2.3 Newton’s Third Law

1. AThe truck's force is larger, as the truck is more massive.
   A student who thinks the more massive object exerts the larger force picks this. The forces of an interaction are equal in magnitude whatever the masses; the truck's larger mass affects how its motion changes, not the size of the force it exerts.
2. BThey are equal in magnitude, as they form a single interaction. Correct
   The two forces are the two forces of one interaction, so by Newton's third law F⃗_{car on truck} = −F⃗_{truck on car} at every instant: equal in magnitude, opposite in direction. This holds while the vehicles speed up, whatever their masses and whichever is doing the pushing.
3. CThe car's force is larger, as it is the one doing the pushing.
   A student who thinks the pusher exerts the larger force picks this. Whichever vehicle 'does' the pushing, the truck pushes back on the car with a force of equal magnitude.
4. DThe truck exerts no force, since its engine is off.
   A student who thinks a passive object exerts no force picks this. The truck is in contact with the car, so it pushes back on the car; no engine is needed to exert a contact force.

CED 2.3.A.1 · Read this in Fix

1. AIt continues along the parabola that the shell was following. Correct
   The forces of the explosion are exerted between the fragments, so they are internal to the system and do not influence the motion of its center of mass. The only external forces are the gravitational forces on the fragments, the same as on the whole shell, so the center of mass continues along the shell's parabola until a fragment lands.
2. BIt falls straight down, as the explosion ends the forward motion.
   A student who thinks the explosion destroys the shell's forward motion picks this. The explosion's forces are internal, so the center of mass keeps the horizontal motion the shell had.
3. CIt follows the path of the largest fragment that the explosion makes.
   A student who thinks the center of mass moves with the largest part picks this. The center of mass is the mass-weighted average position of all the fragments and does not follow any one of them.
4. DIt moves faster than before, as the explosion pushes everything outward.
   A student who thinks internal forces can change the motion of the center of mass picks this. The explosion speeds up individual fragments in different directions, but its forces are internal and leave the motion of the center of mass unchanged.

CED 2.3.A.2 · Read this in Fix

1. AThe student's pull, which weakens along the rope and is weakest next to the wall
   A student who thinks a pull is used up along a rope picks this. In a light rope the tension is the same at every point; the pulls between neighboring segments do not weaken toward the wall.
2. BThe pulls that the parts of the rope on either side of P exert on each other Correct
   Tension is the large-scale result of the forces that neighboring segments of the rope exert on each other in response to the pulls on its ends. At P, the part of the rope on the wall's side pulls the part on the student's side toward the wall, and vice versa; the tension at P is the magnitude of these pulls.
3. CA force stored in the rope, which the rope keeps even after the student lets go
   A student who thinks tension is something a rope stores picks this. Tension exists only while the ends are pulled; when the student lets go, the segments stop pulling on each other and the tension disappears.
4. DThe sum of the student's pull and the wall's pull on the ends of the rope
   A student who adds the magnitudes of the two end pulls picks this. The pulls on the two ends point in opposite directions; the tension at any point is the magnitude of the pull at one end, not their sum.

CED 2.3.A.3 · Read this in Fix

1. AIts peak is below +12 N, as cart 1 has the smaller mass.
   A student who thinks the less massive object exerts the smaller force picks this. The forces of an interaction are equal in magnitude whatever the masses, so the peak is 12 N; the masses affect only how each cart's motion changes.
2. BIts peak is more than +12 N, as cart 1 was the one moving.
   A student who thinks the moving object exerts the larger force picks this. Cart 2 pushes on cart 1 just as hard as cart 1 pushes on cart 2, so the two pulses have the same peak magnitude, 12 N.
3. CIts peak is +12 N, and it is exerted from 0.10 s to 0.30 s. Correct
   By Newton's third law the force on cart 2 by cart 1 is equal in magnitude and opposite in direction to the force on cart 1 by cart 2 at every instant. Its graph is the mirror image of the one shown: zero until 0.10 s, a peak of +12 N at 0.20 s, and zero again from 0.30 s.
4. DIts peak is +12 N, but it acts after the force on cart 1 ends.
   A student who thinks a reaction follows the action picks this. The two forces of an interaction exist at the same time: the force on cart 2 begins at 0.10 s and ends at 0.30 s, exactly as the force on cart 1 does.

