4.1 Linear Momentum

1. A4.5
   A student who treats momentum like kinetic energy squares the speed factor and picks this: (1/2)(3²) = 4.5. That is the factor for the kinetic energy, K = (1/2)mv². Momentum depends on the first power of the speed, so the factor is (1/2)(3) = 1.5.
2. B3.0
   A student who thinks momentum depends on speed alone picks this, using only the factor 3. The halved mass matters just as much as the tripled speed: p = mv, so the factor is (1/2)(3) = 1.5.
3. C0.5
   A student who thinks the more massive ball must have more momentum uses only the mass factor and picks this. Ball 2 is lighter but much faster; the product mv is (1/2)(3) = 1.5 times as large.
4. D1.5 Correct
   Momentum is proportional to both the mass and the speed, p = mv. Halving the mass multiplies the momentum by 1/2 and tripling the speed multiplies it by 3, so ball 2's momentum is (1/2)(3) = 1.5 times ball 1's.

Working p = mv. Ball 1: p₁ = mv. Ball 2: p₂ = (m/2)(3v) = (3/2)mv. Ratio p₂/p₁ = 3/2 = 1.5.

CED 4.1.A.1 · Read this in Fix

1. AIt points downward, the direction of the gravitational force on the ball.
   A student who thinks momentum points along the net force picks this. The gravitational force is downward, and it is slowing the ball, but the ball is still moving upward, and momentum has the direction of the velocity.
2. BIt has no direction, since it is a product of a mass and a speed.
   A student who treats momentum as a scalar picks this. Momentum is the product of the mass and the velocity, a vector, so it has the velocity's direction: upward while the ball rises.
3. CIt points upward, the direction of the ball's velocity at that instant. Correct
   Momentum is a vector with the direction of the velocity, p⃗ = mv⃗. While the ball rises its velocity is upward, so its momentum is upward, even though the only force on it, the gravitational force, is downward.
4. DIt points upward, as the ball still carries the upward force of the throw.
   A student who thinks a moving object carries the force that set it moving picks this. The hand's force ends when the ball leaves the hand; after that only Earth exerts a force on the ball. The momentum is upward because the velocity is upward, and momentum is not a force.

CED 4.1.A.2 · Read this in Fix

1. AIt is not an explosion, since nothing burns and no fuel is used.
   A student who thinks an explosion needs fire or fuel picks this. The physics model needs only that forces internal to the system push objects in the system apart, which the spring does.
2. BIt is not an explosion, since the spring's forces on the carts are external.
   A student who thinks the force that starts the motion must come from outside the system picks this. The spring is part of the chosen system, so the forces it exerts on the carts are internal.
3. CIt is an explosion: the spring, part of the system, pushes the carts apart. Correct
   An explosion is an interaction in which forces internal to the system move objects in the system apart. The spring is in the system, so its forces on the carts are internal, and they push the carts apart.
4. DIt is a collision, since the carts are in contact with the spring throughout.
   A student who thinks any contact interaction is a collision picks this. The carts start at rest and are pushed apart by forces from inside the system, which is the explosion model.

CED 4.1.A.3.iii · Read this in Fix

1. AThe truck's momentum is larger, since the truck has more mass.
   A student who thinks the heavier vehicle must have more momentum picks this. Momentum depends on mass and velocity together: the truck has 4 times the car's mass but only 1/4 of its speed, so the products are equal.
2. BTheir momenta are equal in size, since mv is the same for both. Correct
   Car: (1200 kg)(20 m/s) = 24 000 kg·m/s. Truck: (4800 kg)(5 m/s) = 24 000 kg·m/s. The truck's 4 times larger mass exactly makes up for its 4 times smaller speed.
3. CThe car's momentum is larger, since the car is moving faster.
   A student who thinks momentum is set by speed alone picks this. The car is 4 times as fast, but the truck has 4 times the mass, so mv is 24 000 kg·m/s for each.
4. DThe car's momentum is larger, since the car has more kinetic energy.
   A student who treats momentum and kinetic energy as the same quantity picks this. The car does have 4 times the truck's kinetic energy, because K depends on v², but momentum depends on v, and mv is equal for the two vehicles.

Working Car: p = (1200 kg)(20 m/s) = 24 000 kg·m/s. Truck: p = (4800 kg)(5 m/s) = 24 000 kg·m/s. Equal magnitudes (and both east). Kinetic energies differ: car (1/2)(1200)(20²) = 240 000 J; truck (1/2)(4800)(5²) = 60 000 J.

