Physics 105 Final - Conceptual Questions

Answer: d) Right In this case, the acceleration of the bob is pointing in the opposite direction of velocity because the bob is slowing down.

CQ 33: Consider the bob of a pendulum swinging from right to left. In which direction is the net acceleration of the bob when it is half-way between the bottom and the top of the swing on its way up. a) Up b) Down c) Left d) Right e) There is no net acceleration at the start of the swing.

Answer: b) Green (or blue) Since we are traveling toward the light source, the Doppler shift will be up. Thus, we will see a color that has a higher frequency than yellow. Velocity = Wavelength*Frequency Wavelength = Velocity/Frequency If we are looking for a wave with a higher frequency, that will have a shorter wavelength, which is thus, green (or blue)

** CQ 76: Red light has a longer wavelength than yellow light, and green light has a shorter wavelength than yellow light. Suppose that you are approaching a yellow traffic light. If you go fast enough, you can use the Doppler effect to change the yellow light to a) Red. b) Green.(or blue) c) Bright yellow. d) Faint yellow. e) This question does not make sense, since you cannot change the color of light.

Answer: a) Increase The source of light moving away from us causes the frequency to go down. Since wavelength is equal to velocity/frequency. Thus, if the frequency goes DOWN, the wavelength must INCREASE.

** CQ 88: If some source of light travels away from us, the Doppler effect causes the wavelength of the light we receive from that source to a) Increase. b) Decrease. c) Remain the same. d) Alternate between increase and decrease. e) Change in discrete steps.

Answer: c) Both Bob and Jim are pulling with equal force

**CQ 82: Bob and Jim are engaged in a contest of strength. They clasp right hands and are attempting to pull the other person across a line. Bob is winning this game—he is pulling Jim across the line, but not at constant velocity. Jim is being accelerated across the line. Which of the two men is pulling with the most force? a) Bob (who is winning) b) Jim (who is losing) c) Both Bob and Jim are pulling with equal force. d) The answer depends on the weight for Bob and Jim. e) There are no net forces acting in this problem.

Answer: When those positions are further apart, that means that the object went further in that time interval than in a different time interval, and thus, is going faster.

**CQ 84: The diagram represents a multiflash photograph of an object moving along a horizontal surface. The positions as indicated in the diagram are separated by equal time intervals. The first flash occurred just as the object started to move and the last just as it came to rest. Which of the graphs best represents the object's velocity as a function of time?

Answer: Q(h): When a heat engine absorbs a heat Q(c): When a heat engine gives off heat Note: Some of the Q(h) leaves as work (W) and some leaves as heat Q(c). The engine absorbs more heat than it gives off. Thus, Q(h) is positive.

**CQ 87: For a heat engine, Q(net) is: a) Positive b) Negative c) Zero

Answer: a) Remain in motion forever It is very difficult to find a case where there is absolutely no forces acting on an object, and to also show that the object will remain in morion forever. Most things, when you start them, they will eventually stop. But this is because we live in a world where there are forces (friction, air resistance) that will stop something from continuing to move. It requires a force to change what an object is doing. - When an object is at rest, it requires a force to get it moving. - When an object is moving, it requires a force to get it to stop moving. In outer space, for example, when you reach a desired speed, you can turn off the rockets and coast at the speed forever, or at least until you come near some planet or star where the gravity affects you. Other than that, you don't need to keep the rockets on in order to keep going.

CQ 10: Newton's First Law states that, in the absence of any forces, an object in motion will: a) Remain in motion forever b) Eventually come to rest c) Slowly accelerate over long periods of time. d) Accumulate mass. e) Tend to oscillate about its center of mass.

Answer: c) Both objects have the same inertia Inertia has to do with mass, not velocity. Inertia is the tendency for an object to resist change in motion. It is just as hard to change an object's motion when it is at rest, as when it is moving.

CQ 11: Consider two objects with the same mass. If one object is moving and the other object is not, which object has the greater inertia? a) The object that is moving. b) The object that is not moving. c) Both objects have the same inertia. d) Inertia is a meaningless concept if there is no motion. e) It depends on the speed of the moving object.

Answer: c) Require about the same effort The car has the same mass on the moon as it does on the Earth. And therefore, it has the same inertia. Thus, it is just as difficult to get it moving.

CQ 12: Consider pushing a car on a level road to start it rolling from rest to 5 mi/hr in 10 seconds. Neglecting the effect of friction in the wheels, the same action on the moon would: a) Require much more effort. b) Require much less effort. c) Require about the same effort. d) Cause the wheels to leave the lunar surface. e) Result in a variable moment of inertia.

Answer: a) Up This question asks about a block of wood mounted on a pipe. Since the hole is a little bit tight, the block will slide up and down the pipe, but only with a considerable amount of friction. Thus, it just stays there, and only slides up and down if there is a force. You are then going to hold the pipe with your hand (not touching the block) and hammer on the top of the pipe. The block will move up the pipe. It is important to remember that the block has inertia. Thus, when you strike down on the top of the pipe, the pipe goes downward suddenly and the block tends to stay where it is. In actuality, it is not the block that is moving up, but rather the pipe that is moving down.

