Please note that there is some overlap between exams. Your exams may not cover the exact material included in each of these sample exams.

Exam I - Kinematics- Solutions
Exam II - Dynamics & Statics- Solutions
Exam III - Energy & Momentum - Solutions
Exam IIII - Rotation and Rotational Equilibrium - Solutions
Exam V - Waves & Oscillations - Solutions

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) Which of the following has not been the definition of the metre?

A) 1/10,000,000 of the distance between the equator and the North Pole
B) the distance between scratches on a platinum-iridium bar kept just outside Paris
C) the distance light travels in vacuum in 1/299,792, 458 of a second
D) each of these has been the definition of a metre
E) none of these has been the definition of a metre.

2) Consider vectors A and B. If |A| = 2 and |B| = 5, then |A+B| can not be

A) 2
B) 3
C) 4
D) 5
E) 7

3) See the figure. Which of these statements is true?

A) A = C cos q
B) A = C sin q
C) A = B + C
D) A = B - C
E) A = C - B
 

4) Consider the figure, which shows an aerial view of the path taken by an object. The object starts from rest at I, reaching a speed v_f at II. The object turns right to III, continues straight to IIII, then turns left to V, all at speed v_f. During which of these intervals does the object have zero acceleration?

A) I to II
B) II to III
C) III to IIII
D) IIII to V
E) There is no interval when the acceleration is zero.

5) You're riding a smoothly-turning carousel of diameter 20 m which makes a complete revolution every 15 seconds. At the end of two complete revolutions,

A) your distance traveled is 63 m.
B) your displacement is 126 m.
C) your instantaneous velocity is zero.
D) your average acceleration is zero.
E) your distance traveled is zero.

Using the component method, find the magnitude and direction of vector D.

A = 13 m, q_A = 25^o

B = 20 m, q_B = 225^o

C = 7.5 m, q_C = - 80^o

D = A + B - C

Suppose that an airplane must fly due north a distance of 900 km in exactly two hours. All the way, the plane encounters an easterly wind (which blows to the west) of 100 km/hr. What should be the air speed of the plane, and in what direction should the pilot point the plane to arrive on schedule?

Consider a ball shot from a cannon over a flat, level plain at angle q_o and with muzzle velocity v_o. Neglect air resistance.

A) Show that the range is given by v_o² sin(2q_o)/g.
 

B) Show that the maximum range is obtained when q_o = 45^o.
 

HINT: sin(2q) = 2 sinq cosq

A bolt comes off an elevator that is moving upward at speed 6 m/s. The bolt reaches the bottom of the elevator shaft 3 seconds later.

A) What is the acceleration of the bolt after it has fallen from the elevator? (2 pts)
 
 

B) How high above the bottom of the shaft was the elevator when the bolt came loose? (6 pts)
 

C) What is the speed of the bolt when it hits the bottom of the elevator shaft? (6 pts)
 
 

D) How long did it take the bolt to reach a height of 30m above the bottom of the shaft? Give a physical interpretation of your answer. (6 pts)

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) A force F acts alone on a mass m, giving it an acceleration of 2 m/s². If the same force were to act alone on a mass M (=2m), then the acceleration of M would be

A) 0 m/s²
B) 1 m/s²
C) 2 m/s²
D) 4 m/s²
E) 8 m/s²

2) Consider two automobiles traveling up I-95. The faster one travels at 33 m/s and is 150m behind the slower, which is traveling at 18 m/s. How long will it take the faster car to catch up with the slower?

A) 0.22 s     B) 2.9 s     C) 5.9 s     D) 10 s     E) 17 s
 

3) Consider a block of mass m which is slid along the ceiling with acceleration a by a force F, as shown. The coëfficient of kinetic friction between block and ceiling is µ_K. Which of the following sets of equations follow from Newton's second law?

A) F sinq - mg - N = ma           F cosq - F_f = 0         F_f = µ_KN
B) F sinq - mg - N = 0              F cosq - F_f = ma     F_f = µ_KN
C) F cosq - mg + N = 0         F sinq - F_f = ma         F_f = µ_KN
D) F cosq - mg - N = ma         F sinq - F_f = 0         F_f = µ_KN
E) F sinq - mg + N = 0         F cosq - F_f = ma         F_f = µ_KN
 

4) Choose the answer which best completes the sentence:

If an object is at rest, then

A) no forces act on the object.
B) any forces which form third law pairs cancel each other out.
C) the mass of the object must be very large.
D) the sum of all forces acting on the object must be zero.
E) the weight and the normal force must be equal and opposite.
 

5) You're pushing a heavy crate across a rough floor. To do so, you must apply a force to the box. Which is the 'reaction' force to your 'action' force?

A) the floor pushing up on the box (the normal)
B) the floor pushing backwards on the box (friction)
C) the box pulling upward on the earth (gravity)
D) the box pushing backwards on you
E) you pushing backwards on the floor

Consider a man in a boat who wants to cross a river which is 1 km wide. The speed of the boat in still water (v_B,W) is 8km/h, and the speed of the water (v_W,G) is 10 km/h. Suppose that at first, he steers his boat at right angles to the current.

A) How far downstream will he land from dock B? (7 pts)
 
 

B) How long will it take him to cross (leave answer in hours)? (5 pts)
 
 

 C) Now, on his second try, he wants to land at dock B exactly opposite his starting point. At what angle should he steer his boat? (8 pts)

PROBLEM II (20 points)
 

Two blocks (M₁ and M₂) are on the inclines (q₁ and q₂), as shown in the figure, and are moving to the right. The string is massless and the wheel is frictionless, but the left-hand block/surface has coëfficient of kinetic friction µ_K.

A) Draw free body diagrams for each mass. Be sure to include all information, e.g., axes labels, et c. (4 pts)
 
 

 B) Write a set of equations using Newton's second law which will allow you to determine the motion. Do not solve them yet. (7 pts)
 
 

 C) Now, solve these equations to find the acceleration of the masses. (6 pts)
 
 

 D) Find the acceleration of the masses if there were no friction. (3 pts)

Consider a crate (M = 40 kg) which is sitting on a rough surface (µ_S = 0.3, µ_K = 0.2). A force F is applied at angle q = 40^o.

A) How big must F be to get the crate just to begin to move? (15 pts)
 
 
 

 B) How big must F be to maintain a constant velocity, once the crate is in motion? (5 pts)

PROBLEM IIII (20 points)

A plant is hung from massless wires as shown in the figure. What is the tension on each wire if the plant weighs 20N? Include free body diagrams, et c.

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) See the figure. Consider Newton's Balls, or the 'clackers.' Each of the balls has mass M. Assume that two of the balls (1&2) are withdrawn to the left and allowed to collide with the remaining three. Just before they hit, the balls have speed v_o. Although not all of the following outcomes do actually happen, one can be eliminated on the basis of momentum conservation alone. Which is it?

A) Balls 4 & 5 move to the right with speed v_o.
B) Balls 2, 3, 4, & 5 move to the right with speed ½ v_o.
C) Ball 5 move to the right with speed 2v_o.
D) Balls 3, 4, & 5 move to the right with speed ½ v_o.
E) Ball 5 moves to the right with speed 3v_o, while ball 1 bounces back to the left with speed v_o.

2) An empty gondola car (mass = 10⁴ kg) is moving along the tracks at 5 m/s. As it passes under a chute, the car is filled with 2x10⁴ kg of water. What is the final speed of the car? Neglect friction.

A) zero
B) 1.67 m/s
C) 2.5 m/s
D) 5 m/s
E) 15 m/s

3) Consider the rollercoaster car shown in the figure. The track is frictionless. If the speed of the car at point A is 7 m/s and h = 10m, what will be the speed at point E?

A) zero
B) 7 m/s
C) 14 m/s
D) 16 m/s
E) 250 m/s

4) Nancy Homewrecker and Tonya Thug are playing chicken on very slippery ice. If neither woman gives in and they collide, which of the systems listed below is the smallest system for which horizontal momentum is conserved ?

A) Tonya alone
B) Nancy Alone
C) Tonya and Nancy
D) Tonya, Nancy, and the ice.
E) Tonya, Nancy, the ice, and the earth.

5) Suppose that you whirl a bob of mass 0.25 kg around at the end of a string in a horizontal circle of radius 1.5 m a distance of 2 m above the floor. The tension you need to apply to the string is 73 N. How much work do you do on the bob each second?

A) 0 J (zero)
B) 4.9 J
C) 110 J
D) 688 J
E) The answer depends on how quickly you swing the bob.

PROBLEM I (20 points)
 

Consider two cars which undergo a completely elastic head-on collision. Car 1 has mass M₁ = 1500 kg and an initial speed of 20 m/s. Car 2 has mass M₂ = 1000 kg and initial speed 15 m/s. Use the technique of relative velocities to determine the velocities of each car after the collision.
 

A) What is the initial velocity of Car 1 in a frame in which Car 2 is initially at rest? (4 pts)
 
 

 B) What is the final velocity of Car 1 in that frame? (4 pts)
 
 

 C) What is the final velocity of Car 2 in that frame? (4 pts)
 
 

 D) What is the final velocity of Car 1 in the original frame of reference? (4 pts)
 
 

 E) What is the final velocity of Car 2 in the original frame of reference? (4 pts)

Consider a mass M = 2 kg sliding across a frictionless floor at speed v_o = 8 m/s. It contacts a spring of constant k = 200 N/m at the same time it encounters a rough portion of the floor with coëfficient of kinetic friction µ_K = 0.15.

