2.1 Systems and Center of Mass

1. AEach molecule in the ice is itself solid, while each molecule of water is liquid.
   A student who thinks each particle has the properties of the substance picks this. The molecules in ice and in liquid water are the same molecules; being solid is a property of the system, set by how the molecules interact, not a property of any single molecule.
2. BThe ice has more mass than the water it melts into, so it holds together.
   A student who thinks mass changes with state picks this. Melting neither adds nor removes molecules, so the ice and the water it becomes have the same mass. The difference in behavior comes from the interactions between the molecules.
3. CEach molecule in the ice is held in place by its interactions with neighboring molecules. Correct
   The ice cube is a system of water molecules, and its rigidity is a property of the system that comes from how those molecules interact. In ice, each molecule is held in a fixed position by its neighbors; in liquid water, the same molecules can slide past each other, so the liquid spreads out.
4. DAn ice cube is a single object, so the molecules inside it play no part in its shape.
   A student who thinks a solid object is not a system of parts picks this. An ice cube can be modeled as a single object for some questions, such as how it slides, but its ability to keep its shape is explained only by the interactions between its molecules.

CED 2.1.A.1 · Read this in Fix

1. ANot appropriate: the ship is longer than the distance it moves, so it is not one object.
   A student who thinks only small objects can be modeled as single objects picks this. Size is not the test: the ship's parts all move together toward the dock, and the question is only about that motion, so the object model works.
2. BNot appropriate: the ship is made of many parts, and its crew keep moving about.
   A student who thinks a system with many or moving parts must be analyzed part by part picks this. The crew's walking and the ship's many parts do not affect how long the ship as a whole takes to move 50 m, so the system can be treated as one object for this question.
3. CAppropriate: a system can be modeled as a single object whatever question is asked.
   A student who thinks the object model always works picks this. The conclusion is right for this question, but the reason is wrong: if the student asked how the crew move relative to the deck, the ship could not be treated as a single object. The model is justified only because this question does not involve the parts.
4. DAppropriate: the question concerns only the motion of the ship as a whole. Correct
   The object model can be used when the properties and interactions of the parts do not matter for the behavior being analyzed. The time to move 50 m depends only on the motion of the ship as a whole, so the crew's movements and the ship's structure can be ignored, whatever the ship's size.

CED 2.1.A.2 · Read this in Fix

1. AIts mass stays constant, since mass is conserved and cannot be created or destroyed.
   A student who reads 'mass is conserved' as 'every system keeps its mass' picks this. Mass is conserved overall, but this system is open: the exhaust carries mass out of it, so its mass decreases. Only a system that includes the exhaust keeps a constant mass.
2. BIts mass decreases, as the exhaust gas carries mass across the system's boundary. Correct
   The exhaust gas is matter that was inside the system as fuel and is now outside it. Mass crosses the system's boundary, so the mass of the rocket and its remaining fuel decreases. The mass of the rocket plus all the exhaust it has produced stays constant.
3. CIts mass stays constant, as the exhaust is a gas, and a gas has no mass.
   A student who thinks gases have no mass picks this. A gas is matter made of particles, each with mass; the exhaust carries away the mass of the fuel it was made from, so the system's mass decreases.
4. DIts mass decreases, as the burning fuel is destroyed and its mass ceases to exist.
   A student who thinks burning destroys matter picks this. The mass does leave the system, but it is not destroyed: burning turns the fuel into exhaust gases, which carry that mass across the boundary into the environment.

CED 2.1.A.3 · Read this in Fix

1. AIts center of mass moves at a constant velocity, although neither cart does. Correct
   With equal masses, x_cm = (x₁ + x₂)/2, the midpoint of the two curves. From the graph it is 0.40 m at t = 0, then 0.50, 0.60, 0.70 and 0.80 m at t = 1, 2, 3 and 4 s, and the wiggles cancel in between, so it moves at a constant 0.10 m/s. Each cart's curve changes slope, so each cart speeds up and slows down: the parts behave differently from the system as a whole.
2. BIt speeds up and slows down repeatedly, just as each of the carts does.
   A student who thinks a system moves just as each of its parts does picks this. The carts' wiggles are opposite: when one is ahead of its trend the other is behind by the same amount, so the midpoint of the two positions increases steadily, by 0.10 m each second.
3. CIts motion is the same as the motion of cart 2, the cart in front.
   A student who takes the motion of one part, here the front cart, as the motion of the system picks this. Cart 2 speeds up and slows down, but the system's center of mass, halfway between the equal-mass carts, moves at a constant 0.10 m/s.
4. DIts center of mass stays at rest while the carts move back and forth about it.
   A student who thinks a center of mass is a fixed point picks this. With equal masses the center of mass is midway between the carts, and the graph shows that midpoint moving from 0.40 m at t = 0 to 0.80 m at t = 4 s.

