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

A spacecraft of mass m is in a clockwise circular orbit of radius R around Earth, as shown in the figure above. The mass of Earth is ME .
(a) In the figure below, draw and label the forces (not components) that act on the spacecraft. Each force must be represented by a distinct arrow starting on, and pointing away from, the spacecraft.

(b)
i. Derive an equation for the orbital period T of the spacecraft in terms of m, ME , R, and physical constants, as appropriate. If you need to draw anything other than what you have shown in part (a) to assist in your solution, use the space below. Do NOT add anything to the figure in part (a).
ii. A second spacecraft of mass 2m is placed in a circular orbit with the same radius R. Is the orbital period of the second spacecraft greater than, less than, or equal to the orbital period of the first spacecraft?
____ Greater than ____ Less than ____ Equal to
Briefly explain your reasoning.

(c) The first spacecraft is moved into a new circular orbit that has a radius greater than R, as shown in the figure below.

Is the speed of the spacecraft in the new orbit greater than, less than, or equal to the original speed?
____ Greater than ____ Less than ____ Equal to
Briefly explain your reasoning.

Answer/Explanation

Ans:

(a)

(b) (i)

\(\sum F = ma\)

Fg = ma

\(G\frac{mM_{E}}{r^{2}} = \frac{mv^{2}}{r}\)

\(\frac{GM_{E}}{r} = \left ( \frac{2\pi .r}{T} \right )^{2}\)

\(\frac{GM_{E}}{r} = \frac{4\pi ^{2}.r^{2}}{T^{2}}\)

\(T^{2} = \frac{4\pi ^{2}.r^{2}}{\left ( \frac{GM_{E}}{r} \right )}\)

\(\sqrt{T^{2}} = \sqrt{\frac{4\pi ^{2}.r^{3}}{GM_{E}}}\)

\(T = \sqrt{\frac{4\pi ^{2}.r^{3}}{GM_{E}}}\)

(ii) 

× equal to

The orbital period of the second spacecraft is the same as the first because in the equation \(G\frac{mM_{E}}{r^{2}} = \frac{mv^{2}}{r}\) the mass of the spacecraft cancels on both sides of the equation, which means it does not affect the period.  \(\binom{Not in euation }{T = \sqrt{\frac{4\pi ^{2}.r^{3}}{GM_{E}}}}\)

(c) 

× less than 

The speed of the spacecraft in the second orbit will be slower because in the equation \(G\frac{mM_{E}}{r^{2}} = \frac{mv^{2}}{r}\) , the m’s and r’s cancel and you get the equation \(\frac{GM_{E}}{r} =v ^{2}\) . In the second orbit, r increases and so the number or the left side of the equation is smaller, This means that V2 equals a smaller number than before, so therefore V decreases. (Speed decreases).

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

A group of students prepare a large batch of conductive dough (a soft substance that can conduct electricity) and then mold the dough into several cylinders with various cross-sectional areas A and lengths l. Each student applies a potential difference ΔV across the ends of a dough cylinder and determines the resistance R of the cylinder. The results of their experiments are shown in the table below.

(a) The students want to determine the resistivity of the dough cylinders.
i. Indicate below which quantities could be graphed to determine a value for the resistivity of the dough cylinders. You may use the remaining columns in the table above, as needed, to record any quantities (including units) that are not already in the table.
Vertical Axis: ________________________ Horizontal Axis: ________________________

ii. On the grid below, plot the appropriate quantities to determine the resistivity of the dough cylinders. Clearly scale and label all axes, including units as appropriate.

iii. Use the above graph to estimate a value for the resistivity of the dough cylinders.

(b) Another group of students perform the experiment described in part (a) but shape the dough into long rectangular shapes instead of cylinders. Will this change affect the value of the resistivity determined by the second group of students?
____ Yes ____ No
Briefly justify your reasoning.
(c) Describe an experimental procedure to determine whether or not the resistivity of the dough cylinders depends on the temperature of the dough. Give enough detail so that another student could replicate the experiment. As needed, include a diagram of the experimental setup. Assume equipment usually found in a school physics laboratory is available.