CQ 13: Consider a block of wood with a hole in it so it can slide up and down on a pipe. The block fits rather tightly on the pipe, so that it requires some force to make it slide up and down. If I hold the pipe with my hand and strike the top of the pipe with a hammer, which way will the block slide on the pipe? a) Up b) Down c) The block will not move. d) It depends on the mass of the hammer. e) There is not enough information given to answer this question.

This is a trick question, because the answer depends on "how" I pull downward. If I pull downward on the bottom string with a jerk, then the weight (since it has inertia), does not move very far during that jerk. The bottom string will break. I.e. The weight hasn't moved far enough to stretch the top string to break it. If I pull downward on the bottom string slowly/steadily, then the top string will break. This is because the top string has not only the force applied to it from hand, but also the force of the weight hanging on it.

CQ 14: Consider a weight hanging by a string. If I pull downward on the bottom string, which string will break? (refer to picture)

Answer: a) Yes If a object is moving with constant velocity, then the net force on that object is zero. Now, surely, going up hill, the engine is going to have to provide a larger force than going on a level ground. But, this larger force of the engine is cancelled out by both the air resistance force and gravity force that is pulling the car back. Thus, we still have a net force of zero.

CQ 15: Consider a car going with constant velocity up a hill. Is the net force on the car zero? a) Yes b) No

Answer: b) No F(1) = Force of Ralph on the rope F(2) = Force of Bill on the rope Newton's Third Law is about the force two objects exert on each other. It is a law about TWO objects. >>> Note: In this problem, we have THREE objects (Ralph, Bill and the rope) Newton's Third Law always talks about two forces that are on two different objects. >>> Note: In this problem, the two forces are on the SAME object (the rope) Thus, the answer is NO; Newton's Third Law does not require that the two forces be equal in magnitude. They "may" be equal in magnitude, but Newton's Third Law does not "require" it.

CQ 16: Consider two students (Ralph and Bill) pulling on the two ends of a rope. Does Newton's Third Law require that the two forces on the rope be equal in magnitude? a) Yes b) No

Answer: c) The two forces are equal Note: This answer is only an approximation, however, it is a very good approximation, and we will be using this approximation throughout this course.

CQ 17: Consider a weightless string which passes through a frictionless pulley. We pull on the string to hold up a weight. We hold the string in two different positions, as shown below. Which force must be greater? a) F(a) b) F(b) c) The two forces are equal

Answer: b) No W = Gravitational force of EARTH on ME n = Force of FLOOR on ME In the third law, this is a force of object 1 on object 2 equal to the force of object 2 on object 1. So, you have two objects that just switch places in the two statements. Here, these are both forces on "me". For this reason, it cannot be a third law, as the third law always involves two forces on two different objects. The third-law pair for "n" would be "the force of ME on the FLOOR" The third-law pair for "W" is "the gravitational force of ME on the EARTH"

CQ 18: I am standing on the floor. Gravity exerts a downward force "W" on me, and the floor exerts and upward force "n" on me. Are "W" and "n" a third-law pair? a) Yes b) No

a) Down

CQ 32: Consider the bob of a pendulum swinging from left to right. In which direction is the net acceleration of the bob at the start of the swing when v = 0? a) Up b) Down c) Left d) Right e) There is no net acceleration at the start of the swing.

Answer: b) No, both teams are pulling with the same force. Just consider the free body diagram of the rope. The losing team and winning team are both pulling the two ends of the rope in opposite direction. We will assume that there is an acceleration towards the winning team. If we use the equation (F = m*a), in which F(x) = F(2) - F(1) = 0 ("0" because we neglect the mass of the rope). Thus, the two forces are approximately equal (F(1) = F(2)). Note: Even if we don't neglect the mass of the rope, the difference between F(2) and F(1) is still very small. Thus, it is not the difference in the two forces that determines which team wins or loses; those two forces are more or less EQUAL. The factors that affect which team will win is: - Total mass of each team - Friction

CQ 19: Consider a game of tug-of-war. One team is winning. They are pulling the other team over the line. Is the winning team pulling on the rope with more force than the losing team? Neglect the mass of the rope. a) Yes, the winning team is pulling with more force. b) No, both teams are pulling with the same force.

Answer: (a) For a given amount of time, object A goes further then object B. Therefore, it must be going faster

CQ 1: Consider the graph shown below for the displacement x of two objects as a function of time t. Which object is moving faster?

Answer: a) F > F' (based on the circled, red equation in the picture: F = F' + m*a) The net force on object 1 has to be pointing to the right, therefore, the force F must be greater than the force F', so that their difference is a net force in the direction of the acceleration. Another way to think about this problem: F' is the force that accelerates object 2, whereas F is the force that accelerates both objects. Thus, F must be larger than F' because it is accelerating both objects, while F' is only accelerating object 2.