CED 4.1.A.1 · Read this in Fix

1. A−2.0 kg·m/s Correct
   The velocity is the slope of the position–time graph: (0 − 12 m)/(3.0 s) = −4.0 m/s, the same at every instant because the line is straight. The momentum is mv = (0.50 kg)(−4.0 m/s) = −2.0 kg·m/s: 2.0 kg·m/s directed to the left, the direction of the velocity.
2. B+2.0 kg·m/s
   A student who treats momentum as a size without direction picks this. The magnitude is right, but the position decreases with time, so the cart moves in the negative direction, and its momentum points that way: −2.0 kg·m/s.
3. C−4.0 kg·m/s
   A student who takes momentum to be set by the velocity alone, leaving out the mass, picks this: the slope gives v_x = −4.0 m/s. Momentum is the mass times the velocity, (0.50 kg)(−4.0 m/s) = −2.0 kg·m/s; a more massive cart with the same velocity would have more momentum.
4. D+3.0 kg·m/s
   A student who reads the height of the graph at t = 1.5 s, 6.0 m, as the velocity picks this: (0.50)(6.0) = 3.0. The height of a position–time graph is the position; the velocity is the slope, −4.0 m/s, so p = −2.0 kg·m/s.

Working The graph is a straight line, so the velocity is constant and equals the slope: v_x = (0 − 12 m)/(3.0 s − 0) = −4.0 m/s. p_x = mv_x = (0.50 kg)(−4.0 m/s) = −2.0 kg·m/s (2.0 kg·m/s to the left).

CED 4.1.A.2 · Read this in Fix

1. AAt t = 4 s it is the same as at t = 0, since the cart's speed is then the same.
   A student who treats momentum as a size without direction picks this. The speed at t = 4 s equals the speed at t = 0, but the velocity has reversed, so the momentum has reversed too. Momenta in opposite directions are different.
2. BAt t = 4 s it has the magnitude it had at t = 0, but the opposite direction. Correct
   The graph shows v = +3 m/s at t = 0 and v = −3 m/s at t = 4 s. Momentum is mv with the direction of the velocity, so it is 1.5 kg·m/s up the ramp at t = 0 and 1.5 kg·m/s down the ramp at t = 4 s: equal magnitudes, opposite directions.
3. CAt t = 4 s its magnitude is smaller than at t = 0, since the velocity is negative.
   A student who reads a negative momentum as a smaller one picks this. The minus sign shows only that the cart is moving down the ramp. The magnitudes at the two instants are both (0.50 kg)(3 m/s) = 1.5 kg·m/s.
4. DFrom t = 0 to t = 4 s it points down the ramp, the direction of acceleration.
   A student who gives momentum the direction of the acceleration picks this. The acceleration, the slope of the graph, is down the ramp throughout, but from t = 0 to t = 2 s the velocity is positive, so the momentum then points up the ramp.

Working At t = 0, v = +3 m/s, so p = (0.50 kg)(+3 m/s) = +1.5 kg·m/s (up the ramp). At t = 4 s, v = −3 m/s, so p = −1.5 kg·m/s (down the ramp): the same magnitude, opposite direction. From t = 0 to 2 s the momentum is up the ramp although the acceleration (slope −1.5 m/s²) is down the ramp.

CED 4.1.A.2 · Read this in Fix

1. ANo external forces are exerted on the cars while they are in contact with each other.
   A student who thinks the collision model needs zero external force picks this. Gravity, the normal force and friction from the road all act on the cars during the crash. The model needs only that the net external force is small compared with the forces between the cars.
2. BThe cars exert equal and opposite forces on each other, so each car has zero net force.
   A student who thinks third-law forces cancel picks this. The two forces act on different cars, so they do not cancel on either car. Each car has a large net force exerted on it during the crash, which is why each car's velocity changes.
3. CDuring the crash, the forces of the cars on each other far exceed the road's friction on them. Correct
   A collision is an interaction in which the forces the objects exert on each other are much larger than the net external force on them. External forces act on the cars, but during the brief crash the road's friction is negligible compared with the huge forces between the cars.
4. DThe cars touch each other, and it is contact that makes an interaction a collision.
   A student who defines a collision by contact picks this. Many contact interactions, such as pushing a box slowly across a rough floor, are not collisions. The test is whether the forces between the objects are much larger than the net external force.