CED 2.1.A.4 · Read this in Fix

1. AEach can be treated as a single object throughout, since each stays in one piece.
   A student who thinks any system can be treated as a single object for any question picks this. Staying in one piece is not enough: the balls behave differently at the floor because their internal structures differ, and the object model has no internal structure, so it cannot explain why one ball bounces and the other flattens.
2. BEach can be treated as a single object while falling, but not to explain the impact. Correct
   During the fall, the balls' internal structure does not affect their motion: each falls as a single object with the same acceleration. At the floor, the balls behave differently because their parts interact differently: the rubber springs back and the clay deforms. Explaining that difference requires the internal structure of each ball, so the single-object model cannot do it.
3. CThe clay ball needs a different model while falling, since heavier objects fall faster.
   A student who thinks a heavier object falls with a greater acceleration picks this. With air resistance negligible, both balls fall with the same acceleration whatever their masses and land together; their masses and materials matter only once they reach the floor.
4. DNeither can be treated as a single object at any stage, since each is made of many particles.
   A student who thinks a system of many parts can never be treated as one object picks this. Every ball is made of particles, but during the fall those particles move together, so each ball can be treated as a single object until it hits the floor.

CED 2.1.A.5 · Read this in Fix

1. AIt leaves the system's internal arrangement unchanged, as the same parts make it up.
   A student who thinks the choice of parts fixes a system's arrangement picks this. The system still contains the same box and book, but after the hard push the book is in a different place relative to the box: the arrangement has changed.
2. BIt makes the whole system move exactly as the box does, since the box rides on the cart.
   A student who takes the motion of one part, here the box that rides on the pushed cart, as the motion of the whole system picks this. After the hard push the book slides relative to the box, so the book moves differently from the box and the system as a whole does not move exactly as the box does.
3. CIt changes the system's internal arrangement, so the system no longer moves as one object. Correct
   How hard and how suddenly the cart is pushed is a variable external to the system. With a gentle push the box and book keep their arrangement and move as one object; with a sudden, hard push the arrangement changes, because the book moves relative to the box. The system's parts are the same, but its substructure is not.
4. DIt leaves the system modeled as one object, since the book stays on the box.
   A student who thinks any system can be treated as a single object whatever happens picks this. Staying on the box is not enough: the book moves relative to the box, so a single-object model cannot describe the system during the hard push.

CED 2.1.A.6 · Read this in Fix

1. AIt is at the midpoint, since the rod's shape is symmetric about that point.
   A student who thinks the center of mass is at the geometric center of the shape picks this. The center of mass lies on a line of symmetry only if the MASS is symmetric; here the steel half is more massive, so the center of mass is on the steel side of the midpoint.
2. BIt is on the aluminum side of the midpoint, nearer the lighter section.
   A student who weights each section's position with the other section's mass picks this. Each position is weighted by the mass of the section at that position, so the center of mass is pulled toward the more massive steel section, not toward the aluminum.
3. CIt is at the center of the steel section, the more massive part.
   A student who places the center of mass at the most massive part picks this. The aluminum section's mass also counts in the average, so the center of mass lies between the rod's midpoint and the center of the steel section, not at the steel section's center.
4. DIt lies between the midpoint and the center of the steel section. Correct
   The rod's shape is symmetric about its midpoint, but its mass is not: the steel half has more mass, so the center of mass is shifted from the midpoint toward the steel. The aluminum half still contributes, so the center of mass does not reach the center of the steel section; it lies between the two points, on the rod's axis.

CED 2.1.B.1 · Read this in Fix

1. A2.3 m Correct
   Only A is at y = 0: y_cm = (2.0 × 0 + 1.0 × 2.0 + 3.0 × 4.0) kg·m ÷ (2.0 + 1.0 + 3.0) kg = 14 kg·m ÷ 6.0 kg = 2.3 m. A contributes nothing to the numerator, but its mass counts in the total mass.
2. B3.5 m
   A student who leaves object A, at the origin, out of the calculation entirely gets 14 kg·m ÷ 4.0 kg = 3.5 m. A contributes zero to Σ m_i y_i, but its 2.0 kg still belongs in the total mass, so y_cm = 14 kg·m ÷ 6.0 kg.
3. C2.0 m
   A student who averages the three y-coordinates without weighting them, (0 + 2.0 m + 4.0 m) ÷ 3, gets 2.0 m. Each coordinate must be weighted by its object's mass; C carries half of the total mass, which pulls the center of mass up to 2.3 m.
4. D4.0 m
   A student who puts the center of mass at the most massive object, C, picks this. A and B, with half of the total mass between them, are below C, at y = 0 and y = 2.0 m, so the center of mass lies well below C's y-coordinate, at 2.3 m.