CQ 20: Suppose I push two boxes along a frictionless surface by applying a force F. If box 2 is heavier than box 1, what can we say about the force F which box 1 exerts on box 2? a) F > F' b) F < F' c) F = F'

Answer: c) The weights will experience the same amount of force from the string 1st - The force of the string on each of the weights is just the "tension" in the string. 2nd - Whenever a string passes through a pulley, the tension on the two ends of the string will be equal. This doesn't matter if the two weights are not the same or accelerating. This is only exactly true if we neglect: - The weight of the string - The weight of the pulley - The friction in the pulley However, even if we don't neglect these variables, it is still approximately true, and a good approximation.

CQ 21: Consider two weights connected by a string that passes through a pulley. If m2 > m1, which weight will experience the greater amount of force from the string? (Neglect the weight of the string, the weight of the pulley, and the friction in the pulley.) a) m1 b) m2 c) The weights will experience the same amount of force from the string. d) The answer would depend upon the circumference of the pulley. e) The forces will vary with time.

Answer: a) Negative

CQ 22: If I lift an object, the work done by gravity on that object is: ______. a) Negative b) Positive c) Zero

Answer: a) Negative

CQ 23: If I lift an object, the work done by gravity on my hands is: _______. a) Negative b) Positive c) Zero

Answer: c) Zero

CQ 24: Consider a cart rolling down a ramp. The work done by the normal force of the ramp on the cart is: _______. a) Negative b) Positive c) Zero

Answer: c) The bullet and the gun receive the same impulse.

CQ 25: I fire a bullet from a gun. Which receives the greatest impulse? a) The bullet b) The gun c) The bullet and the gun receive the same impulse. d) It depends on the caliber of the bullet. e) It depends on the mass of the gun.

Answer: c) Equal to p

CQ 26: A ping-pong ball collides with a bowling ball and sticks to the bowling ball. The initial momentum of the ping-pong ball is p, and the bowling ball is initially at rest. What is the momentum of the combined objects after the collision? a) Greater than p b) Less than p c) Equal to p d) The momentum of the ping-pong ball is negligible compared to the combined momentum. e) There is not enough information given to answer this question.

Answer: a) Greater than p

CQ 27: A ping-pong ball collides with a bowling ball and bounces off the bowling ball. The initial momentum of the ping-pong ball is p, and the bowling ball is initially at rest. After the collision, the momentum of the ping-pong ball is also p but in the opposite direction. What is the momentum of the bowling ball after the collision? a) Greater than p b) Less than p c) Equal to p

Answer: c) Same Note: This is true whether the collision is elastic or inelastic. There is always the same change (ex. 2p) between two objects, but a different direction/sign (+ or -).

CQ 28: I bounce a ball on the floor. Which is greater, the magnitude of the change in momentum of the ball or the magnitude of the change in momentum of the Earth? a) Ball b) Earth c) Same

Answer: c) Both horses have the SAME angular velocity. Note: When the merry-go-round has completed a 90 degree turn, the outside horse will have travelled farther, and is therefore traveling at a faster velocity. But the "angular" velocity looks at the angle theta, which in this problem is 90 degrees. Since both horses have travelled 90 degrees, their angular velocity is the SAME.

CQ 29: Which has greater "angular" velocity: a horse near the outside rail of a merry-go-round or a horse near the inside rail? a) The outside horse b) The inside horse c) Both horses have the same angular velocity. d) The horse with the greatest tangential velocity. e) This question cannot be answered without knowing the torque for each horse.

Answer: b) Slowing down Initially, it has a larger velocity than at the end. Therefore, it is slowing down.

CQ 2: Consider the graph shown below for the velocity v of an object as a function of time t. What is the object doing? a) Speeding up b) Slowing down c) Neither

Answer: b) Horizontal Note: The ball is moving in a circle with constant motion and angular velocity. In this case, the acceleration points toward the center of the circle. And the center of the circle is always in the plane of the circle. Thus, the acceleration is horizontal.

CQ 30: Consider a ball on the end of a string going at a constant speed around a horizontal circle. In which direction is the acceleration of the ball? a) In the direction of the string b) Horizontal c) Somewhere between (A) and (B)

Answer: a) Up

CQ 31: Consider the bob of a pendulum swinging from left to right. In which direction is the net acceleration of the bob at the bottom of the swing? a) Up b) Down c) Left d) Right e) There is no net acceleration at the bottom of the swing

Answer: b) At the farthest point from the sun Note: For an object near the surface of the Earth, its gravitational potential energy increases as we move that object away from the surface of the Earth. In this same way, the comet will have its greatest potential energy when it is furthest away from the Sun.