CED 4.1.A.3.i · Read this in Fix

1. AOnly trial 2, since there A's force on B is much larger than the friction on B. Correct
   In trial 2 the force of A on B peaks at 60 N while the friction on B is 2 N, so the external force is negligible during the interaction and the collision model applies. In trial 1 the 3 N peak is comparable to the 2 N friction, so the collision model does not apply, even though that interaction is short.
2. BOnly trial 1, since its interaction is the one lasting less than a second.
   A student who defines a collision by its short duration picks this. Duration is not the test: trial 1 is brief, but there A's force on B (at most 3 N) is not much larger than the 2 N friction on B. In trial 2 the force between the blocks dwarfs the friction, so that interaction is the collision.
3. CBoth trials, since blocks A and B are in contact during each interaction.
   A student who defines a collision by contact picks this. Contact is not the test: in trial 1 the contact force of A on B peaks at 3 N, comparable to the 2 N friction on B, so the external force is not negligible and the collision model does not apply.
4. DNeither trial, since the floor exerts friction on B throughout each interaction.
   A student who thinks the collision model needs zero external force picks this. The model needs the external force to be negligible compared with the forces between the objects, not zero. In trial 2 the 2 N friction is negligible beside the 60 N force of A on B, so that interaction is a collision.

Working Collision model test: the forces between the objects must be much larger than the net external force during the interaction. Trial 1: peak force of A on B = 3 N; friction on B = 2 N. These are comparable, so trial 1 is not a collision (although it lasts only 0.06 s). Trial 2: peak force of A on B = 60 N; friction on B = 2 N, 30 times smaller, so the external force is negligible and trial 2 is a collision (although it lasts 2 s).

CED 4.1.A.3.i · Read this in Fix

1. ANo, since the ball changes shape while it is in contact with the racket.
   A student who thinks any deformation rules out the object model picks this. The flattening happens during the contact, and the analysis uses only the ball's motion just before and just after the hit, when its shape does not matter.
2. BOnly if the force on the ball is known at each instant of the hit.
   A student who thinks every analysis must start from the forces at each moment picks this. Collision analysis compares the initial and final states; how the force varies during the contact is not needed.
3. CYes, since the ball is much smaller than the racket that strikes it.
   A student who thinks the object model is for small objects picks this. Size is not the test: the model may be used because the analysis needs nothing about the ball's size, shape or internal configuration, only its states before and after the hit.
4. DYes, since only its motion just before and just after the hit is compared. Correct
   A collision is analyzed by comparing the initial and final states. The ball's change of shape during the contact does not enter that comparison, so the ball can be treated as a single point with mass.

CED 4.1.A.3.ii · Read this in Fix

1. Ap = √(2K/m)
   A student who treats momentum as set by speed alone finds the speed, v = √(2K/m), and stops there. Momentum is the mass times that speed: m√(2K/m) = √(2mK).
2. Bp = √(2mK) Correct
   From K = (1/2)mv², v = √(2K/m). Then p = mv = m√(2K/m) = √(2mK). Check: p² = m²v² = 2m · (1/2)mv² = 2mK.
3. Cp = √(mK)
   A student who uses K = mv² without the factor 1/2 gets v = √(K/m) and p = m√(K/m) = √(mK). With K = (1/2)mv² the result is √(2mK).
4. Dp = 2mK
   A student who finds p² = 2mK and leaves out the square root picks this. The units show the error: 2mK has units kg²·m²/s², the units of p². Taking the square root gives √(2mK).

Working K = (1/2)mv², so v = √(2K/m). p = mv = m√(2K/m) = √(m² · 2K/m) = √(2mK). Check: p² = m²v² = 2m · (1/2)mv² = 2mK.

CED 4.1.A.1 · Read this in Fix

1. A1.0 mv
   A student who treats momentum and kinetic energy as the same quantity gives the two objects equal momenta because their kinetic energies are equal. Kinetic energy depends on v², momentum on v: B's half speed and four times the mass give it twice A's momentum.
2. B0.5 mv
   A student who thinks momentum depends on speed alone sees that B moves at half A's speed and halves A's momentum. B's mass is four times A's, so p_B = (4m)(v/2) = 2.0 mv.
3. C4.0 mv
   A student who thinks momentum is set by mass, like inertia, multiplies A's momentum by the mass ratio, 4. B moves at only half A's speed, so its momentum is (4m)(v/2) = 2.0 mv.
4. D2.0 mv Correct
   Equal kinetic energies with four times the mass mean B moves at v/2, since (1/2)(4m)(v/2)² = (1/2)mv². B's momentum is (4m)(v/2) = 2.0 mv, twice A's.

Working Equal kinetic energies: (1/2)(4m)v_B² = (1/2)mv², so v_B² = v²/4 and v_B = v/2. The magnitude of B's momentum is p_B = (4m)(v/2) = 2.0 mv, twice A's momentum mv.

CED 4.1.A.1 · Read this in Fix