Working y_cm = (Σ m_i y_i)/(Σ m_i) = [(2.0 kg)(0) + (1.0 kg)(2.0 m) + (3.0 kg)(4.0 m)]/(2.0 kg + 1.0 kg + 3.0 kg) = (14 kg·m)/(6.0 kg) = 2.3 m.

CED 2.1.B.2 · Read this in Fix

1. A0.30 m/s
   A student who uses the midpoint of the two carts, without weighting by mass, gets positions 0.40 m and 1.00 m and an average velocity of 0.30 m/s. Cart B has three times cart A's mass, so the center of mass is the weighted average, 0.50 m and then 1.30 m.
2. B0.40 m/s Correct
   Locate the center of mass at each time: x_cm = (1.0 × 0.20 + 3.0 × 0.60) ÷ 4.0 = 0.50 m at t = 0 and (1.0 × 0.40 + 3.0 × 1.60) ÷ 4.0 = 1.30 m at t = 2.0 s. The object modeling the system moves 0.80 m in 2.0 s, an average velocity of 0.40 m/s.
3. C0.50 m/s
   A student who takes the motion of the more massive cart, B, as the motion of the system picks B's average velocity, (1.60 − 0.60) m ÷ 2.0 s. The lighter cart A also counts: B moves 1.00 m but the lighter A moves only 0.20 m, so the center of mass moves 0.80 m, not 1.00 m.
4. D0.20 m/s
   A student who weights each cart's position with the other cart's mass gets 0.30 m and 0.70 m, an average velocity of 0.20 m/s. Each position must be multiplied by the mass of the cart at that position, which gives 0.50 m and 1.30 m.

Working x_cm at t = 0: [(1.0 kg)(0.20 m) + (3.0 kg)(0.60 m)]/(4.0 kg) = 0.50 m. At t = 2.0 s: [(1.0)(0.40) + (3.0)(1.60)]/4.0 m = 5.20/4.0 m = 1.30 m. Average velocity = (1.30 m − 0.50 m)/(2.0 s) = 0.40 m/s.

CED 2.1.B.3 · Read this in Fix

1. A+2.2 m
   A student who enters P's position as a distance, +2.0 m, gets (8.0 + 3.0) kg·m ÷ 5.0 kg = 2.2 m. P is on the negative side of the origin, so its coordinate is −2.0 m; with the sign kept, the center of mass is at −1.0 m, nearer the more massive P.
2. B−2.5 m
   A student who divides Σ m_i x_i = −5.0 kg·m by the number of objects, 2, picks this. The divisor is the total mass, 5.0 kg; dividing by 2 leaves the unit kg·m and puts the 'center of mass' beyond P, outside the two objects.
3. C−1.0 m Correct
   x_cm = [(4.0 kg)(−2.0 m) + (1.0 kg)(3.0 m)] ÷ (5.0 kg) = (−8.0 + 3.0) kg·m ÷ 5.0 kg = −1.0 m. The result lies between P and Q and nearer P, the more massive object, as it must.
4. D+2.0 m
   A student who weights each position with the other object's mass gets [(1.0)(−2.0) + (4.0)(3.0)] kg·m ÷ 5.0 kg = +2.0 m, nearer the lighter Q. Each position must be multiplied by the mass of the object at that position, which puts the center of mass nearer P.

Working x_cm = (m_P x_P + m_Q x_Q)/(m_P + m_Q) = [(4.0 kg)(−2.0 m) + (1.0 kg)(+3.0 m)]/(4.0 kg + 1.0 kg) = (−8.0 + 3.0) kg·m/(5.0 kg) = −1.0 m.