CQ 34: Consider a comet in orbit around the sun. At what point in its orbit with the potential energy of the comet be greatest? a) At the closest point to the sun b) At the farthest point from the sun c) At an arbitrary point in the comet's orbit d) The concept of potential energy is meaningless when applied to an orbit e) The potential energy would depend upon the mass lost by the comet

Answer: a) At the closest point to the sun Note: Here we use conservation of mechanical energy. Since the gravitational potential energy is greatest at point b, then the kinetic energy is greatest at point a. As the comet gets closer and closer to the sun, its gravitational potential energy decreases, and turns into kinetic energy.

CQ 35: Consider a comet in orbit around the sun. At what point in its orbit with the kinetic energy of the comet be greatest? a) At the closest point to the sun b) At the farthest point from the sun c) At an arbitrary point in the comet's orbit d) The concept of potential energy is meaningless when applied to an orbit e) The potential energy would depend upon the mass lost by the comet

Answer: a) F1 > F2 Note: The tension of F1 will be greater because the weight is closer to this part of the string.

CQ 36: A meter stick is hanging by two strings. The tensions in the two strings are F1 and F2. If we hang a weight as shown, then which of the following is true? a) F1 > F2 b) F1 < F2 c) F1 = F2

Answer: b) Rise Note: The person on the left produces a counterclockwise rotation. The person on the right produces a clockwise rotation. If the person on the right leans forward, then the left side of the seesaw will be favored. The left side of the seesaw favors the counterclockwise rotation, thus pushing the person on the right UP into the air.

CQ 37: Two people sit on opposite ends of a seesaw. They sit in positions such that the seesaw is balanced in a horizontal position. When one person leans toward the center of the seesaw, that person's end of the seesaw will a) Fall b) Rise c) Stay at the same level d) Rise or fall depending on the mass of that person e) Oscillate up and down

Answer: b) The broom part Note: The reason that we are able to balance it on our finger is NOT because the two weights are equal, but because the two torques are equal. And in order for the two torques to be equal, w2 > w1.

CQ 38: A push broom balances at its center of gravity near the bottom of the handle. If you cut the broom into two parts at its center of gravity, you will have a "handle part" and a "broom part". Which part has the greater weight? a) The handle part. b) The broom part. c) The two parts have the same weight. d) More information is needed to answer this question. e) It depends on the respective densities of the handle part and the broom part.

Answer: b) Horizontal axis that passes through the stomach and spinal cord Notes: - Horizontal axis: Parts of body are fairly close to the axis; This results in a smaller inertia - Vertical axis: Parts of body are farther from the axis; This results in a larger inertia

CQ 39: About which axis will the moment of inertia of a person be greater? a) Vertical axis that passes from head to toe b) Horizontal axis that passes through the stomach and spinal cord

Answer: a) Speeding up In this case, the object has a negative velocity, which means that it is moving in the negative x direction. The further you are from the t axis, the faster you are going. So, we start at a small speed and end up at a large speed. Thus, the object is speeding up.

CQ 3: Consider the graph shown below for the velocity v of an object as a function of time t. What is the object doing? a) Speeding up b) Slowing down c) Neither

Answer: a) Tension b > a Notes: - The pulley will experience angular acceleration in a clockwise direction - If there is an angular acceleration, there must be a net torque on the pulley - The only forces that produce a torque on the pulley is the rope, part a (pulling to left), and part b (pulling down) - The torque from the tension pulling down (b) is clockwise and the torque from the tension pulling to the left (a) is counter-clockwise. IF the net torque has to be clockwise, then the tension of b must be greater than the tension in a. (tension b > a).

CQ 40: Consider two boxes connected by a rope that runs through a pulley without slipping. If the boxes are accelerating as shown, which part of the rope has the greater tension? Neglect the mass of the rope, but do not neglect the mass of the pulley. a) Tension b > a b) Tension b < a c) Tension b = a

Answer: b) Up the hill Note: Friction is responsible for the angular acceleration of the ball. Without friction, the ball would not rotate, but would rather just slide down without rotating. The angular acceleration is clockwise, therefore the torque caused by friction must also be clockwise. Thus, the friction must be in a direction UP the hill.

CQ 41: Consider a round object rolling down a hill without slipping. Which direction is the force of friction of the ground on the object? a) Down the hill. b) Up the hill. c) There is no friction of the ground on the object. d) It depends on the direction of the angular acceleration. e) You must know the moment of inertia for the rolling object.

Answer: c) They will both reach the bottom at the same time. Notes: - Mass (m) and radius (r) cancel out - THUS, the answer to this question does not depend on the mass or radius of the two balls. Even though the bowling ball (compared to the pillar ball) has a greater mass, radius and thus, moment of inertia, they will still reach the bottom of the hill at the same time because they have the same velocity.

CQ 42: Suppose a billiard ball and a bowling ball are rolling down a smooth hill. If they both started at rest at the top of the hill, which one will reach the bottom first? a) The billiard ball. b) The bowling ball. c) They will both reach the bottom at the same time. d) You must know the density of each object in order to answer this question. e) You must know the mass of each object in order to answer this question.

