4.1 Linear Momentum

1. Ap = m(Ct² − Dt³)
   A student who takes the velocity from the position itself, rather than from its rate of change, picks this: it is m times x(t). Velocity is dx/dt, the slope of the position–time graph, not its height.
2. Bp = 2m(C − 3Dt)
   A student who computes momentum as mass times acceleration picks this: d²x/dt² = 2C − 6Dt, and m(2C − 6Dt) = 2m(C − 3Dt). That product is the net force; momentum is mass times velocity, dx/dt.
3. Cp = mt(C − Dt)
   A student who divides the position by the time picks this: x/t = Ct − Dt², and m(x/t) = mt(C − Dt). That is an average velocity since t = 0, not the velocity at time t, which is dx/dt = 2Ct − 3Dt².
4. Dp = m(2Ct − 3Dt²) Correct
   Momentum is mass times velocity, and the velocity is the derivative of position: v = dx/dt = 2Ct − 3Dt². So p = m(2Ct − 3Dt²).

Working v = dx/dt = 2Ct − 3Dt², so p = mv = m(2Ct − 3Dt²). Checks of the distractors: mx = m(Ct² − Dt³); m(x/t) = mt(C − Dt); m(dv/dt) = m(2C − 6Dt) = 2m(C − 3Dt).

CED 4.1.A.1 · Read this in Fix

1. ADown and to the right, along the path at point P Correct
   Momentum, p⃗ = mv⃗, has the direction of the velocity. The velocity is tangent to the path in the direction of travel, and at P, past the highest point, the ball is moving down and to the right.
2. BStraight down, the direction of the gravitational force
   A student who thinks an object's momentum points along the net force on it picks this. The gravitational force points down and changes the vertical velocity, but the momentum points along the velocity, which at P is down and to the right.
3. CUp and to the right, parallel to the launch velocity
   A student who thinks a launched ball keeps the motion it was given picks this. Once launched, the vertical velocity component changes while the horizontal one stays constant, so the direction of motion turns along the path.
4. DUp and to the right, from the launch point to P
   A student who takes the direction of velocity to be the direction of the displacement from the start picks this. That line gives where the ball is relative to the launch point, not the direction in which it is moving at P.

Working Momentum has the direction of the velocity, which is tangent to the path in the direction of travel. At P the ball is past the highest point and moving down and to the right.

CED 4.1.A.2 · Read this in Fix

1. AIt is valid, since only the cars' states just before and just after the collision are used. Correct
   A collision analysis compares the states just before and just after the collision, and each car's state is described by its mass and velocity. How the cars deform during the collision does not enter, so each can be modeled as an object.
2. BIt is not valid, since each car's shape changes while the two cars are in contact with each other.
   A student who thinks a body that deforms cannot be modeled as an object picks this. The deformation happens during the collision, but the analysis uses only the states just before and just after it, so the deformation need not be described.
3. CIt is not valid, since each car is large compared with how far it moves during the collision.
   A student who thinks only tiny bodies can be modeled as objects picks this. Whether the object model applies depends on whether a body's internal structure matters to the analysis, not on its size.
4. DIt is valid, since each car's mass really is concentrated at its center of mass.
   A student who takes the object model literally picks this. A car's mass is spread through the car; the object model is a modeling choice that is justified here because only the states just before and just after the collision are analyzed.

CED 4.1.A.3.ii · Read this in Fix

1. A+2.0 kg·m/s
   A student who treats momentum as a size with no direction picks this: 0.50 kg × 4.0 m/s = 2.0 kg·m/s. The position is decreasing, so the velocity, and with it the momentum, is negative.
2. B−2.0 kg·m/s Correct
   At t = 5.0 s the cart is on the straight segment from (2 s, 6 m) to (6 s, −10 m), whose slope is −16 m/4 s = −4.0 m/s. The momentum is mv = (0.50 kg)(−4.0 m/s) = −2.0 kg·m/s: negative because the cart is moving in the −x direction.
3. C−3.0 kg·m/s
   A student who reads the velocity from the height of the graph picks this: at t = 5.0 s, x = −6 m, and 0.50 × (−6) = −3.0. The velocity is the slope of the position–time graph, −4.0 m/s, not its height.
4. D+1.2 kg·m/s
   A student who divides the distance traveled by the elapsed time picks this: the cart has gone 12 m in 5.0 s, an average speed of 2.4 m/s, and 0.50 × 2.4 = 1.2. The cart was at rest for the first 2 s, so that average is not its velocity at t = 5.0 s, which is −4.0 m/s.