Answer/Explanation

Ans:

(a)  (i)

Vertical Axis:  __Resistance_______,    Horizontal Axis:             Length (L) / Area (A)__

(ii)

(iii) 

slope = resistivity             \(\frac{n3c}{tan} = \frac{105-23.c}{263.2 – 61.2} = \frac{81.4}{202}\)

                                                  = 403 Ω m

(b)

   X  No

Resistivity is a property of the material, not the slope it is in. Since the dough does not change, the resisting limit either.

(c) 

A circuit like the one pictorial below will be set up, with the voltage serve being content, perhaps a ev Battery. An computer  should be wired in series and a volt meter should be wired in parallel with the dough. This setup will allow the student to see the Resistance of the dough because voltage will be known (via battery  limit of voltmeter reading) and current will be known (via computer reading), and the relationship v = IR can be used. While maintaining constant  length and cross section area. the dough should be heated or cooled, or both, which can be done in a variety of ways, depending on available resources. An re meter both could cool the dough, or a hot plate could hot it up. The student should get measurements of current and voltage several times after either heating the dough up or cooling it down, as it travers a range of Temp to item to room temp while a the meter could help determine mac pearly resisting at each temp. it is not needed both it is same the dough is changing temp. The relationship Since A and l ve could, any change in R, fenot in the method desubi eaten, with signity a change on resistivity   P(resisting) = \(\frac{RA}{l}\) is used now.

 

Question: (12 points, suggested time 25 minutes)

The disk shown above spins about the axle at its center. A student’s experiments reveal that, while the disk is spinning, friction between the axle and the disk exerts a constant torque on the disk.
(a) At time t = 0 the disk has an initial counterclockwise (positive) angular velocity ω0. The disk later comes to rest at time t = t1.
i. On the grid at left below, sketch a graph that could represent the disk’s angular velocity as a function of time t from t = 0 until the disk comes to rest at time t = t1.
ii. On the grid at right below, sketch the disk’s angular acceleration as a function of time t from t = 0 until the disk comes to rest at time t = t1.

(b) The magnitude of the frictional torque exerted on the disk is τ 0. Derive an equation for the rotational inertia I of the disk in terms of τ 0, ω0 , t1, and physical constants, as appropriate.

(c) In another experiment, the disk again has an initial positive angular velocity ω0 at time t = 0. At time t = \(\frac{1}{2}t_{1}\) , the student starts dripping oil on the contact surface between the axle and the disk to reduce the friction. As time passes, more and more oil reaches that contact surface, reducing the friction even further.
i. On the grid at left below, sketch a graph that could represent the disk’s angular velocity as a function of time from t = 0 to t = 1t , which is the time at which the disk came to rest in part (a).
ii. On the grid at right below, sketch the disk’s angular acceleration as a function of time from t = 0 to t = t1.

(d) The student is trying to mathematically model the magnitude τ of the torque exerted by the axle on the disk when the oil is present at times t > \(\frac{1}{2}t_{1}\) . The student writes down the following two equations, each of which includes a positive constant (C1 or C2 ) with appropriate units.

(1)  τ =  \(C_{1}\left ( t – \frac{1}{2}t_{1} \right )(for t > \frac{1}{2}t_{1})\)

(2)  τ =  \(\frac{C_{2}}{\left ( t + \frac{1}{2}t_{1} \right )}\)                       \((for t > \frac{1}{2}t_{1})\)

Which equation better mathematically models this experiment?
____ Equation (1) ____ Equation (2)
Briefly explain why the equation you selected is plausible and why the other equation is not plausible.

Answer/Explanation

Ans:

(a) 

(ii)

(b)

L  = Ico

0L = τ. Δt

L – ΔL = 0

Iw0 = τ0.t1

I  = τ0 t1 / w0

(c)

(d)

__√_ Equation (2)

According to equation, if the time continues to increase, torque will continue to increase, but torque is supposed to decrease if we are oiling the system. Equation 2 shows that as t increases, the torque gets closer to 0 which is what we want. Equation 1 also impress that torque can go to ∞ as time goes to ∞  which isn’t plansible in the problem’s context. 