Answer: c) Remains the same - Although angular velocity (w) increases, angular momentum (l*w) is conserved, and so the answer is c. - When the student brings the weights in toward him, this decreases the radius (r) and decreases the moment of inertia (I) - If angular momentum (I*w) is conserved and must remain the same, this means that the angular velocity (w) increases. Therefore, when he brings in the weight, he goes faster.

CQ 43: Consider a student sitting on a rotating chair. His arms are extended outward, holding two weights. When the student pulls the weights inward, his angular momentum: a) Increases b) Decreases c) Remains the same *Include the weights in the angular momentum.

Answer: a) Increases Notes: - When he pulls the weights inward, the moment of inertia (i) decreases, but the angular velocity (w) increases. Thus, it is not clear what happens to the kinetic energy. - To solve this problem, re-write the kinetic energy formula so that there is a term for the angular momentum (I*w). - We know that the angular momentum is conserved and we know that the angular velocity (w) increases, and so the angular momentum (I*w) INCREASES.

CQ 44: Consider a student sitting on a rotating chair. His arms are extended outward, holding two weights. In order to answer this question, include the weights in the rotational kinetic energy. When the student pulls the weights inward, his rotational kinetic energy a) Increases b) Decreases c) Remains the same d) Varies according to how quickly he pulls the weights inward e) Slowly moves to zero

Answer: c) The pressure will be the same in both locations.

CQ 45: Where is the pressure greater, one meter beneath the surface of Lake Powell or one meter beneath the surface of a swimming pool? (Neglect any atmospheric pressure difference between Lake Powell and the swimming pool.) a) Lake Powell b) The swimming pool c) The pressure will be the same in both locations. d) It depends on the length of the swimming pool. e) It depends on the current depth of Lake Powell.

Answer: c) Pressure equal in both containers The reason that b doesn't have more pressure at the bottom, even though it has more water in it, the pressure at the bottom is due to the weight of the water ABOVE the bottom. The water on each side is being applied to the sides of the container, not the bottom of the container.

CQ 46: In which container is the pressure of the water greatest? (see picture) a) Container a b) Container b c) Pressure equal in both containers

Answer: b) The volume of the object

CQ 47: Associated question: The buoyant force on an object totally submerged in a fluid depends on a) The mass of the object. b) The volume of the object. c) The volume of the container of fluid. d) The depth of the container of fluid. e) All of the above.

Answer: c) Equal to 500 pounds

CQ 48: Consider a rowboat floating on a lake. The weight of the rowboat, including the weight of the people sitting in it, is 500 pounds. The buoyant force on the rowboat is a) Greater than 500 pounds. b) Less than 500 pounds. c) Equal to 500 pounds. d) Equal to the difference between the weight of the rowboat and the weight of the displaced water. e) Impossible to determine without knowing how much water is displaced.

Answer: c) It will remain where it was

CQ 49: Consider ice floating in a glass of water. When the ice melts, what will happen to the water level? a) It will rise. b) It will fall. c) It will remain where it was. d) It depends on the amount of ice in the glass. e) It depends on the temperature of the water.

Answer: c) Both on its way up and on its way back down. In physics we say that an object is in free fall when the only force acting on it is gravity. When I throw a ball up in the air, as soon as I let go of it, it is in free fall from that point on, until I catch it again.

CQ 4: Associated question: Neglecting the effect of air resistance, if I throw a ball straight up into the air, we say the ball is a "freely falling object" a) On its way up (after I let go of it). b) On its way back down (after I catch it). c) Both on its way up and on its way back down. d) A thrown object can never be considered "freely falling." e)The answer depends on the acceleration of gravity.

Answer: a) Increase The forces on this object are: - Gravity - Normal force of scale on container - Force of finger on water - Water exerts an upward force on the finger Note: The downward force of finger on water is equal and opposite to the upward force that the water exerts on the finger.

CQ 50: A container of water is sitting on a scale. I dip my finger into the water without touching the sides or bottom of the container. The normal force exerted on the scale will (see picture): a) Increase b) Decrease c) Remain the same

Answer: a) Less than The forces on this object are: - Tension of the string (T) - Buoyant force of the water (B) - Weight of the object (mg) Therefore, forces in Y: B + T - mg = 0

CQ 51: An object is hanging on the end of a string and is submerged in water. The tension in the string is ____ ____ the weight of the object (see picture). a) Less than b) Greater than c) Equal to

Answer: b) Decrease In a free body diagram, we see that the forces on the object are: - Gravity (mg) - Normal force of the scale (N) - Buoyancy of air (B = pgv, where p = density of air) Thus, the sum of all forces in the Y is: N + B - mg = 0 or N = mg - B The weight of the object does not change because the mass of all of the carbon dioxide atoms does not change. The Buoyancy of air (B), however, changes because the Volume (V) of the glove changes. The Volume of the glove increases as it is inflated. Since the Volume gets larger, the Buoyancy Force increases, and thus, the Normal force decreases. Thus, the answer is "Decrease".

