AP Physics 1 · Algebra-Based

An object moves along a straight line. Its velocity increases uniformly from 0 to 8.0 m/s during the first 4.0 s, stays at 8.0 m/s for the next 3.0 s, then decreases uniformly to 0 during the final 2.0 s. What total distance does the object travel?

1. A40 m
2. B48 m
3. C56 m
4. D72 m

A person of mass m stands on a bathroom scale in an elevator. The elevator is moving downward and slowing down. What does the scale read?

1. Aless than mg
2. Bexactly mg
3. Cgreater than mg
4. Dzero

A car traveling at speed v brakes to rest in a distance d. Assuming the same constant braking force, what distance is required to bring the same car to rest from a speed of 2v?

1. Ad√2
2. B2d
3. C4d
4. D8d

A 0.50 kg ball moving at 4.0 m/s to the right collides head-on with a 1.5 kg ball moving at 2.0 m/s to the left. The two balls stick together. What is the velocity of the combined object immediately after the collision?

1. A0.50 m/s to the left
2. B0.50 m/s to the right
3. C2.5 m/s to the left
4. D2.5 m/s to the right

A solid disk and a thin hoop have the same mass and the same radius. Both are released from rest at the top of the same incline and roll down without slipping. Which reaches the bottom first?

1. Athe disk, because a smaller fraction of its energy goes into rotation
2. Bthe hoop, because its mass is farther from the axis
3. Cneither — they arrive together, because they have the same mass
4. Dneither — they arrive together, because gravity acts equally on both

A solid block of density 600 kg/m³ floats at rest in water of density 1000 kg/m³. What fraction of the block's volume is below the surface?

1. A0.40
2. B0.60
3. C1.00
4. D1.67

A block of mass m is released from rest at a height h on an incline that makes an angle θ with the horizontal. The coefficient of kinetic friction between the block and the incline is μ. The block slides down the incline and then continues onto a horizontal surface with the same coefficient of kinetic friction.

1. (a)

   Derive Derive an expression for the speed of the block at the bottom of the incline, in terms of m, h, θ, μ and physical constants.

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   1. 1

      1 point: uses energy conservation with a friction term, ΔK = mgh − f·L

   2. 2

      1 point: identifies the distance along the incline as L = h / sin θ and the friction force as f = μmg cos θ

   3. 3

      1 point: friction work = μmg cos θ · (h / sin θ) = μmgh cot θ

   4. 4

      1 point: v = √(2gh(1 − μ cot θ)), with m correctly cancelling

2. (b)

   Calculate Calculate the speed of the block at the bottom of the incline for m = 2.0 kg, h = 1.5 m, θ = 30°, and μ = 0.25.

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   1. 1

      1 point: cot 30° = 1.73, so 1 − 0.25(1.73) = 0.567

   2. 2

      1 point: v = √(2 × 9.8 × 1.5 × 0.567)

   3. 3

      1 point: v = 4.1 m/s, with unit

3. (c)

   Determine Determine the distance the block travels along the horizontal surface before coming to rest.

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   1. 1

      1 point: on the horizontal surface the only horizontal force is friction, μmg

   2. 2

      1 point: ½mv² = μmg·d, so d = v²/(2μg)

   3. 3

      1 point: d = 16.7 / (2 × 0.25 × 9.8) = 3.4 m

4. (d)

   Justify A student claims that doubling the mass of the block would double the distance found in part (c). Justify why the student is incorrect, referring to your expression.

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   1. 1

      1 point: mass appears in both the kinetic energy and the friction force and cancels from d = v²/(2μg)

   2. 2

      1 point: v itself is also independent of m, so the distance is unchanged when the mass is doubled

A cart moves along a straight horizontal track. Its velocity as a function of time is described as follows: from t = 0 to t = 2.0 s the velocity is constant at +3.0 m/s; from t = 2.0 s to t = 5.0 s the velocity decreases uniformly from +3.0 m/s to −3.0 m/s; from t = 5.0 s to t = 7.0 s the velocity is constant at −3.0 m/s.

1. (a)

   Sketch Sketch a graph of the cart's acceleration as a function of time from t = 0 to t = 7.0 s. Label the value of the acceleration on each interval.

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   1. 1

      1 point: a = 0 on 0 ≤ t < 2.0 s and on 5.0 s < t ≤ 7.0 s

   2. 2

      1 point: a constant and negative between 2.0 s and 5.0 s

   3. 3

      1 point: value stated as −2.0 m/s², from (−3.0 − 3.0)/(5.0 − 2.0)

2. (b)

   Determine Determine the time at which the cart is farthest from its starting point, and determine that distance.

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   1. 1

      1 point: the cart is farthest away when the velocity changes sign, at t = 3.5 s

   2. 2

      1 point: area under the graph from 0 to 3.5 s = (3.0)(2.0) + ½(1.5)(3.0)

   3. 3

      1 point: distance = 6.0 + 2.25 = 8.25 m

3. (c)

   Sketch Sketch a graph of the cart's position as a function of time over the same interval, taking the starting position as zero.