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

A transverse wave travels to the right along a string.
(a) Two dots have been painted on the string. In the diagrams below, those dots are labeled P and Q.
i. The figure below shows the string at an instant in time. At the instant shown, dot P has maximum displacement and dot Q has zero displacement from equilibrium. At each of the dots P and Q, draw an arrow indicating the direction of the instantaneous velocity of that dot. If either dot has zero velocity, write “v = 0” next to the dot.

ii. The figure below shows the string at the same in stant as shown in part (a)i. At each of the dots P and Q, draw an arrow indicating the direction of the instantaneous acceleration of that dot. If either dot has zero acceleration, write “a = 0” next to the dot.

The figure below represents the string at time t = 0, the same instant as shown in part (a) when dot P is at its maximum displacement from equilibrium. For simplicity, dot Q is not shown.

(b)
i. On the grid below, draw the string at a later time t = T/4, where T is the period of the wave.
Note: Do any scratch (practice) work on the grid at the bottom of the page. Only the sketch made on the grid immediately below will be graded.

ii. On your drawing above, draw a dot to indicate the position of dot P on the string at time t = T/4 and clearly label the dot with the letter P.
(c) Now consider the wave at time t = T. Determine the distance traveled (not the displacement) by dot P between times t = 0 and t = T.

Answer/Explanation

Ans:

(a)

(i)

(ii)

(b) (i)

(c) 

–  8 → 8 → -8 

16 + 16  32 cm

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

Block P of mass m is on a horizontal, frictionless surface and is attached to a spring with spring constant k. The block is oscillating with period TP and amplitude AP about the spring’s equilibrium position x0. A second block Q of mass 2m is then dropped from rest and lands on block P at the instant it passes through the equilibrium position, as shown above. Block Q immediately sticks to the top of block P, and the two-block system oscillates with period TPQ and amplitude APQ.

(a) Determine the numerical value of the ratio TPQ /T P .

(b) The figure is reproduced above. How does the amplitude of oscillation APQ of the two-block system compare with the original amplitude AP of block P alone?
____ APQ < AP     ____APQ  = AP ____APQ > AP
In a clear, coherent paragraph-length response that may also contain diagrams and/or equations, explain your reasoning.

Answer/Explanation

Ans:

(a)

\(T_{p} = 2\pi \sqrt{\frac{m}{k}}\)

\(T_{PQ} = 2\pi \sqrt{\frac{3m}{k}}\)                         \(\frac{T_{PQ}}{T_{P}} = \frac{2\pi \sqrt{\frac{3m}{k}}}{2\pi \sqrt{\frac{m}{k}}} = \frac{\sqrt{\frac{3m}{k}}}{\sqrt{\frac{m}{k}}}\)

\(\frac{\sqrt{\frac{3m}{k}}\sqrt{\frac{m}{k}}}{\frac{m}{k}} = \frac{\frac{\sqrt{3m}}{k}}{\frac{m}{k}} = \sqrt{3}\)

(b)

_×_ APQ < AP 

Momentum is conserved during the collision, mv0 = 3m,     v ⇒      v0 =  3vf           v    = \(\frac{v_{0}}{3}\)

Final velocity is \(\frac{1}{3}\)  of the initial velocity.

KEPQ =  \(\frac{1}{2}3m \left ( \frac{v_{0}}{3} \right )^{2} = \frac{1}{6} m{v_{0}}^{2} \)                 vs,         \(\frac{1}{2}m{v_{0}}^{2} = KE_{Q}\)

KE at the equilibrium of blocks Q and P is less than with only Block P. 

\(V_{s} = \frac{1}{2}kx^{2}\)                       \(\frac{1}{6}m{v_{0}}^{2} = \frac{1}{2}kx^{2}\)              vs.     \(\frac{1}{2}m{v_{0}}^{2} = \frac{1}{2}k{x_{1}}^{2}\)

         \(X_{PQ} = v_{0} \sqrt{\frac{m}{3k}}\)                                         \(X_{P} = v_{0} \sqrt{\frac{m}{k}}\)

If all KE is converted to spring potential, block P will compress the spring a greater distance. (XPQ < XP)

Therefore APQ < AP

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