AP Physics 1 Quiz: Pressure

What this quiz covers

This quiz focuses on Pressure, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Point A, because it is closer to the surface
2. Point B, because pressure increases with depth (correct answer)
3. They are equal because the container is sealed
4. They are equal because pressure depends only on container shape

Explanation: This question assesses understanding of hydrostatic pressure in fluids, which increases with depth. In a fluid at rest, pressure at a depth h is P = ρgh + P_atm, where ρ is density, g is gravity, and h is depth, showing pressure grows linearly with depth. Point A at 0.20 m has less pressure than Point B at 0.50 m due to the greater overlying fluid weight at B. Thus, Point B experiences greater water pressure. Distractor C suggests equal pressure because the container is sealed, but sealing does not affect the depth dependence. A useful strategy is to compare depths directly, as pressure differences depend solely on Δh in the same fluid.

Question 2

1. $1.0\times10^5\,\text{Pa}$ (correct answer)
2. $4.0\times10^5\,\text{Pa}$
3. $1.0\times10^2\,\text{Pa}$
4. $4.0\times10^2\,\text{Pa}$

Explanation: This question tests pressure due to contact force, defined as force per unit area. Pressure is P = F/A, where F is the perpendicular force on the surface. The 400 N crate is supported by four feet, so each foot bears 100 N, and with area 1.0 × 10^{-3} m², the pressure per foot is 100 / 0.001 = 1.0 × 10^5 Pa. This calculation assumes even weight distribution across the feet. Distractor B (4.0 × $10^5$ Pa) might come from using the total force instead of per foot. When analyzing multi-support contacts, divide the total force equally among supports before applying P = F/A.

Question 3

1. $P_A>P_B$, because A is directly under the surface
2. $P_A<P_B$, because B is farther from the container wall
3. $P_A=P_B$, because pressure depends only on depth in a static fluid (correct answer)
4. Cannot be determined without the container's shape

Explanation: This question tests understanding that pressure depends only on depth in a static fluid. In any static fluid, pressure at a given depth is P = P₀ + ρgh, which depends only on the vertical depth h below the surface, not on horizontal position. Since both points A and B are at the same 0.30 m depth, they experience identical pressure regardless of their horizontal separation or position relative to container walls. This principle allows pressure to be transmitted equally throughout a fluid at the same depth. Choice A incorrectly suggests that being "directly under" matters, confusing vertical depth with horizontal position. When analyzing fluid pressure, only the vertical depth below the surface determines the pressure value.

Question 4

1. Point R, because it is closer to the surface
2. Point S, because pressure increases with depth (correct answer)
3. They are equal, because the container could be wide or narrow
4. They are equal, because oil is less dense than water

Explanation: This question tests understanding of pressure increasing with fluid depth. Pressure in a static fluid follows P = P₀ + ρgh, where ρ is the oil's density and h is depth. Point S at 0.40 m depth has four times the gauge pressure (ρgh) compared to Point R at 0.10 m depth, regardless of the oil's specific density. The absolute pressure includes atmospheric pressure P₀ at both points, but the pressure difference depends only on the depth difference. Choice D incorrectly implies that oil's lower density compared to water affects which point has higher pressure, when depth alone determines the ranking. For any fluid, deeper points have higher pressure following the same P = P₀ + ρgh relationship.

Question 5

1. four times larger, because there are four feet
2. the same, because the total weight is unchanged
3. four times smaller, because the force per foot is smaller while area per foot is the same (correct answer)
4. dependent on the spacing between the feet, not on force and area

Explanation: This question tests understanding of pressure distribution across multiple contact points. When the 300 N machine rests on four identical feet, each foot supports 300 N / 4 = 75 N. The pressure at each foot is P = 75 N / (2.0×10⁻³ m²) = 37,500 Pa. If one foot supported the full weight, pressure would be 300 N / (2.0×10⁻³ m²) = 150,000 Pa, which is four times larger. The four-foot configuration reduces pressure to one-fourth because each foot carries one-fourth the force while maintaining the same contact area. Choice A incorrectly multiplies by the number of feet. When weight is distributed across multiple supports, divide the total force by the number of supports to find force per support.