A) Find the distance the spring will be compressed. (10 pts)
 
 
 

 B) Find the final speed of the mass as it slides back to the left. (10 pts)
 

A) What is the equation definition for kinetic energy? (4 pts)
 
 

 B) Derive the Work-Energy Theorem, W_total = DKE. (12 pts)
 
 

 C) Why is it necessary that the left hand side be the total work done on the object? Be specific! (4 pts)

Consider the time-varying force F(t) shown in the graph. This force acts (in one dimension) on a mass of 15 kg which is at rest at t=0.

A) What impulse has acted on the mass during these 12 seconds? (6 pts)
 
 
 

 B) What is the final velocity of the object? (8 pts)
 
 
 

C) How much work did the force do on the object? (6 pts)
 

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) In class, we discussed the moments of inertia for several shapes which had the same mass M and radius, R: I_HOOP = MR², I_DISC = ½ MR², I_{SOLID SPHERE} = ²/₅ MR². Now consider a hollow sphere of radius R and mass M. Which of the following choices places these four objects in order of increasing moment of inertia about the axes shown in the figures? HINT: use the argument given in class.

A) hollow sphere, solid sphere, disc, hoop
B) solid sphere, hollow sphere, disc, hoop
C) solid sphere, disc, hollow sphere, hoop
D) solid sphere, disc, hoop, hollow sphere
E) There is no Choice E.

2) Consider a rigid body, upon which three forces act. In order for the body to be in translational equilibrium,

A) the forces must be equal in magnitude.
B) the forces must all act in a single plane.
C) at least two of the forces must be equal in magnitude.
D) any two of the forces must sum to zero.
E) one of the forces must be zero.

3) Suppose that you want to remove a rusty nut with a wrench. You are already applying as large a force as you can in the manner shown in the figure. To unscrew the nut, you should

A) apply the same force at point A, perpendicular to the handle.
B) apply the same force at point B, but along the wrench handle.
C) apply half the force at point A, perpendicular to the handle.
D) apply half the force at point B, perpendicular to the handle.
E) get a bigger wrench.

4) Consider a tire (which rolls without slipping) on a car traveling 10 m/s along a straight flat road. What is the speed of the point on the outer edge of the tire shown, as measured by a stationary observer alongside the road?

A) 0 m/s
B) 1.6 m/s
C) 10 m/s
D) 14 m/s
E) 20 m/s

5) What is the angular velocity of the second hand on the clock at the front of the room?

A) 0.105 rad/s, into the wall
B) 0.105 rad/s, out of the wall
C) 6.28 rad/s, into the wall
D) 6.28 rad/s, out of the wall
E) 6.28 rad/s, up

Consider two discs. The larger of the two has mass M and radius R, while the smaller has mass m<M and radius r<R. If the two discs are placed at the top of an incline (height h and length L), released from rest, and roll without slipping down the incline, which will reach the bottom first? Shown all work or otherwise justify your answer.

HINT: Use conservation of energy.

Consider a banked highway curve (q = 9^o) with radius of curvature r = 750m. The curve is designed for trafic traveling at 120 km/h. Now, Daryl Dimwit takes the curve in his 1600 kg Porsche 928S at 230 km/h (=64 m/s). If the coëfficient of static friction between tire and road is µ_S = 1.1, will Daryl make it the curve?

A) Draw a free body diagram for the car as it rounds the curve.
 
 

B) Calculate the centripetal force necessary to keep the car moving in a circle around the curve.
 
 

C) Write Newton's second law for the x and y directions, as shown in the figure above.
 
 

D) Now, solve these equations to find the maximum possible centripetal force which could act on the car. Remember that F_f < or = µ_S N.
 
 

E) Will Daryl make the curve?

PROBLEM III (20 points)
 

A horizontal 4m long bar with weight 400 N is joined to a wall with a hinge. A 5m long wire connects the other end of the bar with the wall, as shown. In addition, a balloon is tied to the end of the bar, exerting a buoyant force of 200 N.

A) Find the tension, T, in the wire.
 

B) Find the vertical component of the force of the hinge on the bar, F_V.
 

C) Find the horizontal component of the force of the hinge on the bar, F_H.

Prove that the rotational kinetic energy of an object as it turns about some axis with angular velocity is KE_rot = ½ Iw², where I (=Sm_i r_i²) is the moment of inertia of the object about that axis.

HINT: Break the object up into many very small pieces.

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) A transverse wave moving along a compound string goes from a region of high mass density to a region of low mass density. Which figure best represents the orientations of the transmitted and reflected waves?

2) Suppose that a particular sound has an intensity level of 50 dB. If the intensity of the sound is doubled, the new intensity level will be

A) 25 dB
B) 47 dB
C) 50 dB
D) 53 dB
E) 100 dB

3) Suppose that you use a Spectrum Analyzer like the one in lecture to analyze the frequencies produced by an organ pipe (see figure).

From this spectrum, you can say that

A) the pipe must be open at both ends.
B) the pipe must be closed at both ends.
C) the pipe must be open at one end, only.
D) the pipe could be either open at both ends, or closed at both ends.
E) the pipe could be either open at one end only, or open at both ends.

4) Suppose that you have a grandfather's clock (with pendulum) which keeps perfect time. Now you take it with you when you vacation on the moon, where gravity is only one-sixth that on the earth. If you set the clock correctly at 1:00, what time will the clock read one hour later?

A) 1:10
B) 1:25
C) 2:00
D) 3:27
E) 7:00

5) Suppose that a mass M is placed on a frictionless surface, between two massless springs (k₁ and k₂), as shown. The springs are not actually connected to the mass, and when the mass is at its equilibrium point, neither spring is compressed. Find the period of oscillation of this mass if it is set into motion.

 

Consider a taut string of length L, fixed at both ends. Let the tension be T and the mass be M.

A) Write an expression for the velocity of transverse waves on this string, in terms of the given quantities. You do not have to derive the expression, merely write it. (4 pts)
 
 

B) Write an expression for the allowed resonant frequencies on this string. (10 pts)
 
 

C)Write an expression for the allowed resonant frequencies if someone were to grab the mid-point of the string. (6 pts)
 

Assume that a particle is known to exhibit Simple harmonic motion, so that its position as a function of time is

where A is the amplitude and w = (k/m)^½ is the natural frequency of the system.
 

A) Using Hooke's Law, prove that the acceleration as a function of time is given by

B) Using conservation of energy, prove that the velocity as a function of time is given by

Consider a human leg (still attached, of course!). When humans walk, the leg acts essentially like a pendulum; this minimizes the amount of effort the walker must expend.

A) If the period of oscillation of a leg is actually 0.8 that of a simple pendulum of the same length, write an expression for the period of oscillation of a leg of length L. (4 pts)
 
 

B) Estimate the length of a stride, in terms of L, if the leg makes a maximum angle of 10^o with respect to the vertical (both in front and behind). (4 pts)
 
 

C) Write an expression for the speed of the walker in terms of L. (6 pts)
 
 

D) Suppose that Sarah's leg is 1 metre long from hip to heel. At what speed would Sarah walk? (4 pts)
 
 

E) Suppose that Jesse's leg is only 0.7 metres long. How fast will Jesse walk? (2 pts)

PROBLEM IIII (20 points)

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) Which of the following statements is true?

A) Gauss' law holds true only for symmetric charge distributions.
B) If the net charge on a conductor is zero, then the charge density must be zero at every point on the surface of the conductor.
C) If there is no charge in a region of space, then the electric field must be zero.
D) Electric field lines never diverge from a point in space.
E) None of the statements above is true.
 

2) Consider the dipole and external electric field shown in the figure.  Which of the following statements best describes this situation?

A) The net force on the dipole is zero, the net torque on the dipole is zero.
B) The net force on the dipole is zero, the net torque on the dipole is not zero.
C) The net force on the dipole is not zero, the net torque on the dipole is zero.
D) The net force on the dipole is not zero, the net torque on the dipole is not zero.
E) There is no choice E.

3) Which of the following charges is not possible?

A) 1.6x10^-19 C
B) 2.4x10^-19 C
C) 3.2x10^-19 C
D) 4.8x10^-19 C
E) 6.4x10^-19 C
 

4) Find the electric field magnitude at the centre of a circular ring of radius R on which charge Q is uniformly disributed.

A) zero
B) kQ/R²
C) kQ/R
D) kQ/2pR
E) kQ/pR²

5) Find the electric potential at the centre of a circular ring of radius R on which charge Q is uniformly distributed.

A) zero
B) kQ/R²
C) kQ/R
D) kQ/2pR
E) kQ/pR²

Starting from Gauss' Law and showing all work, find an expression for the electric field at a distance r from the centres of the concentric, charged, spherical shells shown

A) for 0 < r < r_A.
B) for r_A < r < r_B.
C) for r_B < r < r_C.
D) for r > r_C.

HINT: You can do most of the work for Parts A-D all at once.
 
 

The hydrogen may be described as a positively charged (+e), stationary proton orbited by a negatively charged electron (-e).  Given the radius of the circular orbit (r = 5.3x10^-10 m), the mass of the electron (m_e = 9.11x10^-31 kg), and the fact that e = 1.6x10^-19 C,

A) find the force acting between electron and proton.
B) find the orbital frequency, w, of the electron in its orbit.
C) find the speed of the electron as it moves about the proton.

Consider two infinite, flat sheets of charge (one with charge density s, the other -s) which intersect each other at a 90^o angle.  Find the magnitude and direction of the electric field at point P.  You do not have to start from Gauss'Law, but can use the results of our in-class derivation.

Describe, using words, figures, and/or equations, what happens when a block of metal is placed in a uniform, external electric field, E_o.  Include descriptions of the final distribution of charge within the metal, the mechanisms by which this state is achieved, the net E-field within the metal, et c.