CED 2.1.B.2 · Read this in Fix

1. A5.0 cm Correct
   Treat the U as three uniform rectangles of equal area (20 cm² each), each located at its own center: the base at 1.0 cm and each arm at 7.0 cm above the bottom edge. y_cm = (20 × 1.0 + 20 × 7.0 + 20 × 7.0) ÷ 60 = 5.0 cm. This point is on the line of symmetry, in the gap between the arms, where there is no material.
2. B6.0 cm
   A student who puts the center of mass at the geometric center of the U's outline, halfway up its 12 cm height, picks this. The mass is not symmetric about that height: more of the plate lies below 6.0 cm (the whole base and the lower parts of the arms) than above it, so the center of mass is lower, at 5.0 cm.
3. C1.7 cm
   A student who places each arm's mass where the arm is attached to the base, 2.0 cm above the bottom edge, instead of at the arm's own center, gets (20 × 1.0 + 20 × 2.0 + 20 × 2.0) ÷ 60 = 1.7 cm. Each uniform arm enters the calculation at its own center, 7.0 cm above the bottom edge.
4. D1.0 cm
   A student who thinks the center of mass must lie within the material looks along the line of symmetry, finds material there only in the base, and picks the base's center. The center of mass is an average position and can lie in the empty gap: here it is 5.0 cm above the bottom edge.

Working The plate is uniform, so each rectangle's mass is proportional to its area and its center of mass is at its center (symmetry). Base: 10 cm × 2.0 cm = 20 cm², center 1.0 cm above the bottom edge. Each arm: 2.0 cm × 10 cm = 20 cm², center 2.0 cm + 5.0 cm = 7.0 cm above the bottom edge. y_cm = (20 × 1.0 + 20 × 7.0 + 20 × 7.0) cm³/(60 cm²) = 300 cm³/60 cm² = 5.0 cm. By symmetry the center of mass is on the dashed line, in the empty gap between the arms.

CED 2.1.B.2 · Read this in Fix

1. A(5/6)L
   A student who weights each position with the other object's mass writes [(m)(L/2) + (2m)(L)] ÷ (3m) = (5/6)L. Each position must be multiplied by the mass located there: the rod's 2m at L/2 and the ball's m at L.
2. B(1/3)L
   A student who places the rod's mass at its end, at the origin, writes [(2m)(0) + (m)(L)] ÷ (3m) = (1/3)L. A uniform rod enters the calculation at its own center of mass, its midpoint, L/2 from the bare end.
3. C(3/4)L
   A student who averages the two positions without weighting them, (L/2 + L) ÷ 2, gets (3/4)L. The rod has twice the ball's mass, so the positions must be weighted: x_cm = [(2m)(L/2) + (m)(L)] ÷ (3m) = (2/3)L.
4. D(2/3)L Correct
   The uniform rod acts as a mass 2m at its midpoint, L/2 from the bare end, and the ball is a mass m at L. x_cm = [(2m)(L/2) + (m)(L)] ÷ (3m) = 2mL ÷ 3m = (2/3)L.

Working Put the origin at the bare end of the rod. The uniform rod's center of mass is at its midpoint, x = L/2 (symmetry); the ball is at x = L. x_cm = [(2m)(L/2) + (m)(L)]/(2m + m) = (mL + mL)/(3m) = (2/3)L.

CED 2.1.B.2 · Read this in Fix

1. A×2.0
   A student who sees m₂ in the numerator of x_cm = m₂d/(m₁ + m₂) and doubles x_cm picks this. The total mass in the denominator also increases, from 3m₂ to 4m₂, so x_cm grows only from d/3 to d/2.
2. B×1.5 Correct
   From x_cm = m₂d/(m₁ + m₂) = d/3, block 1 has twice block 2's mass: m₁ = 2m₂. With block 2's mass doubled, x_cm = 2m₂d/(2m₂ + 2m₂) = d/2. The coordinate changes from d/3 to d/2, a factor of 1.5.
3. C×1.0
   A student who thinks the center of mass depends only on where the blocks are picks this. Neither block moves, but the center of mass is a mass-weighted average: giving block 2 more mass pulls the center of mass toward it, from d/3 to d/2.
4. D×0.6
   A student who weights each position with the other block's mass reads x_cm = m₁d/(m₁ + m₂) = d/3, so m₂ = 2m₁; doubling m₂ then gives m₁d/(5m₁) = d/5, a factor of 0.6. Each position is weighted by its own block's mass, so adding mass to block 2 moves the center of mass toward block 2, not away from it.

Working Before: x_cm = m₂d/(m₁ + m₂) = d/3, so m₁ + m₂ = 3m₂ and m₁ = 2m₂. After: x_cm' = (2m₂)d/(m₁ + 2m₂) = 2m₂d/(4m₂) = d/2. Factor: (d/2)/(d/3) = 3/2 = 1.5. The mass of block 2 appears in the denominator as well as the numerator, so doubling it does not double x_cm.