Question 6

1. Same pressure in both orientations because the force is the same
2. Greater pressure on the $0.10\,\text{m}^2$ face (correct answer)
3. Greater pressure on the $0.20\,\text{m}^2$ face because it has more contact area
4. Greater pressure depends on the crate's shape, not the contact area

Explanation: This question tests understanding of pressure as force per unit area. Pressure is defined as P = F/A, where F is the perpendicular force and A is the contact area. Since the crate's weight (600 N) remains constant regardless of orientation, the force on the floor is always 600 N. When the crate sits on the 0.10 m² face, pressure = 600 N / 0.10 m² = 6000 Pa, while on the 0.20 m² face, pressure = 600 N / 0.20 m² = 3000 Pa. Choice C incorrectly assumes larger area means greater pressure, but pressure is inversely proportional to area when force is constant. To solve pressure problems involving contact forces, identify the perpendicular force and divide by the contact area.

Question 7

1. Point $G$
2. Point $H$ (correct answer)
3. Same, because the container is sealed.
4. Same, because pressure is uniform in a closed container.

Explanation: This question assesses hydrostatic pressure in AP Physics 1, due to fluid depth. In a static fluid, pressure increases with depth because of the overlying fluid's weight, P = ρgh. Greater depth results in higher pressure from the fluid. Point H is three times deeper than G, so it has greater pressure from the oil. Choice D is a distractor, mistakenly claiming uniform pressure in closed containers, but pressure varies with depth. A transferable strategy is to measure depth from the fluid surface when evaluating pressure in sealed or open containers.

Question 8

1. Point G, because it is closer to the shore
2. Point H, because pressure increases with horizontal distance
3. They are equal, because both points are at the same depth (correct answer)
4. They are unequal, because pressure depends on the container's shape

Explanation: This question assesses hydrostatic pressure, focusing on its dependence on vertical depth rather than horizontal position. In fluids, pressure increases with depth h as P = ρgh + P0, but is independent of horizontal distance. Points G and H are both at 2.0 m depth, so they have equal pressure despite the 10 m horizontal separation. The pressures are the same because only vertical depth matters. Choice B is a distractor that incorrectly claims pressure increases with horizontal distance, which is not true in static fluids. A transferable strategy is to ignore horizontal positions and base comparisons solely on vertical depths from the surface.

Question 9

1. Case A, because the area is larger
2. Case B, because the same normal force is distributed over less area (correct answer)
3. They are equal, because the crate's weight is unchanged
4. Case A, because friction increases pressure

Explanation: This question tests understanding of pressure as force per unit area. Pressure is calculated as P = F/A, where F is specifically the normal force (perpendicular to the surface) and A is the contact area. Since the normal force is the same in both cases but Case B has half the contact area (0.25 m²) compared to Case A (0.50 m²), Case B produces twice the pressure. The horizontal friction force from pulling does not contribute to vertical pressure. Choice D incorrectly suggests friction affects pressure, when only the perpendicular normal force matters. For pressure calculations, always identify which force component is perpendicular to the surface.

Question 10

1. Trial 1, because the area is larger
2. Trial 2, because the same force acts on a smaller area (correct answer)
3. They are equal, because the force is the same
4. Cannot be determined without the block's mass

Explanation: This question tests understanding of pressure as force per unit area. Pressure is defined as P = F/A, where F is the perpendicular force and A is the contact area. Since the block's weight (force) remains constant but the contact area changes between trials, the pressure will be different. In Trial 2, the same force acts on a smaller area (0.010 m²) compared to Trial 1 (0.020 m²), resulting in pressure that is twice as large. Choice C incorrectly assumes that equal force means equal pressure, ignoring the crucial role of area. When solving pressure problems involving contact forces, always identify both the force and the area, then apply P = F/A to compare pressures.