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) Consider a parallel plate capacitor with air between the plates (k_air = 1).  A potiential difference V_o is applied to the plates so that the plates acquire a charge Q_o.  Then, with the capacitor still connected to its voltage source, a slab of dielectric material (k_new = 4) is inserted.  This results in a charge Q_f  and a potential difference V_f.  Which of the following statements is true?

A) V_f = V_o and Q_f < Q_o
B) V_f = V_o and Q_f > Q_o
C) V_f = V_o and Q_f = Q_o
D) V_f < V_o and Q_f = Q_o
E) V_f > V_o and Q_f = Q_o

2) Consider a wire with length L_o and cross-sectional area A_o, made of a metal of resistivity r, which then has a resistance R_o.  If the wire were to be extruded out to three times its original length, what would be its new resistance?

A) R_o/9
B) R_o/3
C) R_o
D) 3R_o
E) 9R_o

3) Suppose that a parallel plate capacitor has capacitance C_o.  Suppose that we insert a third plate exactly between and parallel to the other two.  Then, the new capacitnace will be

A) C_o/4
B) C_o/2
C) C_o
D) 2C_o
E) 4 C_o

4) Which of the following relationships is applicable to the circuit shown?

A) - R₁I₁ + R₂I₂ = E₁ + E₂
B) - R₁I₁ - R₂I₂ = E₁ + E₂
C)   R₁I₁ + R₂I₂ = - E₁ + E₂
D)   R₁I₁ + R₂I₂ = E₁ + E₂
E) - R₁I₁ + R₂I₂ = E₁ - E₂

5) Consider the circuit at right.  What fraction of the total current I_TOT will pass through the 30 W resistor?

A) ¹/₄ I_TOT
B) ¹/₃ I_TOT
C) ¹/₂ I_TOT
D) ²/₃ I_TOT
E) ³/₄ I_TOT

Using any means you see fit (e.g., words, graphs, formulas, diagrams, et c), prove that the energy stored in a capacitor is given by U = ¹/₂QV.

PROBLEM II (20 points)

Consider a pair of parallel plates (like a capacitor), each of area 1 m² and separated by 1 cm, which are charged up so that there is a potential difference of 5000 V between them.  A proton (mass = 1.67x10^-27 kg, charge = 1.6x10^-19 C) is placed at rest next to the positely charged plate.

A) What is the electric field between the plates?  (4 pts)

B) What is the electric force acting on the proton?  (4 pts)

C) The proton will move toward the negatively charged plate.  Find the change in its potential energy as it moves from one plate to the other.  (4 pts)

D) Find the proton's kinetic energy just before it hits the negative plate.  (4 pts)

E) Find the proton's velocity just before it hits the negative plate.  (4 pts)

A) Prove that three resistors (R₁, R₂, & R₃) in series are equivalent to a single resistor, R_eq, where R_eq = R₁ + R₂ + R₃.  (10 pts)

B) Prove that three resistors (R₁, R₂, & R₃) in parallel are equivalent to a single resistor, R_eq, where 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃.  (10 pts)

PROBLEM IIII (20 points)

A fuse is a device composed of a thin band of metal whose purpose is to protect your home from fire.  Typically, a current of more than 20 A through a fuse causes it to melt.  Explain how you can arrange to have more than 20 A pass through the fuse, why this is a fire hazard, and how the fuse acts to protect your home. 

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) Consider the three currents, +I, -I, and +i, as shown in the diagram. In which of the following directions will the force on +i act?

A) in the +x direction
B) in the -x direction
C) in the +y direction
D) in the -y direction
E) There will be no net force on +i.

2) Consider an infinitely long wire with a circular loop as shown.  The loop has radius 2 cm and the wire carries 7 amps.  What is the magnitude of the magnetic field at the centre of the loop?

A) zero
B) 7x10^-5 T
C) 1.5x10^-4 T
D) 2.2x10^-4 T
E) 2.9x10^-4 T

3) Consider the figure, in which there is a magnetic field (out of the page).  Three particles enter the region and follow paths as shown.  Which of the following statements is true?

A) X is positive, Y is negative, Z is neutral (no charge)
 B) X is negative, Y is positive, Z is neutral
 C) X is positive, Y is positive, Z is positive
 D) X is negative, Y is negative, Z is neutral
 E) X is positive, Y is positive, Z is negative

4) Consider a current-carrying circular loop in a magnetic field, as shown.  Which of the following figures shows correctly the orientation of the loop's magnetic moment (µ) and the torque acting on the loop (t)?

5) Two circular hoops (radius R) are parallel, co-axial, almost in contact with one another (separation d) , and carry equal currents (I) in opposite directions.  How much force does the upper loop exert on the lower loop?  HINT: Think about what one loop looks like to a bug perched on the other loop.

A)  m_oI²R/d
B) 2pm_oIR/d
C) m_oI²R/2pd
D) m_oI/2pd
E) 2m_oI²R/d

Consider a mass spectrometer.  Particles are ionized (charged) and shot through a velocity selector, a region of crossed electric and magnetic fields. The selected particles then encounter a magnetic field which bends the trajectory into a semi-circle.  The radius of this path is measured with a detector.

A) What will be the velocity of the particles which pass through the velocity selector undeflected if E = 5000 V/m and B = 0.1 T?  (7 pts)

B) If the radius of the curved trajectory were 1.6 cm, which of the choices listed below are these particles most likely to be?  Assume B is still 0.1 T.  (13 pts)
 
 

Consider the arrangement of conductors shown in the figure.  At the centre is an extremely thin, infinitely long wire carrying current I₁.  An infinitely long, extremely thin shell is concentric with the wire, has radius R, and carries current I₂ (<I₁).  Starting from either the Biot-Savart Law or Ampere's Law, which ever you think is more appropriate, find the magnetic field, including direction, for

A) r < R.  (10 pts)

B) r > R.  (10 pts)

HINT:  You can do most of parts A & B together.

You're stranded on a planet where there is no test equipment or minerals around.  The planet does not have any magnetic field.  An alien hands you two pieces of iron, one of which is a magnet and the other of which is not; otherwise, they're identical.  How can you tell which is the magnet? You may not use any method which will destroy the magnet.

Consider the material shown below.  The orientations of the magnetic moments of the atoms are given by the arrows (point = N, tail =S).  There is no external magnetic field in the vicinity of the material.  Answer the following questions as completely as you can.

A) The magnetic moments of these atoms appear to exhibit what type of ordering?  Explain.

B) Is this piece of material a magnet?  Explain why or why not.

C) How could one destroy this ordering of the magnetic moments?  Why would that succeed?

D) What is a magnetic domain?

E) Draw in lines between the domains in the diagram above.

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) An ideal (no losses) step-down transformer at the end of a transmisison line reduces the effective voltage from 2400 V to 120 V.  The average power output of the transformer is 5000 W.  The primary coil has 10,000 turns.  What is the average power input to the transformer?

A) 0.5 W
B) 250 W
C) 5000 W
D) 100,000 W
E) 1,000,000 W

2) Consider an infinitely long straight wire carrying current I_o laid across along a diameter of a loop of wire of radius R .  If the current in the wire is reduced from I_o to 0 in time t_o seconds, then what is the magnitude of the average induced emf in the loop?

A) I_opR²/t_o
B) m_oI_oR/2t_o
C) m_oI_opR/2t_o
D) m_oI_oR²/t_o
E) zero (no emf is produced)

3) Suppose that you're given a black box which contains a series combination of either LR, LC, RC, or LRC components.  If at f = 400 Hz, E lags the current by 27^o, then what's in the box?

A) RC
B) LC
C) LR
D) LRC
E) Either RC or LRC, there is not enough information to tell which.

4) Inductance is measured in Henrys, while capacitance is measured in Farads.  What are the SI units of LC?

A) ohms
B) ohms^-1
C) seconds
D) seconds²
E) seconds^-1

5) What is I_max provided by the power supply if the frequency is extremely low?

 A) 0 (zero)
 B) 0.2 amps
 C) 0.3 amps
 D) 0.6 amps
 E) "infinite"

A helicopter has blades of length 9 m which turn at a frequency of 5 revolutions per second.  If the vertical component of the earth's magnetic field is 3x10^-5 T, what emf is generated between the hub of the rotor and the ends of each blade?

Hint:  How much area do the blades sweep out each second?

Consider the circuit shown below.  The switch was thrown long ago to include the battery in the circuit.  Now, however, at time t = 0, we change the switch to the other position, as shown.  Let E = 6 V, R = 20 W, and L = 30 mH.

A) Write Kirchhoff's Loop Law (using symbols) for the circuit as shown.  DO NOT SOLVE.  (4 pts)

B) At time t = 0, the instant after the switch is changed, what is the potential drop across the inductor?  (2 pts)

C) At time t = 0, the instant after the switch is changed, what is the current in the circuit?  (2 pts)

D) Write the solution for I(t>0) for the equation in part A.  Include numerical values.  (4 pts)

E) How much time is required for the current to fall to one-half of the initial value (given in part C)?  (8 pts)

Starting with Faraday's Law of Induction, E = ^(-)NDF_M/Dt, and comparing to E = ^(-)L DI/Dt, find the self inductance of a toroid of radius R and coil radius r.  Assume that R >> r.

Consider a series LRC circuit for which L = 0.01H, C = 2x10^-6 F, R = 4 W , and  E_max = 40V.