Working Third law at every instant: F⃗_{1 on 2}(t) = −F⃗_{2 on 1}(t). The graph shows F_x on cart 1 going from 0 at 0.10 s to −12 N at 0.20 s and back to 0 at 0.30 s; so F_x on cart 2 is the mirror image: a pulse from 0.10 s to 0.30 s with peak +12 N at 0.20 s.

CED 2.3.A.1 · Read this in Fix

1. AThe upward gravitational force exerted on Earth by the book Correct
   The gravitational force on the book is exerted by Earth, so its partner is the force of the same interaction the other way round: the book's gravitational pull on Earth, equal in magnitude and directed upward, toward the book.
2. BThe upward normal force exerted on the book by the table
   A student who takes the normal force as the partner of the gravitational force picks this. Both forces are exerted on the book, and they come from different interactions, book–table and book–Earth; third-law partners are exerted on different objects.
3. CThe downward push exerted on the table by the book
   A student who thinks the book's weight is the force it exerts on the table picks this. The book's push on the table is a contact force and the partner of the table's push on the book, not of Earth's gravitational pull.
4. DNone: a book is too small to exert a force on all of Earth
   A student who thinks a much less massive object exerts little or no force on a massive one picks this. The book pulls on Earth with a force exactly equal in magnitude to Earth's pull on the book; because Earth's mass is so large, that force has no noticeable effect on Earth's motion.

CED 2.3.A.1 · Read this in Fix

1. AThe forces are equal only in equilibrium; while the cart speeds up the horse's is larger.
   A student who thinks the third law holds only when nothing accelerates picks this. The horse's pull on the cart and the cart's pull on the horse are equal in magnitude at every instant, including while the cart speeds up; the argument fails for a different reason.
2. BThe cart exerts no force on the horse, since a cart cannot pull on anything by itself.
   A student who thinks a passive object exerts no force picks this. The cart is attached to the horse, so it does pull back on it, with a force equal in magnitude to the horse's pull; the argument is right about that.
3. CThe cart's pull starts after the horse's pull, so the two forces do not overlap in time.
   A student who thinks a reaction follows the action picks this. The two forces of an interaction begin and end together, so they do act at the same time; they do not cancel because they are exerted on different objects.
4. DThe two forces are exerted on different objects, so they cannot cancel each other. Correct
   The horse's pull is exerted on the cart and the cart's pull is exerted on the horse. Forces cancel only if they are exerted on the same object, so these two never cancel. Whether the cart starts moving depends only on the forces exerted on the cart: the horse's pull and the ground's forces on it.

CED 2.3.A.1 · Read this in Fix

1. AmL/M
   A student who takes L as the person's displacement relative to the water writes mL + MΔx_boat = 0 and gets mL/M. L is measured along the moving boat, so the person's displacement relative to the water is L minus the distance the boat moves backward.
2. BL/2
   A student who thinks equal forces make the person and the boat move equal distances, which add up to L, picks this. The center of mass stays fixed, so the displacements are in the inverse ratio of the masses; they are equal only if m = M.
3. CmL/(m + M) Correct
   The forces between the person and the boat are internal, so the center of mass of the person–boat system stays at rest: mΔx_person + MΔx_boat = 0. The distance L is measured along the boat, so Δx_person = Δx_boat + L. Solving gives Δx_boat = −mL/(m + M): the boat moves a distance mL/(m + M) backward.
4. D0
   A student who thinks internal forces cannot move any part of a system picks this. The center of mass cannot move, but the person moves forward, so the boat must move backward to keep the mass-weighted average position fixed.

Working System person + boat: no horizontal external force, so the forces between the person's feet and the deck are internal and the center of mass stays where it was: mΔx_p + MΔx_b = 0. The walk is measured along the boat: Δx_p = Δx_b + L. So m(Δx_b + L) + MΔx_b = 0, Δx_b = −mL/(m + M). The boat moves mL/(m + M) backward. Checks: M ≫ m gives 0; M = m gives L/2.