CQ 52: A container of water is sealed off with a latex glove containing alka seltzer. The container is sitting on a scale. If we drop the alka seltzer into the water, CO2 is released, inflating the glove. The normal force exerted on the scale will (see picture): a) Increase b) Decrease c) Remain the same Note: Do not neglect the buoyancy of air.

Answer: c) Equal to its velocity at the top of the hill.

CQ 53: Water flows downhill through a pipe with constant diameter. At the bottom of the hill, the velocity of the water is a) Greater than its velocity at the top of the hill. b) Less than its velocity at the top of the hill. c) Equal to its velocity at the top of the hill. d) Dependent on the height of the hill. e) Random at different sections of the pipe.

Answer: b) Less than the pressure in the large pipe.

CQ 54: Air flows from a pipe with large diameter into a pipe with smaller diameter. The pressure of the air in the small pipe is a) Greater than the pressure in the large pipe. b) Less than the pressure in the large pipe. c) Equal to the pressure in the large pipe. d) Dependent on the length of the pipe. e) Independent of the size of the pipe.

Answer: a) Becomes larger

CQ 55: Imagine a hobbit tossing a certain gold ring into a fire. As the ring is heated, the size of the hole in the ring a) Becomes larger. b) Becomes smaller. c) Remains the same. d) Is in general not predictable. e) The direction of the change cannot be given without more information.

The object doesn't move at first. Then it moves backwards and finally stops.

CQ 83: Below is a graph of an object's motion. Which sentence is a correct interpretation

Answer: c) The two jars will contain the same number of gas molecules.

CQ 56: Suppose we have two jars of gas, one of helium and one of neon. Consider both gases to obey the ideal gas law, and remember that the mass of a neon molecule is greater than the mass of a helium molecule. If both jars have the same volume, and the two gases are at the same pressure and temperature, which jar contains the greater number of gas molecules? a) The jar of helium b) The jar of neon c) The two jars will contain the same number of gas molecules. d) This question cannot be answered without knowing the shape of the jars. e) This is a trick question because noble gases do not obey the ideal gas law.

Answer: c) Equal to

CQ 57: I mix some Helium and Neon gas together in the same container. The average kinetic energy per Helium molecule is ____ ____ the average kinetic energy of the molecule. a) Greater than b) Less than c) Equal to Note: Consider both gases to obey the ideal gas law. Also note that the mass of a neon molecule is greater than the mass of a helium molecule.

Answer: b) Small

CQ 58: Desert sand is very hot during the day and very cold during the night. From this, we can conclude that the specific heat of sand is a) Large. b) Small. c) Variable with the seasons. d) Unpredictable in general. e) Comparable to the specific heat of water.

Answer: c) Closer to the initial temperature of the water

CQ 59: The specific heat of water is greater than that of lead. Suppose we have 1 kg of hot lead shot. We pour 1 kg of cold water over the lead shot. The final temperature of the water will be a) Halfway between the initial temperatures of the lead shot and the water. b) Closer to the initial temperature of the lead shot. c) Closer to the initial temperature of the water. d) The direction of the change is dependent on the initial temperature difference between the lead shot and the water. e) There is not enough information to answer this question.

Answer: b) Get farther apart As drop leaks from a water faucet, they are equal time apart, NOT equal distance apart. As they fall faster and faster, the distance that they travel during that time gets farther and farther.

CQ 5: Consider drops of water leaking from a water faucet. As the drops fall, they a) Remain at a relatively fixed distance from each other. b) Get farther apart. c) Get closer together. d) The motion of falling bodies is unpredictable. e) he answer would depend on the mass of the drops

Answer: c) Zero The work done on the gas is given by: W = -P*deltaV Since deltaV is zero, volume is constant during this process. Thus, the work done is zero.

CQ 60: The path shown below is vertical. The work done on the gas is (see picture): a) Positive b) Negative c) Zero

Answer: a) Path (a) As you can see, the area under the curve for path "a" is larger than the area under the curve for path "b". Thus, the most work is done in path "a".

CQ 61: In which of the two paths below is the most work done on the gas? (See picture) a) Path (a) b) Path (b) c) Equal

Answer: a) Positive In the first segment of this process, the gas is being compressed, and so the work is positive. In the third segment of this process, the gas is expanding, and therefore the work done on the gas is negative. Note that amount of negative work done is less in magnitude than the amount in positive work done. When you add those two work values together, you will get a positive answer. Thus, the answer is "positive".

CQ 62: In the path below, the gas returns to its original state. The net work done on the gas along this path is (see picture): a) Positive b) Negative c) Zero

Answer: b) Negative The path is horizontal going from right to left. From the ideal gas law, we see that we are keeping the pressure constant, compressing the gas (thus, the volume is decreasing). Therefore, from the ideal gas law, you can see that the temperature must be decreasing. From the rule, involving changes in internal energy, if the temperature is decreasing, the internal energy must also be decreasing. From the first law of thermodynamics, a gas that is being compressed, the amount of work will be positive. So, if the work is positive and the change of internal energy is negative, this means that Q must be negative. When you compress a gas, usually the pressure goes up. If you want to compress a gas, but not make the pressure go up, you must cool the gas while compressing it.