A) Find the resonant (natural) frequency. (4 pts)

B) Find the impedance at f = 2000 Hz. (6 pts)

C) Find I_max at f = 2000 Hz. (2 pts)

D) Find V_R, the maximum voltage across the resistor. (4 pts)

E) Find the phase angle between the applied emf and the current. (4 pts)

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) Often, prisms, rather than mirrors, will be used to 'reflect' light by using the phenomenon of total internal reflection (see figure).  If the glass prism is submerged in water (n_W = 1.33), what is the minimum index of refraction of the glass necessary for total internal reflection to occur for the situation shown?

A) 0.94 
B) 1.00001
C) 1.33
D) 1.41
E) 1.88

2) A horizontal ray of light is incident on a vertical flat mirror at an angle of 23^o.  If we rotate the mirror about a vertical axis by 17^o, while keeping the incident ray fixed, by what angle shall we rotate the reflected ray?

A) 0^o
B) 17^o
C) 23^o
D) 34^o
E) 46^o

3) One has a concave spherical mirror with a focal length of +0.25 m.  An object placed in front of this mirror produces an image which has a magnification of -8.  What is the object distance?

A) 0.03 m
B) 0.22 m
C) 0.28 m
D) 2 m
E) 4 m

4) Suppose that you have three ideal polarizers, as shown in the figure.  The orientation of the transmission axes of each of the three sheets is labeled relative to the vertical.  A beam of unpolarized light with intensity I_o is incident on the first sheet.  Calculate the intensity of the light transmitted through the third sheet if  q₁ = 60^o,  q₂ = 40^o, and  q₃ = 60^o.

 A) 0.125I_o
 B)  0.15I_o
 C) 0.23I_o
 D) 0.39I_o
 E) 0.44I_o
 

5) Suppose that two loudspeakers are placed side by side a distance d apart in a large grassy field where sound is not strongly reflected from the ground.  They each produce sound with wavelength l.  If the speakers are 180^o out of phase with one another, then which of the expressions below describes the angles at which no sound can be heard, measured with respect to the perpendicular bisector of the line joining the two speakers?

 A) d sinq = (m+1/2)l      m = 0, ±1, ±2...
 B) d sinq  = (m+1/2)l      m = ±1, ±2, ±3...
 C) d sinq  = ml       m = 0, ±1, ±2...
 D) d cosq  = ml       m = 0, ±1, ±2...
 E) d cosq  = (m+1/2)l      m = ±1, ±2, ±3...

Consider a series of layers of transparent materials arranged as shown in the figure.  The interfaces between layers are all horizontal and parallel and the indices of refraction for each are given.  If a light ray  is incident from air onto the first layer with an angle measured from the normal of 11.2^o, what will be the angle at which the ray exits the stack back into air? Let n₁ = 1.0, n₂ = 1.3, n₃ = 1.1, n₄ = 1.25, and n₅ = 2.2.

The  BTSC standard 20" colour television picture comprises 512 horizontal lines and is 0.33 m high.

A) What is the vertical distance between adjacent lines? (4 pts)

B) What are the two extreme values of the wavelengths visible to humans?  (6 pts)

C) If a typical diameter of the pupil of a human eye is 3 mm, what is the minimum distance one should sit from the TV screen?  (10 pts)

Below is shown a mirror with radius of curvature R = 12 cm (centre of curvature is marked with a black dot), along with its (perpendicular) optical axis.  Also shown is an object.
Note: Even though the mirror is drawn flat to aid in ray tracing, it is curved.  Also, the figure is not to scale.

A) Is the mirror concave or convex?  Explain briefly.  (4 pts)

B) Indicate the position of the focal point.  (2 pts)

Determine the answers to the questions below, using raytracing.  Use a straightedge!

C) Trace at least three rays to determine the position of the image.  (2+2+2 pts)

D) What is the image distance?  Show on the diagram.  (2 pts)

E) Is the image inverted or erect?  Explain briefly.  (2 pts)

F) What is the magnification of the image?  Show work.  (2 pts)

G) Is the image real or virtual?  Explain briefly.  (2 pts)

PROBLEM IIII (20 points)

Fill in the chart below.  Each column represents a different mirror.  All distances are in cm.  Be certain to include negative signs, where appropriate.  There are three boxes which need not be filled in.

NO PARTIAL CREDIT CAN BE GIVEN!!!
 
 

This is a closed book exam. The relations and constants on the last page may be useful; please feel free to detach it from the exam. Please answer multiple choice questions below. Show all work or otherwise justify each answer to receive full credit. IF YOU HAVE A QUESTION DURING THE EXAM, ASK IT!

SCORE:

1) Engine A, which operates on the Carnot cycle, takes Q_H  = 300 cal and produces W = 200 J each second.  Engine B, which operates on the Loh cycle, takes Q_H = 600 cal and produces W = 500 J each second.  Which of the following statements is true?

A) A has a higher thermodynamic efficiency than does B.
B) B has a higher thermodynamic efficiency than does A.
C) A and B have the same thermodynamic efficiencies.
D) It is impossible to tell which has the higher thermodynamic efficiency.
E) A exhausts more thermal energy Q_C than does B.

2) Consider the gas molecule for which a model is shown in the figure.  According to the Equi-partition of Energy Theorem, the molar heat capacity C_v of this material should be

A) 1.5 R
B) 2.5 R
C) 3 R
D) 3.5 R
E) 4.5 R

3) The headline on next month's Physics Today reads 'Man invents engine which turns heat directly and completely into work.'  Such an engine would

A) violate the First Law but be consistent with the Scond Law.
B) violate the Second Law but be consistent with the First Law.
C) violate both the First and Second Laws.
D) be consistent with both the First and Second Laws.
E) There is no Choice E.

4) Two moles of an ideal gas goes through the cycle shown, part of which is irreversible.  The change in the entropy of the gas for the cycle is

A) 5x10⁵ J/K
B) 3x10⁵ J/K
C) 2x10⁵ J/K
D) 1x10⁵ J/K
E) 0 J/K

5) Consider a gas which undergoes an isothermal process whereby the pressure is tripled (P_f = 3P_o).  Correspondingly, the final volume is

A) V_f = 9 V_o
B) V_f = 3 V_o
C) V_f = V_o
D) V_f = V_o/3
E) V_f = V_o/9

A 200 kg block of ice at 0^oC is placed in a large lake.  The temperature of the lake is just slightly higher than 0^oC, and so the ice melts.

A) Do you expect the entropy of the universe to increase, decrease, or remain the same? Justify briefly your answer.  (4 pts)

B) What is the entropy change of the ice?  (4 pts)

C) What is the entropy change of the lake?  (4 pts)

D) What is the entropy change of the universe, according to these calculations?  (4 pts)

E) Account for any disagreement in the answers to Parts A and D.

Consider an engine which runs on the cycle shown in the figure.  Assume that the working fluid is a diatomic gas in which molecules rotate, but do not vibrate.

A) Find the work done by the gas in the cycle.  (8 pts)

B) Find the heat into the gas in each step of the cycle (some terms may be negative).  (8 pts)

C) Find the thermodynamic efficiency of this engine.  (4 pts)

Consider an ideal gas at temperature T₁ in a closed container.

A) Sketch a diagram showing the approximate probability of any molecule having a given speed.  (5 pts)

B) Explain qualitatively why the curve has this shape.  (5 pts)

C) Draw a similar curve if the temperature of the gas is raised to T₂ >T₁.  (5 pts)

D) Explain qualitatively any differences in the shapes of the curves.  (5 pts)

Given that, for a process at constant volume,

Answers to Exam I

MC1 - Each of these has been a definition of a metre. (D)

MC2 - The maximum value of |A+B| will be when they point in the same direction (2+5=7). The minimum value will be when they point in opposite directions (|2-5|=3). So, 2 is not possible. (A)

MC3 - Since A + C = B, the only correct answer is (D), A = B - C. Answer A is incorrect, since it implies that A and C are in the same direction.

MC4 - (C) In all other intervals, there is a change in either the magnitude or the direction of the velocity.

MC5 - Check each quantity - distance = 2x circumference = 128m - displacement = 0 (you end where you start) - instantaneous velocity = 63m/15 sec = 4.2 m/s - ave acceleration = (v_f - v_o)/Dt = 0 since v_f = v_o so correct answer is (D)

PROBLEM I

In general, A_x = A cosq and A_y = A sinq. So,

A_x = 13 cos25^o = 11.78m
B_x = 20 cos225^o = -14.14m
C_x = 7.5 cos(-80^o) = 1.30m
D_x =A_x + B_x - C_x = (11.78) + (-14.14) - (1.30) = -3.66 m

A_y = 13 sin25^o = 5.5m
B_y = 20 sin225^o = -14.14m
C_y = 7.5 sin(-80^o) = -7.4m
D_y = A_y + B_y - C_y = (5.5) + (-14.14) - (-7.4) = -1.25 m

Then, D = (D_x² + D_y²)^½ = 3.86 m and = arctan(D_y /D_x) = arctan(-1.24/-3.66) = 18.7^o  However, we check the signs of the components and discover that the signs are incompatible with this answer. We need to add 180^o to obtain the correct answer of 198.7^o.

Use the notation, v_p,g to represent the velocity of the plane wrt the ground, et c.
We know that v_p,g = 900 km/ 2 hrs = 450 km/h North and that v_w,g = 100 km/h West.
We also know that v_p,g = v_p,w + v_w,g , therefor,
v_p,w = v_p,g - v_w,g = 450 N - 100 W = 450 N + 100 E
Then |v_p,w| = (450² + 100²)^½ = 460 km/h
Direction is arctan(450/100) = 77.5^o N of E where I have treated East as the x direction and North as the y direction.