Working From t = 2 s to t = 6 s the graph is a straight line from x = 6 m to x = −10 m, so the velocity throughout that segment, including at t = 5.0 s, is v = (−10 m − 6 m)/(4 s) = −4.0 m/s. p = mv = (0.50 kg)(−4.0 m/s) = −2.0 kg·m/s.

CED 4.1.A.2 · Read this in Fix

1. A4.8 kg·m/s
   A student who uses the cosine for the y-component picks this: (2.4)(2.5)(0.80) = 4.8. The angle is measured from the x-axis, so the y-component is the side opposite it, mv sin 37°.
2. B6.0 kg·m/s
   A student who takes a component to be the whole magnitude picks this: mv = (2.4)(2.5) = 6.0. The momentum points at 37° to the x-axis, so its y-component is only the part mv sin 37°.
3. C3.6 kg·m/s Correct
   The momentum points along the velocity, so its y-component is m times the velocity's y-component: p_y = mv sin 37° = (2.4 kg)(2.5 m/s)(0.60) = 3.6 kg·m/s.
4. D7.5 kg·m/s
   A student who calculates kinetic energy in place of momentum picks this: (1/2)(2.4)(2.5)² = 7.5, which is in joules, not kg·m/s. Momentum is mv, and its y-component is mv sin 37°.

Working Momentum has the direction of the velocity, so p_y = mv_y = mv sin 37° = (2.4 kg)(2.5 m/s)(0.60) = 3.6 kg·m/s.

CED 4.1.A.2 · Read this in Fix

1. AX > Z > W > Y
   A student who treats momentum as a size with no direction picks this: it ranks the magnitudes 12, 10, 8.0 and 1.0 kg·m/s. W and Y move to the left, so their momenta are negative, and W's, −8.0 kg·m/s, is the most negative.
2. BX > Z > Y > W Correct
   With +x to the right, p = mv gives X: +12 kg·m/s, Z: +10 kg·m/s, Y: −1.0 kg·m/s and W: −8.0 kg·m/s. The carts moving left have negative momenta, and W's is the most negative.
3. CZ > X > W > Y
   A student who ranks kinetic energies in place of momenta picks this: (1/2)mv² gives Z 25 J, X 24 J, W 8.0 J and Y 0.50 J. Momentum is mv, a vector, so X (+12 kg·m/s) has more than Z (+10 kg·m/s), and the left-moving carts have negative momenta.
4. DZ > X > Y > W
   A student who judges momentum by velocity alone picks this: it ranks the velocities +5.0, +4.0, −1.0 and −2.0 m/s. Momentum is mv, so X's larger mass gives it more momentum than the faster Z.

Working p = mv with +x to the right: W (4.0)(−2.0) = −8.0 kg·m/s; X (3.0)(+4.0) = +12 kg·m/s; Y (1.0)(−1.0) = −1.0 kg·m/s; Z (2.0)(+5.0) = +10 kg·m/s. Ranking: X > Z > Y > W. (Magnitudes: X > Z > W > Y. Kinetic energies: Z 25 J, X 24 J, W 8.0 J, Y 0.50 J. Velocities: Z > X > Y > W.)

CED 4.1.A.1 · Read this in Fix

1. A2.00 Correct
   Writing K = p²/(2m) gives p = √(2mK). With equal kinetic energies, p₁/p₂ = √(m₁/m₂) = √4 = 2.00. Equivalently, cart 1 moves at half the speed of cart 2, and 4 × 1/2 = 2.
2. B1.00
   A student who treats momentum and kinetic energy as the same quantity picks this: equal kinetic energies would mean equal momenta. Since K = p²/(2m), equal K gives p proportional to √m, so the heavier cart has the larger momentum.
3. C4.00
   A student who judges momentum by mass alone picks this: four times the mass, four times the momentum. With equal kinetic energies the heavier cart moves more slowly, at half the speed, so its momentum is only twice as large.
4. D0.50
   A student who judges momentum by speed alone picks this: cart 1 moves at half the speed of cart 2. Momentum is mv, and cart 1's four-times-larger mass more than makes up for its lower speed.

Working K = (1/2)mv² = p²/(2m), so p = √(2mK). With equal K, p₁/p₂ = √(m₁/m₂) = √4 = 2.00. (Equivalently, v₁/v₂ = √(m₂/m₁) = 1/2, and p₁/p₂ = 4 × 1/2 = 2.)