CQ 63: We compress a gas, keeping the pressure constant. Draw a PV diagram of the gas during this process. The heat Q put into the gas is (see picture): a) Positive b) Negative c) Zero

Answer: a) Is positive

CQ 64: We raise the pressure of a gas, keeping the volume constant. A PV diagram of the gas during this process shows that the change of internal energy of the gas a) Is positive. b) Is negative. c) Is zero. d) Cannot be predicted. e) Is variable with time during the process.

Answer: a) Is positive

CQ 65: We compress a gas adiabatically (Q = 0). A PV diagram of the gas during this process shows that the change of the internal energy of the gas a) Is positive. b) Is negative. c) Is zero. d) Fluctuates rapidly between positive and negative. e) Cannot be determined during an adiabatic compression.

Answer: c) Remain the same Normally when a gas expands like this, its kinetic energy decreases because the gas is doing work against the moving piston. If I move the piston instantaneously, the gas doesn't do any work against that piston because it is not exerting a force on it while its moving. Therefore, the kinetic energy of the gas molecules will remain the same.

CQ 66: If we move the piston to the left instantaneously, the kinetic energy of the gas molecules will (see picture): a) Increase b) Decrease c) Remain the same

Answer: c) Is zero

CQ 67: We compress a gas isothermally (constant temperature). A PV diagram of the gas during this process shows that the change in internal energy of the gas a) Is positive. b) Is negative. c) Is zero. d) Is dependent on the composition of the gas. e) Cannot be measured for this case.

Answer: d) Less than the pressure of the water at the bottom of the hill

CQ 85: Water is flowing up a hill through a pipe of constant diameter. At the top of the hill, the pressure of the water is a) Zero. b) Dependent on the height of the hill. c) Greater than the pressure of the water at the bottom of the hill. d) Less than the pressure of the water at the bottom of the hill. e) Equal to the pressure of the water at the bottom of the hill.

Answer: c) Zero From the ideal gas equation: PV = nRT (R = 8.314 J/molK) At the point, which represents both the initial point and final point, which has the same pressure and same volume, we know that P(i) = P(f) and V(i) = V(f). If this is true, we know that T(i) must equal T(f). Thus, the net change in temperature must equal zero. If deltaT (temperature change) = 0, then deltaU (internal energy change) = 0. Based on the first law of thermodynamics (deltaU = Q + W) net work done around this loop here is positive. And as we discovered, internal energy is zero. THUS, Q must be negative. In each step, we either have heat going into the gas or heat going out of the gas, but by the time we have done all 4 steps, we collectively have more heat flowing out of the gas than heat flowing into the gas. Thus, Q must be negative.

CQ 68: In the path shown below, the gas returns to its original state. The change in internal energy of the gas is (see picture): a) Positive b) Negative c) Zero

Answer: c) Is zero

CQ 69: Suppose that some process were really reversible. When this process takes a system from one state to another, the net change of entropy a) Is positive. b) Is negative. c) Is zero. d) Is meaningless in this context. e) Would always violate the second law of thermodynamics

Answer: (a) If you want to add vector a to vector b, you have to move vector b so that it begins where a ends. Then, the sum of those two vectors starts where vector a begins and ends where b ends.

CQ 6: In the figure below, which is the vector sum "A + B"?

Answer: b) Is irreversible

CQ 70: Consider a block sliding across the floor and finally coming to rest because of friction between the block and the floor. This process a) Is reversible. b) Is irreversible. c) Cannot be used to illustrate principles of physics. d) Is deterministic and therefore cannot illustrate causality. e) Always yields irreproducible results.

Answer: b) At the end points of motion First, draw y components of vectors. The y component of the horizontal vector is 0 (a = 0). Thus, maximum acceleration is at the end points. In the middle, the acceleration is equal to zero. At the bottom, you have an upward acceleration, in which the object goes faster and faster until it gets to the midpoint. From that point on, the object is slowing down. When you're in between speeding up and slowing down, you have zero acceleration. When you go above the zero, the object is slowing down (i.e. the acceleration is downward) the object until you get to the top where the velocity is zero and you have maximum acceleration.

CQ 71: Some object is moving up and down in harmonic motion. Where is the acceleration of that object greatest (see picture)? a) At the midpoint of motion b) At the end points of motion

Answer: c) Remain the same

CQ 72: Consider an object of mass m hanging on a spring. We pull the object down and then release it so that it oscillates up and down. If we repeat this on the moon with the same object and the same spring, the period of the oscillation will a) Be larger. b) Be smaller. c) Remain the same. d) Go to zero on the moon. e) Be impossible to estimate.

Answer: c) The speed of sound will be equal in both cities

CQ 73: Atmospheric pressure is greater in Los Angeles than in Provo. In which city will the speed of sound be greater at room temperature (20° C)? a) Provo b) Los Angeles c) The speed of sound will be equal in both cities. d) The speed of sound varies randomly from location to location. e) The speed of sound is different for different frequencies.