A) R = x_o + v_oxt (since a_x = 0, & let x_o=0)         y = y_o + v_oyt + ½ a_yt² (For complete trip, y = y_o = 0, and of course, a_y = a_g = -g)

Solve second Eq for t: 0 = v_oyt - ½ gt² => t = 0 or t = 2 v_oy /g

Substitute this second solution into the x equation:

R = v_ox (2 v_oy /g) = (2/g) v_ox v_oy = (2/g) (v_o cosq_o) (v_o sinq_o) = (v_o²/g) (2 sinq_ocosq_o) = (v_o²/g)(sin 2q_o)

B) R goes as sin 2q_o. R will be a maximum when sine is a maximum, which will occur when 2q_o = 90^o, which will happen when q_o is 45^o.

A) a_y =a_g = -9.8 m/s², obviously due to gravity alone.

B) y = y_o + v_oyt + ½a_yt²     But y_f = 0 at bottom of the shaft, v_oy = +6 m/s, a_y = -9.8 m/s², and t = 3 sec.
So, y_o = y_f - v_oyt - ½a_yt² = 0 - 6(3) - ½(-9.8)3² = 27 m

C) v = v_o + at = 6 +(-9.8)3 = -24 m/s where (-) means downward.

D) y = y_o + v_oyt + ½a_yt²            solve for t when y = 30.
30 = 27 + 6t + ½(-9.8)t²     =>     5t² - 6t + 3 = 0 (quadratic eq)
t = (- ^-6 ± (36 - 4(5)3)^½)/(2×5) = 0.4 ± 0.49i         Answer is not real => bolt never reaches 30 m!

MC1) F = ma. If F is constant, then doubling m halves a, so a = 1 m/s² (B)

MC2) Use relative velocities v_Rel = 33-18 = 15 m/s. At 15m/s, it will take the faster car 10 seconds to cover the 150m between the cars. (D)

MC3) See figure. Note that the normal force is down.     Fsinq - mg -N = 0     Fcosq - F_f = ma     F_f = µ_KN (B)

MC4) Correct answer is (D).     F = ma.   Consider the other answers:
A- see D. There can be forces on the object, but they must cancel.
B- third law pairs never cancel each other, even though equal and opposite, since they act on different objects.
C- nonsense answer
E- Not necessarily true, see MC2!

MC5) If A acts on B, then B acts on A. (D)

A) Two solutions: first, use similarity of the displacement and velocity vector triangles (see figure) to write that 10/8 = x/1 so x = 1.25 km. OR, do part B first and see that the time to cross is 0.125 hrs. Then, since the velocity component downstream is simply the speed of the water, d = v_W,G*t = 10*0.125 = 1.25 km.

B) To find the time to cross, only the transverse speed matters: t = d/v = 1 km / 8 km/h = 0.125 hr

C) Redraw the velocity vectors as shown in the figure. He can never cross the river this way. Even if he heads directly upstream, he will float downstream at 2 km/h. OR, find the angle such that sinq = v_W,G/v_B,W = 10/8 = 1.25, and there is no such angle.

A) see figures

B) T - F_f - m₁gsinq₁ = m₁a (1)
N₁ - m₁gcosq₁ = 0 (2)
F_f - µ_KN₁ = 0 (3)
m₂gsinq₂ - T = m₂a (4)
N₂ - m₂gcosq₂ = 0 (5)

C) Substitute (2) & (3) into (1) to obtain: T - µ_K(m₁gcosq₁) - m₁gsinq₁ = m₁a
 Then, add (4) to eliminate the Ts:m₂gsinq₂ - µ_Km₁gcosq₁ - m₁gsinq₁ = (m₁ + m₂)a
 Therefor, a = g [m₂sinq₂ - µ_Km₁cosq₁ - m₁sinq₁]/[m₁ + m₂]

D) take the answer to Part C, but let µ = 0. a = g [m₂sinq₂ - m₁sinq₁]/[m₁ + m₂]

See figure.

F sinq + N - Mg = 0
F cosq -F_f = Ma = 0 (not moving yet, so a = 0)
F_f = µ_SN (just about to move, so use equality)
So, F cosq = F_f = µ_SN = µ_S(Mg - F sinq) => F(cosq + µ_S F sinq) = F(cosq + µ_S sinq) = µ_SMg => F = µ_SMg / [cos q+ µ_S sinq]
F = 0.3*40*9.8/[0.77 + 0.3*.68] = 122 N

B) Use result of part A, but replace µ_S with µ_K
F = 0.2*40*9.8/[0.77 + 0.2*.68] = 87 N

See figure. Clearly, T₃ is 20N, since it must support the weight of the plant. We could justify this with another free-body diagram of the plant and use of the third law.

y: T₂sin53^o + T₁sin30^o - T₃ = 0
x: T₂cos53^o - T₁cos 30^o = 0

Easiest to substitute x eq into the y eq: T₂ = T₁cos30^o/cos53^o
[T₁cos30^o/cos53^o]*sin53^o + T₁sin30^o = T₃ = 20 => T₁[1.65] = 20 => T₁ = 20/1.65 = 12.1 N
T₂ = 12.1[cos30^o/cos53^o] = 17.5N

MC1 - Look for one scenario for which p is not conserved. (D)
p_o = mv_o + mv_o = 2 mv_o         p_f = m(½ v_o) + m(½ v_o) + m(½ v_o) = 3/2 mv_o

MC2 - p_o = 10⁴*5 = 5x10⁴ kg m/s     p_f = (10⁴ + 2x10⁴)*v_f = 3x10⁴ v_f     =>     v_f = 5x10⁴/3x10⁴ = 1.67 m/s (B)

MC3 - KE_o + PE_o = KE_f     =>     ½ m 7² + m*9.8*10 = ½ mv_f²     =>     24.5 + 98 = 0.5 v_f²     =>     v_f = 15.6 m/s (D)

MC4 - Tonya & Nancy (C)

MC5 - W = 0 since tension is perpendicular to the displacement at any instant. (A)

Use the chart system discussed in class.
 

initial frame	convert	new frame	new frame	convert back	initial frame
v10 = +20	add 15	v10' = +35	v1f' = +7	subtract 15	v1f = -8
v20 = -15	add 15	v20' = 0	v2f' = +42	subtract 15	v2f = +27

A) In order to make v_2o = 0, we must add 15 m/s to each car's velocity, so v_1o' = 20 + 15 = 35 m/s.

B) Use the relationships given: v_1f' = v_1o'*[m₁-m₂]/[m₁+m₂] = 35*[1500-1000]/[1500+1000] = 7 m/s

C) v_2f' = v_1o'*2m₁/[m₁+m₂] = 35*2*1500/[1500+1000] = 42 m/s

D) Now we need to revert to the original frame, so subtract 15 m/s from each of the final velocities found above.
v_1f = 7 - 15 = -8 m/s

E) v_2f = 42 - 15 = 27 m/s

A)
The frictional force is obtained using NII for the vertical direction: N - mg = ma_y = 0.  Then, F_f = µ_KN = µ_Kmg.
Consider teh two points: the mass is moving to the right and the spring is relaxed, and the mass is at rest with the spring compressed a distance x_f.
W_friction = DKE + DPE_spring = ½ mv_f² - ½ mv_o² + ½ kx_f² - ½ kx_o² = - ½ mv_o² + ½ kx_f²
but also equals more directly F_f*d*cos180^o = -µ_Kmgx_f.
So, -µ_Kmgx_f = - ½ mv_o² + ½ kx_f²     =>     ½ kx_f² + µ_Kmgx_f - ½ mv_o² = 0 (quadratic eq)
plug in numbers: 100 x_f² + 2.94x_f - 64 =0     We assume that x_f >0 so take the positive root: x_f = 0.785m

B)
Now consider teh two points: the mass initially moving to the right and the spring relaxed, and the mass moving (finally) to the left with the spring relaxed (DPE = 0 in this case).
½ mv_f² - ½ mv_o² = 2(-µ_Kmgx_f)     The two is there because the mass makes two passes over the rough region.
v_f² = v_o² -4µgx_f = 59.4     =>     v_f = 7.7 m/s

A) KE = ½ mv²
 

B)
Use the convention such that the sign of each vector reflects its direction:
W_total = SW_i = S(F_iDx) = (SF_i) Dx = F_totalDx = (ma) Dx = m (a Dx)
Since we know that v_f² = v_o² + 2aDx, we can write that a Dx = (v_f² - v_o²)/2.
Then,
W_total = m (a Dx) = m (v_f² - v_o²)/2 = ½ mv_f² - ½ mv_o²
Define the KE as ½ mv².  Then
= KE_f  - KE_o = DKE.

C) It is essential that this be the total work, since that requires us to use the total force, and only by using the total force can we invoke Newton's second law and substitute ma.

A) J = FDt = area under curve = ½ (10+30)*3 + 30*5 + ½ 30*2 - ½ 30*2 = 210 kg m/s

B) J = Dp = mv_f - mv_o but v_o = 0, so v_f = J/m = 210/15 = 14 m/s

C) W = ½ mv_f² - ½ mv_o² = ½ 15*14² = 1470 J

MC1 - The method used in class compared the distributions of mass within each of the objects. The farther from the axis the mass is, the higher the moment of inertia. The hollow sphere has more mass near the centre than does the hoop, so I_HS<I_H. The hollow sphere has less mass near the centre than does the solid sphere, so I_HS>I_SS. If we flatten the hollow sphere into a pancake, we see that there is more mass at the edges than for the disc, so I_HS>I_D. Therefor, the order is given by (C).