Question 11

1. Setup 1, because the area is larger
2. Setup 2, because the same force acts on a smaller area (correct answer)
3. They are equal, because the applied force is $20\,\text{N}$ in both
4. Cannot be determined without the piston's shape

Explanation: This question tests understanding of pressure as force per unit area. Pressure is defined as P = F/A, where F is the applied force and A is the piston area. Both setups have the same 20 N force, but Setup 2 has half the piston area (2.0×10⁻³ m²) compared to Setup 1 (4.0×10⁻³ m²). Using P = F/A, Setup 2 produces 10,000 Pa while Setup 1 produces 5,000 Pa, making Setup 2's pressure twice as large. Choice C incorrectly assumes equal force means equal pressure, missing the inverse relationship with area. When comparing pressures, always calculate P = F/A for each case rather than relying on force or area alone.

Question 12

1. At the tip, because the same force acts on a smaller area (correct answer)
2. At the head, because the applied force is closer to the hand
3. Equal at head and tip because the force is $20\,\text{N}$ everywhere in the tack
4. Greater at the head because larger area produces larger pressure

Explanation: This question tests understanding of pressure as force divided by area. The thumbtack transmits the same 20 N force through both the head and tip due to Newton's third law. At the head, pressure = 20 N / (2.0×10⁻⁴ m²) = 1.0×10⁵ Pa, while at the tip, pressure = 20 N / (1.0×10⁻⁶ m²) = 2.0×10⁷ Pa. The tip pressure is 200 times greater because its area is 200 times smaller. Choice C incorrectly assumes pressure is constant throughout the tack, confusing force (which is constant) with pressure (which varies with area). To maximize pressure with a given force, minimize the contact area.

Question 13

1. Pressure is greater under the large piston because its area is larger.
2. Pressure is greater under the small piston because the force is applied there.
3. Pressure is the same throughout the connected fluid at the same height. (correct answer)
4. Pressure depends on the shape of the connecting tube.

Explanation: This question assesses pressure transmission in fluids for AP Physics 1, using contact force over area. Pressure is force per area, and in a confined, incompressible fluid, it is transmitted equally throughout at the same height. The applied force on the small piston creates a pressure that is the same under both pistons, per Pascal's principle. Despite different areas, the fluid pressure remains uniform. Choice A distracts by suggesting larger area means greater pressure, but pressure is independent of area in connected fluids. A transferable strategy is to apply Pascal's principle for equal pressures in hydraulic systems at equal heights.

Question 14

1. The student in snowshoes
2. The student in boots (correct answer)
3. They exert the same pressure because their weights are equal.
4. Cannot be determined without the snow density.

Explanation: This question assesses pressure from contact force over area in AP Physics 1. Pressure is the force per unit area, so P = F/A, with smaller areas leading to higher pressures for equal forces. The students have the same weight, but boots have less total contact area than snowshoes, increasing pressure. Snowshoes distribute weight over more area, reducing pressure. Choice C distracts by equating equal weights to equal pressures, ignoring area differences. A transferable strategy is to calculate total contact area and divide weight by it when comparing pressures on surfaces.

Question 15

1. Block 1
2. Block 2 (correct answer)
3. They experience the same pressure because the forces are equal.
4. Cannot be determined without the blocks' shapes.

Explanation: This question assesses the concept of pressure in AP Physics 1, specifically how it relates to contact force and area. Pressure is defined as the force applied per unit area, so P = F/A. For solid objects pressing on a surface, a smaller contact area results in higher pressure when the force is constant. Both blocks exert 40 N, but Block 2 has half the area of Block 1, doubling its pressure. Choice C is a common distractor because it confuses equal forces with equal pressures, ignoring the role of area. A transferable strategy is to always calculate P = F/A explicitly when comparing pressures from contact forces.