Answer: c) 100 Each 10 dB is a factor of 10 Thus, 20 dB is a factor of 10^2 = 100 (Also, 30 dB would be a factor of 10^3 = 1000)

CQ 74: Consider music played from a radio. If I turn up the volume so the sound level is 20 dB louder, by what factor has the intensity increased? a) 10 b) 20 c) 100 d) 200 e) 1000

Answer: b) 10 m 80 dB - 60 dB = 20 dB I(2) = 1/100 * I(1) d(2) = 10 * d(1), where d(1) is given as 1 m d(2) = 10 * 1 d(2) = 10 m

CQ 75: When I sit one meter from a radio, the sound level is 80 dB. How far from the radio should I sit to reduce the sound level to 60 dB? a) 3m b) 10m c) 30m d) 100m e) 300m

Answer: a) Increase The velocity of waves in the cord is given by: V = sq rt [(F/u)] - F = tension in the cord - u = linear mass density (mass per unit length) Since the tension (F) increases, the velocity (V) must also increase.

CQ 77: Consider a 3-loop standing wave on an elastic cord. If I increase the tensioning the cord, the velocity of the waves in the cord will (see picture): a) Increase b) Decrease c) Remain the same

Answer: a) 2-loop standing wave The equation for frequency of a standing wave is given by: f = n*(V/2L) - n = number of loops - L = length of cord (constant; does not change) - f = frequency (problem says this is constant) - V = volume Therefore, for this equation to remain valid, if the velocity (V) increases, the number of loops (n) must go down.

CQ 78: Consider a 3-loop standing wave on an elastic cord. If I increase the tension in the cord, keeping the frequency the same, I can produce a (see picture): a) 2-loop standing wave b) 4-loop standing wave

Answer: c) 500 Hz The period of the fundamental frequency for this sound here is given by the repeating unit in this figure. If we pick a given period length (such as 0.1, 2.1, 4.1, 6.1, and 8.1), we can see that in each of these intervals, the wave looks identical. This time is the period of the fundamental frequency. You can see that the period amount from the figure is about 2 milliseconds. Thus, fundamental frequency of this wave is given by 1/period = 1/2 ms = 1/(2x10^-3 s) = 500 s^-1 = 500 Hz Note: 1 s^-1 = 1 Hz f = 1/t f = 1/(2 ms) f = 1/(2 x 10^-2 s) f = 500 Hz

CQ 79: Calculate the fundamental frequency of the wave below (see picture): a) 100 Hz b) 200 Hz c) 500 Hz d) 1000 Hz

Answer: Refer to picture

CQ 7: Find the x and y components for Vector A (based on the picture).

Answer: c) Both balls will reach the same height.

CQ 80: Two balls are launched upward with the same initial velocity. One ball is heavier than the other. If we neglect air resistance, which ball will reach the greatest height? a) The heavier ball. b) The lighter ball. c) Both balls will reach the same height. d) The mass of both balls is required to answer this question. e) The greatest height will vary from trial to trial.

Answer: a) Is zero

CQ 81: Consider a real car traveling up a hill with constant velocity. If we do not neglect air resistance and friction, then the net force on the car a) Is zero. b) Is positive. c) Is negative. d) Is dependent on the height of the hill. e) Depends on the mass of the car.

Answer: b) Copper Since copper has a HIGHER specific heat than lead, in order to cool down copper, more heat has to flow from copper into the water. Thus, the water will get hotter as it tries to cool down the copper.

CQ 86: The specific heat of lead is 128 J/kg-°C. The specific heat of copper is 387 J/kg-°C. Consider a container of water at room temperature. If I drop a hot piece of metal into the water, the temperature of the water rises. If the mass of the metal is 10g and its initial temperature is 100°C, which type of metal will cause the temperature of the water to rise more? a) Lead b) Copper c) Both lead and copper will cause an equal rise in temperature. d) There will be no temperature increase, since the specific heat of water is so large. e) You do not have enough information to solve this problem.

Answer: c). At the top of its path The speed of the ball is given by the magnitude of its velocity, which is just the square root of the sum of the squares of its x and y components. When we throw a ball, the x component of velocity is a constant; it does not change. We will get a minimum speed when the y component is minimum. At the top of its path, the y component of velocity is zero. And so, the answer is c.

CQ 8: If I throw a baseball across an open field, at what part of its path does the ball have a minimum speed? a) Right before it hits the ground. b) Halfway to the top of its path. c) At the top of its path. d) Right after it leaves my hand. e) There is not enough information to say.

Answer: Refer to picture

CQ 9: Plot the motion of a baseball when it is thrown (Note: Plot only during the free fall motion -- i.e. after it leaves my hand and just before it strikes an object) x vs. t v(x) vs. t a(x) vs. t y vs. t v(y) vs. t x(y) vs. t