MC2 - To be in equilibrium, F_tot =0   =>   F ₁ + F ₂ + F₃ = 0     =>    F₃ = - (F ₁ + F ₂)
i.e, F ₃ is in the plane of F ₁ & F ₂. (B)

MC3 - t = rFsinq.   F is as big as possible, sinq is as big as possible, need to increase r. (E)

MC4 - If v_tire = 10 m/s, then v_tangential as seen from the frame of the tire is 10 m/s. So for point P, add the tangential velocity and the translational velocity as shown to obtain a speed of (10²+10²)^½ = 14 m/s (D)

MC5 - w = #rad/sec = 2/60 = 0.105 rad/sec. Use RHR for direction - into the wall (A)

The one with the higher velocity at the bottom (v_F) has the higher average velocity, and therefor the quickest time to arrive at the bottom.
Use conservation of energy: mgh + 0 = ½ mv_f² + ½ Iw_f²
But, I = ½ mr² for a disk and w=v/r if there's no slipping.     Substitute.
mgh = ½ mv_f² + ½ (½ mr²)(v_f²/r²)     Simplify.
mgh = 3/4 mv_f²     =>     v_f = (4gh/3)^½
Since the final speed is independent of both the masses and the radiuses of the disks, they arrive at the same time!

A) See figure.

B) F_C = mv²/r = 1600*64²/750 = 8738N

C)Use NII:    y: -mg +Ncosq - F_fsinq = 0     x: Nsinq +F_fcosq = ma_y =ma_C = mv²/r         F_f <µ_SN (not equal)

D) Now to find the maximum possible centripetal force, we assume that the frictional force is maxed out at F_f = µ_SN. Substitute these eqs:
Nsinq  + F_f cosq = F_Cmax     =>     Nsinq  + µ_SN cosq = F_Cmax     =>     N[sinq  + µ_S cosq] = F_cmax
From the y-equation:  Ncosq = F_fsinq + mg = µ_SNsinq + mg     Solve for N.
N[cosq - µ_Ssinq] = mg     =>     N = mg/[cosq - µ_Ssinq]     Substitute into the x-equation.
F_cmax = N[sin q + µ_S cosq] = {mg/[cosq - µ_Ssinq]}*[sinq  + µ_S cosq] = 1600*9.8*(1.24/0.81) = 23,838N

E) He will easily make the curve.

A) Place the pivot point at the hinge to eliminate as much calculation as possible. Let L = the length of the bar. The angle q =  arccos (4/5) = 37^o .

S t = F_V*0 + F_H*0 + BL + TLsinq -wL/2 = 0
Then T = [wL/2 - BL]/Lsinq = [400*4* ½ - 200*4]/ 4*0.6 = 0 (purely a coincidence!)

B) SF_y = F_V - w + B + Tsinq = 0     =>     F_V = w - B - Tsinq = - 400 - 200 - 0 = 200 N

C) S F_x = F_H - Tcosq = 0     =>     F_H = 0

KE_total = SKE_i (for all the tiny little masses m_i )= S ½ m_i v_i ² = S ½ m_i (w_i r_i)²
but since the object rotates as a rigid body, all of the ws are the same.
= (S ½ m_i r_i²)w ² = ½ (Sm_i r_i²) w² = ½ (S I_i ) w² = ½ I_tot w²     Q.E.D.

MC1 - The transmitted wave is always in the same orientation as the incident wave. If we're going from high Z to low Z, there is no inversion upon reflection, so (B) is correct.

MC2 - The new sound must be louder, so (A), (B), & (C) are out. Since the intensity only doubled, and we know that the decibel scale is logarithmic, the new intensity level should be only a little higher than the original. (D)

MC3 - Since only odd multiples of the fundamental appear, the pipe is closed at one end only. (C)

MC4 - T = 2p(L/g)^1/2. If g_moon is ¹/₆ g_earth, then T_moon = 6^1/2 T_earth. OR, the clock runs 2.45 times more slowly on the moon than it does on the earth. So one hour would show up as 1hr/2.45 = 0.41 hr = 24.5 minutes on the clock. So, the clock would read 1:25 (B)

MC5 - The period due to spring 1 would be T₁ = 2p(M/k₁)^½. The period due to spring 2 would be T₂ = 2p(M/k₂)^½. But each spring acts on the mass through only half a cycle, so T_total = ½ T₁ + ½ T₂ = p(M/k₁)^1/2 + p(M/k₂)^1/2. (E)

A) v = (T/µ)^1/2 = (TL/M)^1/2

B) Fixed at each end     =>     node at each end => f_n = nv/2L     n = 1, 2, 3, ...

C) If someone grabs the mid-point, only frequencies which have a node there will survive: f'_n = nv/2L     n = 2, 4, 6, ...

A) Write Newton's second law: SF = ma     =>     -kx = ma     =>     a = -(k/m)x = -(k/m)Asin(wt) = -w²Asin(wt).

B) Consider that point at which the spring has reached maximum extension (x = A) and the block is at rest. Then the total energy is ½ kA². When the block is at some other position, the total energy is ½ mv² + ½ kx². So,
½ kA² = ½ mv² + ½ kx²
mv² = kA² - kx² = k (A² - x²)
v = (k/m)^1/2 (A² - x²)^1/2 = (k/m)^1/2 (A² - [Asin(wt)]²)^1/2 = (k/m)^1/2 A (1 - sin²(wt))^1/2 = wAcos(wt)

A) T_pend = 2p(L/g)^1/2 so T_leg = 0.8*2p(L/g)^1/2 = 1.6 L^1/2

B) d ~ 2Lsin10^o = 0.35L

C) v = d/t = d/(0.5T) = 2d/T = 2*0.35L/1.6L^1/2 = 0.44 L^1/2

D) v_Sarah = 0.44 (1)^1/2 = 0.44 m/s  ~ 1 mph (reasonable)

E) v_Jesse = 0.44(0.7)^1/2 = 0.37 m/s

A) For the mass & spring, PE ~ x². For gravity, PE ~ y, so we want y ~ x². Wire should be parabolic.

B) F(x) is of form Ax² + Bx + C. The condition that F(0) = 0 sets B = C = 0. The condition that F(1) = a sets A= a. So F(x) = ax².

C) PE = mgh = mgy = mgF(x) = mgax².

D) Compare this situation to one we understand well, the mass on a spring.
Spring case: PE = ½ kx², KE = ½ mv², T = 2p(m/k)^1/2
Bead on wire case: PE = mgax², KE ~ ½ mv_x² (we're assuming v_y ~ 0) T = ?
Comparison tells us that k <-> 2mga, so by substitution, T = 2p (m/2mga)^1/2 = p(2/ga)^1/2

Solutions to Exam I

MC1) (E)
A) Gauss' Law holds true for all closed surfaces, containing any charge configuration.
B) The charge could be distributed into positive and negative regions, with the total being zero.
C) Consider a point charge.  The field is non-zero everywhere, even where there is no charge.
D) Electric field lines diverge from positive charges.
E) Correct answer.

MC2) (D)
The net torque is clearly non-zero, like for the example we did in class.  However, the net force is also non-zero, because the field is stronger on the left side of the figure, where the field lines are closer together.

MC3) (B)
All charges are integer multiples of e = 1.6x10^-19 C.  Choice B corresponds to 1.5e.

MC4) (A)
For every bit of charge q in the ring, which produces a field at the centre of magnitude Kq/R², there is another bit of charge 180^o away which produces the same field at the centre but in the opposite direction.  Since we can add these fields in pairs, the net field at the centre will be zero.

MC5) (C)
Each bit of charge q_i in the ring will produce a potential kq_i/R at the centre of the ring.  Since potential is a scalar quantity, we need only add all of the potentials generated, which are all at distance R from the centre, to obtain a total potental of kQ/R.
V = SV_i = S kq_i/R = k/R Sq_i = [k/R]Q = kQ/R.

Gauss' Law: for a closed surface, f_E  = S E_perp DA = 4pkq_enclosed
Look at what the E-field looks like, using symmetry.  Since the spheres look the same as viewed from any angle, there is a spherical symmetry which implies that the E-field must be radial, and that the magnitude of E is the same for any given distance from the centre.
Choose the guassian surface to be a sphere of variable radius r concentric with the charged spheres.  In that case, E is perpendicular to the gaussian surface everywhere, so E_perp = E and
S E_perp DA = S E DA
and E is constant on the surface everywhere, so
S E DA = E SDA = EA = E 4pr² = 4pkq_enclosed
Solve for E:
E = (kq_enclosed)/r²
Now, look in each region to find E:
For 0 < r < r_A, q_enclosed = 0 and E = 0
For  r_A < r < r_B, q_enclosed = +Q and E = kQ/r² radially outward (since q is +)
For  r_B < r < r_C, q_enclosed = +Q + (-2Q) = -Q and E = k|Q - 2Q|/r² = kQ/r² radially inward (since q is -)
For  r_C < r, q_enclosed = +Q + (-2Q) + 3Q = +2Q and E = 2kQ/r² radially outward (since q is +)

A)
F_e = kq₁q₂/r² = 9x10⁹*(1.6x10^-19)²/(5.3x10^-10)² = 8.2x10^-10 N
B)
F_e = F_C = mw²r
w = [F_C/mr]^1/2 = [8,2x10^-10/9.11x10^-31*5.3x10^-10]^1/2 = 1.3x10¹⁵ sec^-1
C)
v = wr = 1.3x10¹⁵*5.3x10^-10 = 6.9x10⁵ m/s

Use the results from class that an infinite, flat sheet of charge density s produces a uniform field of magnitude E = 2pk_es.  The contribution to the total field at P from the positive sheet is 2pk_es up and to the left.   The contribution to the total field at P from the negative sheet is 2pk_es down and to the  left.   Add these contributions as vectors (taking the diagonal of the square as the magnitude) to obtain:
E_total = [(2pk_es)² + (2pk_es)²]^1/2 = 2.82pk_es to the left, regardless of the exact location of point P.

