lower table, team 2 constructs the right end of the slide to rise above the tabletop so that the block leaves the slide horizontally at the same height h above the floor as does team 1’s block (see figure above).
(a) Both blocks are released from rest at the top of their respective slides. Do block 1 and block 2 land the same distance from their respective tables?
____ Yes ____ No
Justify your answer.

In another experiment, teams 1 and 2 use tables and low-friction slides with the same height. However, the two slides have different shapes, as shown below.

(b) Both blocks are released from rest at the top of their respective slides at the same time.
i. Which block, if either, lands farther from its respective table?
____ Block 1 ____ Block 2 ____ The two blocks land the same distance from their respective tables.
Briefly explain your reasoning without manipulating equations.
ii. Which block, if either, hits the floor first?
____ Block 1 ____ Block 2 ____ The two blocks hit the floor at the same time.
Briefly explain your reasoning without manipulating equations.

Answer/Explanation

Ans:

 √   No

For team I, all of the block’s potential energy, gets converted into kinetic energy. This is also true for team 2, However with team 2, the block has less initial potential energy than team t, b/c it travels a shorter not vertical distance to therefore it has less kinetic energy at height h. This results in less velocity at height h for team so a shorter distance travelled b/c both blocks are in the air for the same amount of time.

(b)

i.   √    The two blocks land the same distance from their respective tables.

Even though the slides have different heights, the changes in height are the same so they have the same potential energy at the start, and kinetic energy at the and, since they launch from the same height h, they are in the air for the same amount of time and go the same distance.

ii.  √    Block 1

Block 1 travels a shorter horizontal distance on the table before reaching its maximum speed than block 2. This means it reaches the end of the table before block 2 and hits the ground first because they are in the air for the same amount of time.

5. Question: (7 points, suggested time 13 minutes)

Two wave pulses are traveling in opposite directions on a string. The shape of the string at t = 0 is shown above. Each pulse is moving with a speed of one unit per second in the direction indicated.
(a) Between time t = 0 and t = 5 seconds, the entire left-hand pulse approaches and moves beyond point P on the string. On the coordinate axes below, plot the velocity of the piece of string located at point P as a function of time between t = 0 and t = 5 seconds.

(b) At t = 5 s, the pulses completely overlap. On the grid provided below, sketch the shape of the entire string at t = 5 s.
Note: Do any scratch (practice) work on the grids on the following page. You will only be graded for the sketch made on the grid on this page.

The grids below are provided for scratch work only. Sketches made below will NOT be graded.

Answer/Explanation

Ans:

(a)

(b)

Scroll to Top