CED 4.1.A.1 · Read this in Fix

1. A0.00mv₀
   A student who thinks a projectile is momentarily at rest at the top of its path picks this. Only the vertical component of the velocity is zero there; the ball still moves horizontally at v₀ cos 60° = 0.50v₀.
2. B1.00mv₀
   A student who thinks the ball keeps the motion it was given at launch picks this. The vertical component of the velocity decreases to zero on the way up, so at the top the ball moves more slowly than at launch, at v₀ cos 60° = 0.50v₀.
3. C0.50mv₀ Correct
   At the highest point the vertical velocity component is zero, while the horizontal component keeps its launch value, v₀ cos 60° = 0.50v₀, because no horizontal force acts. The momentum is m times that velocity: 0.50mv₀, directed horizontally.
4. D0.87mv₀
   A student who takes the horizontal component to be v₀ sin 60° picks this. The angle is measured from the horizontal, so the horizontal component is v₀ cos 60° = 0.50v₀; v₀ sin 60° is the launch velocity's vertical component, which is zero at the top.

Working At the highest point v_y = 0. The horizontal component is unchanged from launch: v_x = v₀ cos 60° = 0.50v₀. So the speed there is 0.50v₀ and p = m(0.50v₀) = 0.50mv₀.

CED 4.1.A.2 · Read this in Fix

1. AIt does not apply, since a friction force is exerted on cart 1 throughout the contact.
   A student who thinks the collision model needs zero external force picks this. The model requires only that the forces between the objects be much larger than the net external force, and 0.50 N is tiny compared with the 60 N force between the carts.
2. BIt applies, since no friction force is exerted on cart 1 while the carts are in contact.
   A student who thinks external forces stop acting during a collision picks this. The dashed line shows friction acting on cart 1 for the whole time, including the contact; the model neglects it because it is small, not because it is absent.
3. CIt does not apply, since the force exerted by cart 2 on cart 1 varies during the contact.
   A student who thinks only constant forces can be analyzed picks this. The forces in real collisions rise and fall rapidly; the collision model needs them to be much larger than the net external force, not constant.
4. DIt applies, since cart 2's force on cart 1 is far larger than the friction force. Correct
   During the contact the force between the carts reaches about 60 N, more than 100 times the 0.50 N friction force on each cart. The forces between the objects are much larger than the net external force, which is what the collision model requires.

Working Peak interaction force ≈ 60 N; friction on each cart ≈ 0.50 N, so the net external force on the carts is about 1 N or less, roughly 1% of the peak force between them. The collision model applies.

CED 4.1.A.3.i · Read this in Fix

1. AThe spring pushes on cart 2 from outside that cart, so an external force pushes them apart.
   A student who calls a force external whenever it comes from something other than the object it acts on picks this. For the two-cart system the spring is inside the boundary, so its forces are internal, and the explosion model requires internal forces.
2. BThe spring stores momentum while compressed and gives that momentum to the carts on release.
   A student who thinks momentum can be stored like energy picks this. The compressed spring stores elastic potential energy; the carts at rest have zero momentum, and the spring's push gives them momenta in opposite directions.
3. CThe spring is part of the two-cart system, so the forces that push the carts apart are internal. Correct
   An explosion is modeled as an interaction in which forces internal to the system move objects within it apart. The spring is attached to cart 1, so it is inside the two-cart system, and the forces it exerts on the carts are internal forces.
4. DThe carts end up apart, and any interaction that ends with objects apart is an explosion.
   A student who classifies an interaction by its outcome picks this. Objects also move apart after many collisions, such as a bounce; what makes this an explosion is that internal forces push apart carts that were together.

CED 4.1.A.3.iii · Read this in Fix

1. A−12 kg·m/s
   A student who calculates momentum as mass times acceleration picks this: the slope of the graph is −3.0 m/s², and 4.0 × (−3.0) = −12. That product is the net force on the cart; momentum is mass times velocity, the height of the graph.
2. B+24 kg·m/s
   A student who treats momentum as a size with no direction picks this: (4.0 kg)(6.0 m/s) = 24 kg·m/s. At t = 4.0 s the velocity is negative, so the momentum is negative.
3. C+72 kg·m/s
   A student who calculates kinetic energy in place of momentum picks this: (1/2)(4.0)(6.0)² = 72, which is in joules. Momentum is mv = −24 kg·m/s.
4. D−24 kg·m/s Correct
   On a velocity–time graph the height gives the velocity: at t = 4.0 s, v = −6.0 m/s. The momentum is mv = (4.0 kg)(−6.0 m/s) = −24 kg·m/s, negative because the cart is then moving in the −x direction.

Working From the graph, v = 6 m/s − (3 m/s²)t, so v(4.0 s) = −6.0 m/s. p = mv = (4.0 kg)(−6.0 m/s) = −24 kg·m/s.

CED 4.1.A.1 · Read this in Fix