Start by assuming that the metal is a conductor, i.e., that the electrons are very free to move within the body of the metal.  The external E-field E_ext will apply a force to each type of charge, but since the positives are lock inside the atoms, which in turn are locked in a lattice, we can ignore the force on the positives.  The electrons, however, will begin to move in a direction opposite to the external field.  An internal field, E_int, will be set up by virtue of the separation of the charges, so that the total field in the metal will be of magnitude ||E_ext| - |E_int||.  Charges will continue to move until the net field inside the metal becomes zero, at which time there will be no more forces to move the charge.  The final state of the metal will have negative charges one one side and positive charges on the other side, with zero net field on the inside.

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Solutions to Exam II

MC2) (E)
Since the volume of the metal must remain constant, tripling the length correspondingly cuts the cross-sectional area by three.
Given that R_o = rl_o/A_o, then
R_new = rl_new/A_new = r3l_o/(A_o/3) = 9rl_o/A_o = 9R_o.

MC3) (C)
Let the original capacitor have area A, plate spacing d, and therefor capacitance C_o = e_oA/d.
Insertion of the third plate makes two series capacitors each of value C' = e_oA/(d/2) = 2C_o.
Series capacitors combine by adding the inverses: 1/C_eq = 1/2C_o + 1/2C_o = 2/2C_o = 1/C_o,
so the equivalent capacitance is once again C_o.

MC4) (B)
This corresponds to going around the small left-hand loop counter-clock-wise.

MC5) (D)
The 30W and the 60W resistors are in parallel, and so experience the same potential diference, V.  Assume that Ohm's relationship is correct, then
R₃₀I₃₀ = V = R₆₀I₆₀
30I₃₀ = 60I₆₀
I₃₀ = 2I₆₀
So, 2/3 of the current goes through the 30W resistor.

A)
E = ^(-) DV/Dx = 5000/0.01 = 500,000 V/m
B)
F = qE = 1.6x10^-19*500,000 = 8x10^-14 N
C)
DPE = qDV = 1.6x10^-19*(-5000) = -8x10^-16 J
D)
KE_o + PE_o = KE_f + PE_f     KE_o = 0 (starts from rest)
KE_f = ¹/₂mv_f² = - DPE = +8x10^-16 J
v_f = [-2*DPE/m]^1/2 = 9.8x10⁵ m/s  (less than 0.1c, so calculation is O.K.)

Let each resistor R_i have a potential difference V_i and a current I_i.  The three resitors are equivalent to a single resistor R_eq with potential difference V_eq and current I_eq.  Assume that 'Ohm's Relationship' holds true.

A)
We know that I₁ = I₂ = I₃ = I_eq  because charge is conserved.
We know that V_eq = V₁ + V₂ + V₃ because the electric field is a conservative field.
So,
V_eq = V₁ + V₂ + V₃
I_eqR_eq = I₁R₁ + I₂R₂ + I₃R₃
I_eqR_eq = I_eqR₁ + I_eqR₂ + I_eqR₃
R_eq = R₁ + R₂ + R₃

B)
We know that I₁ + I₂ + I₃ = I_eq  because charge is conserved.
We know that V_eq = V₁ = V₂ = V₃ because the electric field is a conservative field.
So,
I₁ + I₂ + I₃ = I_eq
V_eq/R_eq =V₁/R₁ + V₂/R₂ + V₃/R₃
V_eq/R_eq =V_eq/R₁ + V_eq/R₂ + V_eq/R₃
1/R_eq = 1/R₁ + 1/R₂ + 1/R₃

Although each individual device may not draw more than 20 A, putting them is parallel and turning them on may very well draw a total current over 20A.  This is dangerous, because the wires in the wall of the house through which the current runs will get hot, possibly hot enough to set fire to materials within the wall.  Being metal, the fuse element will warm up also as more current passes through it.  It is carefully designed to melt and break the connection (thus shutting down all the devices on that line) when a particular current is exceeded.  The procedure after a fuse blows is to check the devices on the line, remove the offenders so that 20A will not be exceeded, and replace the fuse.  Occasionally, when the home-owner was out of spare fuses, a copper penny was inserted into the fuse socket to re-establish the connection, but the penny does not melt, even at very high currents.  This led to a good number of house fires, and also to the expression, 'putting a penny in the fusebox,' in reference to a dangerous, short term solution to a problem.

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Solutions to Exam III

MC3) (D) F_m = qvBsinq_v,B (RHR).  Z is undeflected, even though sinq = 1, so Z must be uncharged.  If X were positive, the RHR would indicate that the magnetic force would be down-and-to-the-right-ish; the force must however act toward the centre of the circular path, which is the other way, up-and-to-the-left-ish.  We conclude therefor that X is really negative.  Similarly for Y, which must also be negative.

MC4) (E) Use RHR to find m: curl fingers around with the current (CCW), and your thumb points out of the page.  Now use the relationship that t = mBsinq_m,B (RHR) and use the RHR to see that the torque points toward the top of the page.

MC5) (A) A bug would see two parallel, essentially straight wires (they curve over a distance much greater than their separation).  Let's use the field due to an infinite wire as an approximation.  So, the field that the lower loop 'feels' is m_oI/2pd.  The force acting on a short piece of length Dl of the lower wire is then DF  = BIDl = [m_oI²/2pd]Dl downward.  Add up all the forces on all the little bits to get:
F_tot = SDF = S[m_oI²/2pd]Dl = [m_oI²/2pd] SDl = [m_oI²/2pd] 2pR = m_oI²R/d

A)
In the velocity selector, an electric force of magnitude qE acts transversely to the velocity of the charged particle, while a magnetic force of magnitude qvB acts in the opposite direction.  These forces will cancel, allowing the particle to continue through undeflected, if
qE = qvB
E = vB
v = E/B = 5000/0.1 = 50,000 m/s
B)
In the spectrometer proper, the particle will be deflected into a circular path because the magnetic force, qvB, acts as a centripetal force, mv²/r.
qvB = mv²/r
q/m = v/Br = 5x10⁴/[0.1*.016] = 3.1x10⁷ C/kg
Convert this to e/amu:
(3.1x10⁷ C/kg)(1.667x10^-27 kg/amu)(1e/1.6x10^-19 C) = 0.33 e/amu
The closest choice of those listed is tritium.

From the Biot-Savart Law, we have a notion that the B-field forms circular loops around the centres of the conductors.
Use Ampere's law to find the magnitude of B as a function of distance from the central axis:
S_loop B_|| Dl = m_oI_enclosed
Choose an Amperian Loop along one of the B-field lines, so that B is always parallel to the loop (i.e., B_|| = B) :
S_loop B Dl = m_oI_enclosed
Because of the rotational symmetry of the situation, we can assume that the magnitude of B is constant all around the loop:
B S_loopDl = m_oI_enclosed
We see that S_loopDl = 2pr, the circumference of the loop:
B2pr  = m_oI_enclosed
B  = m_oI_enclosed/2pr
How much current is enclosed by the Amperian Loop?
A)
For 0 < r < R, only I₁ is enclosed by the Amperian Loop, so
B = m_oI₁/2prCCW as seen in the figure.
B)
For r > R₂, the enclosed current is I₁ - I₂:
B  = m_o(I₁ - I₂)/2pr    CCW
You are not able to use Biot-Savart for the whole problem without resorting to calculus.

Realize that placing the magnet near the iron bar makes the bar temporarily a magnet.
Place the objects as shown, where the magnet is black and the bar is red:

In the left hand side, the induced south poles in the bar will have a very strong attraction for the north pole of the magnet, and the N-N repulsion will be weaker.  In the right hand case, there will be very weak attractions between unlike poles, and even weaker repulsions from like-pole pairs.

A)
Ferro-magnetic: neighbouring atoms' magnetic moments tend to line up in the same direction.
B)
It may be so slightly.  It appears that the total magnetic moment is directed toward the top of the page and to the right.  It's certainly not zero.
C)
One method to destroy the magnetic ordering is to heat the material above the Curie Temperature.  Then, the thermal energy available is more than sufficient to randomize the order imposed by the magnetic interactions.
D)
A domain is a region is which the vast majority of magnetic moments are oriented in the same direction.
E)

 

MC1) (C) If the transformer is indeed loss-less, then the power in equals the power out.
MC2) (E)  Since the wire is placed symmetrically over the loop, there are equal numbers of flux lines going into the loop through the lower half as there are coming out of the loop through the upper half.  Therefor, no matter what the current does, the total flux through the loop is constant at zero, and no emf is induced.
MC3) (E)  There must be a capacitor, otherwise E could not lag I.  There must be a resistor, or else the phase angle would be either +90^o or -90^o.  So, there must be at least R and C.  We do not know if there is an inductor; if there were, 400 Hz must be lower than the resonant frequency.
MC4) (D)  Consider the relationship at resonance for an LC circuit, that wo = (LC)^-1/2.  Then, LC = w^-2 = (seconds^-1)^-2  = seconds².
MC5)  (C)  At very low frequencies, the inductor looks like a wire (and therefor shorts the 100W resistor) and the capacitor looks like a break in the circuit (and so no current runs through it):

So I_rms = E_rms/R = 60/200 = 0.3 A

E = ^(-)DF/Dt = B DA/Dt = B*[N*pR²]/Dt = B[N/Dt]pR² =3x10^-5*5*p*9² = 0.038 volts