CED 2.1.B.2 · Read this in Fix

1. AIt is at R, since R is halfway along the length of the hammer.
   A student who puts the center of mass at the geometric midpoint picks this. The heavy head puts more mass at one end, so the center of mass is nearer the head than the midpoint is; the diagram confirms this, since R's horizontal spacing changes from image to image.
2. BIt is at P, as P is at the head, the most massive part of the hammer.
   A student who places the center of mass at the most massive part picks this. The handle's mass also counts, so the center of mass is on the handle, a little way from the head. The diagram shows it: P's positions are not equally spaced horizontally, so P does not move as a projectile.
3. CIt cannot be found, since a spinning hammer is not a projectile.
   A student who thinks a spinning object cannot be modeled as a projectile picks this. The spin affects how the points move about the center of mass, not the center of mass itself, which moves as a projectile. The diagram shows exactly one point, Q, moving equal horizontal distances in equal times along a parabola.
4. DIt is at Q, as Q moves along the same kind of path as a thrown ball. Correct
   Modeled as a single object at its center of mass, the hammer is a projectile: its center of mass has zero horizontal acceleration, so it moves equal horizontal distances in equal time intervals while following a parabola. Only Q does this: its seven positions are equally spaced horizontally and lie on a symmetric arch. P's and R's horizontal spacings change from image to image as those points turn about Q. Q, on the handle near the heavy head, is where the center of mass must be.

CED 2.1.B.3 · Read this in Fix

1. ADive 2's is greater, since pulling into a tuck lifts her body higher.
   A student who thinks rearranging the body in flight changes the path of the center of mass picks this. Pulling in her legs moves her body parts relative to her center of mass, but the center of mass keeps to the projectile path fixed at takeoff.
2. BThey cannot be compared, since a rotating body is not a projectile.
   A student who thinks a spinning body cannot be modeled as a projectile picks this. The diver rotates, but her center of mass still moves as a projectile, with the same takeoff position and velocity in both dives, so the heights can be compared: they are equal.
3. CThey are equal, since her center of mass moves as a projectile in both dives. Correct
   Modeled as a single object at its center of mass, the diver is a projectile once she leaves the board. The path of her center of mass is set by its position and velocity at takeoff, which are the same in both dives, so its maximum height is the same. Tucking changes where the center of mass is within her body, not its path.
4. DDive 1's is greater, since her outstretched body reaches higher.
   A student who takes the motion of one part of the body, its top, as the motion of the whole diver picks this. Her outstretched hands may reach higher in dive 1, but the height of her center of mass is the same in both dives.

CED 2.1.B.3 · Read this in Fix

1. A5.00d
   A student who takes the center of mass as the plain average of the positions, (0 + d + x₃)/3 = 2d, picks this. The masses are not equal, so each position must be weighted by its mass; object 3 carries half of the total mass, so it need not be placed as far out.
2. B2.00d
   A student who thinks the center of mass is at the most massive object puts object 3 at x = 2d. Objects 1 and 2 both lie to the left of 2d, so with object 3 there the center of mass would be at (0 + md + 2m·2d)/(4m) = 1.25d; object 3 must be farther right to balance them.
3. C3.50d Correct
   Weight each position by its object's mass and divide by the total mass, 4m: x_cm = (0 + md + 2m·x₃)/(4m). Setting this equal to 2d gives d + 2x₃ = 8d, so x₃ = 3.50d.
4. D2.50d
   A student who leaves out object 1 because it is at the origin, and drops its mass from the total, writes (md + 2m·x₃)/(3m) = 2d and picks this. Object 1 adds nothing to Σm_ixi, but its mass still belongs in the total mass, 4m.

Working x_cm = (m·0 + m·d + 2m·x₃)/(m + m + 2m) = (d + 2x₃)/4. Setting x_cm = 2d: d + 2x₃ = 8d, so x₃ = 7d/2 = 3.50d. Check: (0 + md + 2m(3.50d))/(4m) = 8md/(4m) = 2d. (Plain average of the positions: (0 + d + x₃)/3 = 2d gives x₃ = 5.00d. Object 1 and its mass left out: (md + 2m·x₃)/(3m) = 2d gives x₃ = 2.50d. Most massive object placed at the center of mass: x₃ = 2.00d, which would put the center of mass at 1.25d.)

CED 2.1.B.2 · Read this in Fix