CED 2.3.A.2 · Read this in Fix

1. A6.0 m/s²
   A student who thinks objects joined by a string always move together gives B the same acceleration as A. Two segments of string support B, so when A moves down a distance d, B rises only d/2, and B's acceleration is half of A's.
2. B4.5 m/s²
   A student who takes the velocity at an instant as the acceleration computes B's velocity, v_A/2 = (1.0 m/s³)(3.0 s)²/2 = 4.5 m/s, and gives that number. The acceleration is the rate of change of velocity: a_B = (1/2)dv_A/dt = 3.0 m/s².
3. C1.5 m/s²
   A student who divides B's velocity at t = 3.0 s by the elapsed time gets 4.5 m/s ÷ 3.0 s = 1.5 m/s². That is B's average acceleration since t = 0; because v_A grows as t², the acceleration grows with time, and its value at 3.0 s is the derivative, 3.0 m/s².
4. D3.0 m/s² Correct
   The ideal string keeps its length: with A one segment below the fixed pulley and B supported by two segments, y_A + 2y_B is constant, so B's velocity and acceleration are half of A's. a_A = dv_A/dt = (2.0 m/s³)t = 6.0 m/s² at t = 3.0 s, so a_B = 3.0 m/s².

Working The string does not stretch. Measuring each block's distance below the ceiling, the string's length is y_A + 2y_B + constant (one segment to A, two segments supporting the movable pulley and B). So Δy_A = −2Δy_B: v_B = v_A/2 and a_B = a_A/2. a_A = dv_A/dt = (2.0 m/s³)t = 6.0 m/s² at t = 3.0 s, so a_B = 3.0 m/s² (B moves upward, speeding up).

CED 2.3.A.3.i · Read this in Fix

1. ASetup 1 reads 60 N, and setup 2 reads 120 N.
   A student who adds the two pulls picks this. The two 60 N pulls are exerted in opposite directions on the two ends; the tension at every point is the 60 N pull at one end. In setup 1 the wall also pulls with 60 N, so the two setups are identical for the rope.
2. BSetup 1 reads 60 N, and setup 2 reads 0 N.
   A student who thinks equal pulls cancel picks this. The two pulls give the rope zero net force, but they stretch it, and the tension throughout is 60 N; setup 1 is in fact the same, with the wall pulling instead of a student.
3. CSetup 1 reads 60 N, and setup 2 also reads 60 N. Correct
   In each setup a student pulls one end with 60 N, so the tension at that end is 60 N, and in a light rope and ideal scale the tension is the same at every point: the scale reads 60 N. In setup 1 the wall pulls on its end of the rope with 60 N, exactly as the second student does in setup 2.
4. DBoth read less than 60 N, as pulls fade along a rope.
   A student who thinks a pull weakens along a rope picks this. In a light rope the tension is the same at every point, so the scale in the middle reads the full 60 N.

Working Each student's hand pulls the rope with 60 N, so the rope pulls the hand with 60 N (third law) and the tension at that end is 60 N. Ideal (light) rope and scale: same tension at all points, so the scale reads 60 N. Setup 1: the wall end of the rope also has tension 60 N, so the wall pulls on the rope with 60 N, just as the second student does in setup 2. Both read 60 N.

CED 2.3.A.3.ii · Read this in Fix

1. A(1/2)w₀L
   A student who thinks the tension is the same at every point of any rope gives the midpoint the full weight of the rope, ∫ from 0 to L of w dy = (1/2)w₀L. That is the tension at the top only; this rope's mass is not negligible, and its tension falls to zero at the bottom.
2. B(1/4)w₀L
   A student who multiplies the weight per unit length at the midpoint, w₀/2, by the length below it, L/2, gets (1/4)w₀L. The weight per unit length grows toward the bottom, so the weight below the midpoint must be found by integrating w(y).
3. C(1/8)w₀L
   A student who thinks the tension at a point supports the rope between that point and the ceiling integrates from 0 to L/2 and gets (1/8)w₀L. The tension at the midpoint holds up the part of the rope below it, from L/2 to L.
4. D(3/8)w₀L Correct
   Cut the rope at the midpoint. The part below is at rest, and the only upward force on it is the tension at the cut, which therefore equals that part's weight: ∫ from L/2 to L of (w₀y/L) dy = (w₀/L)(L²/2 − L²/8) = (3/8)w₀L.