A)
IR = ^(-)L DI/Dt
B)
From relationship sheet:  E_L = ^(-) I/R e^-Rt/L = E_o e⁰ = E_o = 6V
C)
I(t) = E_o/R e^-Rt/L = 6/20 e⁰ = 0.3 A
D)
I(t) = E_o/R e^-Rt/L = 6/20 e^-20t/0.03 = 0.3e^-667t
E)
0.15 = 0.3e^-667t
0.5 = e^-667t
ln(0.5) = -667t
667t  = -ln(0.5) = ln(0.5^-1) = ln(2)
t = ln2/667 = 0.001 seconds

Since R>>r, we can assume that the field in the toroid follows closely that of an infinite solenoid:
B = m_onI = m_oNI/2pR.
E = ^(-)N DF/Dt =  ^(-)N D[BA]/Dt =  ^(-)NA DB/Dt =  ^(-)NA D[m_oNI/2pR]/Dt =
        ^(-)[NAm_oN/2pR[ DI/Dt =  ^(-)m_oN²pr²/2pR DI/Dt = ^(-)m_oN²r²/2R DI/Dt
Now, we know that E = ^(-)L DI/Dt, so comparison to the result above yeilds that
L = m_oN²r²/2R

A)
w_o = [LC]^-1/2 = 7071 sec^-1
B)
f = 2000 Hz  => w = 2p*2000 = 12,566 sec^-1
Z = [R² +(wL - 1/wC)²]^1/2 = 86 W
C)
I_max = E_max/Z = 40/86 = 0.465 A
D)
(V_R)_max = I_maxR = 0.0465*4 = 1.86 V
E)
tanF = [wL - 1/wC]/R = [12,566*0.01-1/(12,566*2x10^-6)]/4 = 21.5
F = arctan(21.5) = 87.3^o

MC1) (E) - sinq_C = n_W/n_G     n_G = n_W/sinq_C = 1.33/sin45^o = 1.88
MC2) (D) - q_i = q_r.  We increase (or decrease) the reflected ray angle by the same amount as the incident beam.  The angle between this rays is the sum of q_i and q_r, so that value changes by 2x17^o = 34^o.
MC3) (C) - m = -i/o = -8,  so i = 8o  1/f = 1/i + 1/o = 1/8o + 1/o = 9/8o
Then, o = 9f/8 = 9(.25)/8 = 0.28.
MC4) (D) - For the first polarizer, with unpolarized light, I₁ = ¹/₂I_o.
For the second polarizer, I₂ = I₁cos²q =  [¹/₂I_o]cos²20^o
For the third polarizer, I₃ = I₂cos²q =  [¹/₂I_ocos²20^o]cos²20^o = 0.39 I_o
MC5) (C)  - If the two sources were in phase, the relationship would be choice (A), but since they are one half cycle out of phase already, we remove the l/2 term to get choice (C)

Use Snell's Law for Refraction at each interface, but do not calculate the individual angles.
n₁sinq₁ = n₂sinq₂
Since the surfaces are all parallel, the angle of refraction at the first interface equals the angle of incidence for the second interface, and so on.
n₂sinq₂ = n₃sinq₃
n₃sinq₃ = n₄sinq₄
n₄sinq₄ = n₅sinq₅
 n₅sinq₅ = n₁sinq₆
Back substitution shows then that
n₁sinq₁ = n₁sinq₆
sinq₁ = sinq₆
And since all angles are under 90^o:
q₆ = q₁ = 11.2^o

A)
511 (not 512, since we measure from the first line, counting it as zero) in 0.33 m gives a separation of x = 0.33/511 = 6.45x10^-4 m
B)
We know this from class to be from violet (4x10^-7 m) to red (7x10^-7 m).
C)
Use Rayleigh's criterion: x/y = tanq = q = 1.22l/D, where x is the separation between lines and y is the distance the viewer is from the screen.
Calculate the minimum distance not to see distinct lines for each colour:
y_red = xD/1.22l_red = 6.45x10^-4*0.003/1.22*7x10^-7 = 2.27 m
y_violet = xD/1.22l_violet = 6.45x10^-4*0.003/1.22*4x10^-7 = 3.97 m
So, a viewer must be at least 3.97 m = 13 ft away from the screen in order not to see the individual horizontal lines.

A)
Since the centre of the radius of curvature is on the shiny side of the mirror, it is a concave mirror.

B)
The focal length f = r/2, so the focal point should be 6 cm in front of the mirror.

C)

D)
The image forms a distance 18 cm in front of the mirror.

E)
Since the image is upside down compared to the object, it is inverted.

F)
h_o = 9.5 mm and h_i = 19 mm (at least on my printout).  |m| = h_i/h_o = 19/9.5 = 2

G)
The image is real, since these three rays actually cross.

First Column:
Plane mirror, so f = infinity
1/i + 1/o = 1/f = 0, so i = -o = -30   OR   m = -i/o = +1, so i = -mo = -(1)(30) = - 30
Since m is positive, the image is upright.
Since i is negative (inside the mirror), the image is virtual

Second Column:
1/i + 1/o = 1/f ;   ¹/₁₆ + ¹/_(-12) = 1/f ;   f = -48
r = 2f = -96
Negative r means that the mirror is convex
m = -i/q = -(-12)/16 = + 0.75
m is positive, so the image is upright
i is negative, so the image is virtual (inside the mirror)

Third Column:
m = -i/o = -2;  o = i/2 = 42.5
1/i + 1/o = 1/f ;   ¹/₈₅ + ¹/_42.5 = 1/f ;   f = 28.3
r = 2f = 56.7
r is positive, so mirror is concave
i is positive, so image is real
m is negative, so image is inverted

Column Four:
r is positive, so mirror is concave
f = r/2 = 13.5
1/i + 1/o = 1/f ; + 1/o = 1/f - 1/i  = 1/13.5 - 1/(-71) ;  o = 11.3
Since i is negative, the image is virtual
m = -i/o = -(-71)/11.3 = +6.25
Since m is positive, the image is upright.

MC2) translation + rotation + vibration  =>  C_V = U/nT = N_A(³/₂ + ²/₂ + ²/₂)k_B = 7/2R  (D)

MC3) The First Law is conservation of energy, and that is not violated here if the total energy only changes form.  The Second Law specifically forbids this conversion without the production of waste heat, so it is violated.  (B)

MC4) Since entropy is a state variable, its value at the beginning and the end of a closed cycle must be equal, so DS = 0.  (E)

MC5) Isothermal => T is constant, so from the Ideal Gas Law, V ~ P^-1,   So if P -> 3P, then V -> V/3.  (D)

A) We expect the total entropy to increase, since we have a closed system moving toward thermal equilibrium.

B) DS = Q/T = mL_f /T_ICE = 200*80,000/273 = 58,610 calories

C) Since the same thermal energy was removed from the water as was provided to the ice, DS = Q/T_WATER ≈ - 58,610 calories, actually a bit less because T_WATER > T_ICE.  The exact answer depends on how much warmer the water is than the ice.

D) DS_total = sum of answers to Parts B and C, i.e., close to zero , but positive.

E) The important consideration is that the lake had to cool ever so slightly in order to provide the energy to warm the ice, so the average temperature of the lake during this process was actually above 0^o.  Using that correct value makes the answer to Part C less negative than we thought, and so the net change in S is really positive, as we initially expected.

A) The work is the area enclosed by the PV curve, W = (3P_o - P_o)(2V_o - V_o) = 2P_oV_o.

B) We'll need the temperatures at each state; find them with the Ideal Gas Law: P = (nR/V) T for isochoric processes, and V = (nR/P) T for isobaric processes.
Then, if T = T_o at the lower left corner,
T = 3T_o at the upper left (P tripled, V same)
T = 2T_o at the lower right (V doubled, P same)
T = 6T_o at the upper right (V doubled, P tripled)
Also make use of the facts that C_V = ⁵/₂R and C_P = ⁷/₂R for this type gas.
Then, in CW order from the lower left of the cycle diagram:
Q = nC_VDT = n⁵/₂R(2T_o) = 5nRT_o = 5P_oV_o (in)
Q = nC_PDT = n⁷/₂R(3T_o) = ²¹/₂nRT_o = ²¹/₂ P_oV_o (in)
Q = nC_VDT = n⁵/₂R(-4T_o) = -10nRT_o = -10P_oV_o (out)
Q = nC_PDT = n⁷/₂R(-T_o) = - ⁷/₂nRT_o = -⁷/₂P_oV_o (out)

C) eff = W/Q_in = 2P_oV_o/[5P_oV_o +²¹/₂ P_oV_o] = ⁴/₃₁ = 13%

A)

B) Few molecules are seen at high temperatures, since they tend to lose energy to slower ones through collisons.  There are also few molecules at low speeds, since they tend to pick up energy from collisions with faster ones.

C)

D) The peak is lower, since the same number of molecules are spread over a greater range of speeds.  The curve shifts to higher velocities, since T is a measure of the rms speed.

Start with the First Law:
Q = W + DU
Consider the addition of thermal energy through an isochoric process: Q = nC_VDT and W = 0, so that the First law becomes
 nC_VDT =  0 + DU.
Now, consider an isobaric process: Q = nC_PDT and W = PDV, so that the First Law becomes
nC_PDT = PDV + DU.
From the Ideal Gas Law, PDV = nRDT for an isobaric process, and we've already shown that  nC_VDT = DU.
Substitution of these terms results in
nC_PDT = nRDT + nC_VDT
C_P = R + C_V