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   1. 1

      1 point: straight line of positive constant slope from the origin to (2.0 s, 6.0 m)

   2. 2

      1 point: curve that flattens to a maximum at t = 3.5 s and then falls, with curvature downward throughout 2.0–5.0 s

   3. 3

      1 point: straight line of constant negative slope after t = 5.0 s

4. (d)

   Describe Describe the motion of the cart in words, making clear the difference between the interval in which it is slowing down and the interval in which it is speeding up.

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   1. 1

      1 point: moves in the positive direction at constant speed for the first 2.0 s

   2. 2

      1 point: from 2.0 s to 3.5 s it still moves in the positive direction but slows, because velocity and acceleration have opposite signs

   3. 3

      1 point: from 3.5 s onward it moves in the negative direction and speeds up until 5.0 s, because velocity and acceleration now have the same sign, then travels at constant speed

A student is given a wooden block, a horizontal board of the same material, a set of known masses, a spring scale, a meter stick, and a stopwatch. The student wants to determine the coefficient of kinetic friction μ between the block and the board.

1. (a)

   Describe Describe an experimental procedure the student could use to collect the data needed. Include the quantity that is deliberately varied, the quantity measured in response, and at least two quantities held constant.

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   1. 1

      1 point: pull the block along the board with the spring scale at constant velocity, so that the applied force equals the friction force

   2. 2

      1 point: vary the total mass by adding known masses on top of the block — this is the independent variable

   3. 3

      1 point: record the spring-scale reading for each total mass — this is the dependent variable

   4. 4

      1 point: hold constant the surfaces in contact and the surface area, and keep the pull horizontal for every trial

2. (b)

   Describe Describe how the student should graph the data so that the coefficient of friction can be determined from a straight line, and state what quantity the slope of that line represents.

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   1. 1

      1 point: plot the friction force (spring-scale reading) on the vertical axis against the normal force, or against the total mass, on the horizontal axis

   2. 2

      1 point: the relationship f = μN is linear through the origin

   3. 3

      1 point: the slope equals μ if the horizontal axis is the normal force, or μg if the horizontal axis is the mass

3. (c)

   Determine The student's best-fit line has a slope of 0.32 when friction force is plotted against normal force, and a vertical intercept of +0.15 N rather than zero. Determine the coefficient of kinetic friction, and explain what the non-zero intercept suggests about the experiment.

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   1. 1

      1 point: μ = 0.32, the slope, which is dimensionless

   2. 2

      1 point: the model predicts an intercept of zero, since zero normal force means zero friction

   3. 3

      1 point: a positive intercept indicates a systematic error — for example the spring scale reads high at zero, or the pull was not horizontal, adding a constant component to every reading

4. (d)

   Explain Explain why taking the slope of the best-fit line gives a more reliable value for μ than calculating f/N from a single pair of readings.

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   1. 1

      1 point: the slope uses every data point, so random errors in individual readings tend to average out

   2. 2

      1 point: a systematic offset such as the intercept above shifts a single-point ratio but does not change the slope, so the slope is insensitive to it

A solid sphere and a block have the same mass. Both are released from rest at the same height at the top of an incline. The sphere rolls down without slipping. The block slides down a frictionless section of the same incline. The moment of inertia of a solid sphere about its center is (2/5)MR².

1. (a)

   Indicate Indicate which object reaches the bottom of the incline with the greater translational speed, or whether they arrive with the same speed. No justification is required in this part.

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   1. 1

      1 point: the block

2. (b)

   Derive Derive expressions for the translational speed of each object at the bottom of the incline, in terms of g and the height h.

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   1. 1

      1 point: block — Mgh = ½Mv², so v = √(2gh)

   2. 2

      1 point: sphere — energy is shared between translation and rotation, Mgh = ½Mv² + ½Iω²

   3. 3

      1 point: applies the rolling condition ω = v/R and substitutes I = (2/5)MR², giving Mgh = (7/10)Mv²

   4. 4

      1 point: v = √(10gh/7)

3. (c)

   Justify Use your expressions to justify the answer you gave in part (a).

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   1. 1

      1 point: compares the coefficients — 10/7 = 1.43 is smaller than 2, so the sphere's speed is smaller

   2. 2

      1 point: states the ratio explicitly, v_sphere / v_block = √(5/7) = 0.85

4. (d)

   Explain Explain, without using equations, why the sphere arrives more slowly, and explain why the answer does not depend on the mass or the radius of the sphere.

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   1. 1

      1 point: both objects convert the same gravitational potential energy per unit mass

   2. 2

      1 point: the sphere must also spin as it moves, so part of that energy goes into rotational kinetic energy and less is left for translation

   3. 3

      1 point: the mass appears in every energy term and cancels, so it cannot affect the result

   4. 4

      1 point: the radius cancels because the rolling condition ties ω to v through the same R that appears in the moment of inertia — only the shape factor 2/5 survives