Working The part of the rope below the midpoint (y from L/2 to L) is at rest; the only forces on it are its weight (down) and the tension at the cut (up), so T(L/2) = ∫_L/2^L (w₀y/L) dy = (w₀/L)[y²/2]_L/2^L = (w₀/L)(L²/2 − L²/8) = (3/8)w₀L. Check: total weight ∫₀^L w dy = w₀L/2 = tension at the top; tension at the bottom = 0.

CED 2.3.A.3.iii · Read this in Fix

1. ATension is the same at all points of any rope or chain, so the chain's sensors are faulty.
   A student who applies the ideal-string model to every rope picks this. The model assumes negligible mass; the chain's mass is not negligible, and its readings show the tension that this predicts, increasing toward the top.
2. BThe chain's tension varies along it since its mass is not negligible; the string's does not. Correct
   The string's three readings are equal, as the ideal-string model predicts for a string of negligible mass. The chain's readings increase from 10 N at the bottom to 30 N at the top: each point of the chain also holds up the chain below it, so in a chain with nonnegligible mass the tension is not the same at all points.
3. CTension decreases along any rope or chain away from its support, as both sets of data show.
   A student who thinks a pull fades along a rope picks this. The string's readings are equal at all three points, so the data do not show a decrease for the string; only the chain's tension varies, because of its mass.
4. DEach tension equals the weight of the block, because the block is what is being held up.
   A student who sets every tension equal to the hanging block's weight picks this. That fits the string, but the chain's top and middle readings are larger than 10 N, because those points also hold up part of the chain.

CED 2.3.A.3.iii · Read this in Fix

1. AThe pulley is not an ideal pulley: its mass or friction is not negligible. Correct
   For a light string over an ideal pulley, one of negligible mass and with negligible axle friction, the tension is the same on both sides, whether or not the blocks accelerate. Different tensions on the two sides mean that the pulley itself is not ideal: its mass, or friction (for example at its axle), accounts for the difference while it speeds up.
2. BA string over any pulley has more tension on the side of the heavier block.
   A student who links each tension to the weight on its side picks this. Over an ideal pulley the tension in a light string has one value on both sides whatever the two weights; only a pulley with mass or friction lets the heavier block's side carry more tension, as here.
3. CEvery pulley, even an ideal one, passes on less force than it receives.
   A student who thinks a single pulley reduces the force picks this. An ideal pulley changes only the direction of the string; with an ideal pulley both sensors would give the same reading.
4. DTension is the same throughout a string only when nothing is speeding up.
   A student who thinks the tensions must differ for the blocks to speed up picks this. Over an ideal pulley the tension is the same on both sides even while the blocks speed up; what differs is the tension and the weight on each block.

CED 2.3.A.3.iv · Read this in Fix

1. A2.00T
   A student who adds the magnitudes of the two segments' pulls picks this. The pulls are perpendicular, so they add as vectors: √2T, not 2T. The total would be 2T only if the two segments were parallel.
2. B1.41T Correct
   The ideal string has tension T on both sides of the ideal pulley. Where it wraps around the pulley it pulls on it along both straight segments: with a force T horizontally toward block 1 and a force T straight down toward block 2. These are perpendicular, so the total force has magnitude √(T² + T²) = √2T ≈ 1.41T.
3. C0.00T
   A student who thinks a string pulls only on the objects tied to its ends picks this. The string presses on the pulley where it wraps around it, and the pulley must be held by its bracket against that force.
4. D1.00T
   A student who thinks a pulley only redirects one force T picks this. Each of the two segments pulls on the pulley with a force T, and the two perpendicular pulls add to √2T.

Working Ideal string over an ideal pulley: tension T in both straight segments. The string touches the pulley along a quarter turn and pulls it along each segment, away from the pulley: T toward block 1 (horizontal) and T toward block 2 (straight down). Sum: (−T, −T), magnitude √2T ≈ 1.41T, directed 45° below the horizontal toward the table.

CED 2.3.A.3.ii · Read this in Fix