Question 16

1. At the head, because it touches the thumb
2. At the tip, because the same force acts on a much smaller area (correct answer)
3. Equal at head and tip, because the force is $10\,\text{N}$ throughout
4. Depends only on the tack's shape, not the area

Explanation: This question examines contact pressure as force per unit area in a pointed object. Pressure is P = F/A, with higher values for smaller areas under the same force. The 10 N force is applied to the head (area 1.0 × $10^{-4}$ m²) but transmitted to the tip (area 1.0 × $10^{-6}$ m²), resulting in much higher pressure at the tip. Thus, the tip experiences greater pressure due to its smaller area. Choice C is a distractor asserting equal pressure because the force is the same, but it neglects the area difference. For tapered objects, calculate pressures at each end using the respective areas, recognizing force transmission.

Question 17

1. $P_E > P_F$ because the wide beaker holds more liquid.
2. $P_E < P_F$ because the narrow beaker concentrates the liquid.
3. $P_E = P_F$ because depth and fluid are the same. (correct answer)
4. Cannot be determined without the beakers' volumes.

Explanation: This question assesses hydrostatic pressure in AP Physics 1, based on fluid depth. Pressure increases with depth in a fluid due to the weight of the fluid column above, independent of container width. At the same depth in the same fluid, pressures are equal regardless of beaker shape. Points E and F are both 0.30 m deep, so P_E equals P_F. Choice A is a distractor, wrongly assuming more liquid in a wide beaker affects pressure, but only depth matters. A transferable strategy is to ignore container volume and focus on depth for pressure comparisons in static fluids.

Question 18

1. Brick Y, because it has a larger contact area
2. Brick X, because the same force acts on a smaller area (correct answer)
3. They exert equal pressure because both weigh $20\,\text{N}$
4. Pressure depends on the brick material, not area

Explanation: This question evaluates knowledge of contact pressure as force divided by area. Pressure in such cases is P = F/A, with higher pressure resulting from smaller areas for the same force. Both bricks weigh 20 N, but Brick X has a smaller contact area of 0.010 m², leading to greater pressure than Brick Y with 0.020 m². Therefore, Brick X exerts more pressure on the floor. Distractor C claims equal pressure because both have the same weight, but this ignores the difference in contact areas. For similar problems, remember to apply P = F/A directly and note that pressure inversely depends on area for constant force.

Question 19

1. Lying flat, because the force is spread out more
2. On its side, because the same force acts on a smaller area (correct answer)
3. Both orientations produce the same pressure because the weight is $12\,\text{N}$
4. Pressure depends on the book's shape, not the contact area

Explanation: This question assesses understanding of pressure as force per unit area in contact scenarios. Pressure is defined as the force applied perpendicular to a surface divided by the area over which it is distributed, P = F/A. For the book weighing 12 N, when lying flat with a larger contact area of 0.060 m², the pressure is lower compared to when it is on its side with a smaller area of 0.020 m². Since the force (weight) remains constant, the orientation with the smaller area produces greater pressure, which is the case when the book is on its side. A common distractor is choice C, which incorrectly states that both orientations produce the same pressure because the weight is unchanged, ignoring the role of contact area. To approach similar problems, always calculate pressure using P = F/A and compare values based on varying areas for constant force.

Question 20

1. $p_Q>p_P$ because pressure increases with depth in a fluid at rest. (correct answer)
2. $p_Q=p_P$ because water pushes equally in all directions.
3. $p_P>p_Q$ because the water above $P$ has less area to push on.
4. The deeper point has lower pressure because its volume is smaller.

Explanation: This question tests understanding of pressure variation with depth in fluids. In a static fluid, pressure increases linearly with depth according to P = P₀ + ρgh, where ρ is fluid density, g is gravitational acceleration, and h is depth below the surface. Since point Q is 2.0 m deeper than point P, the pressure at Q must be greater than at P due to the additional weight of water above it. Choice B incorrectly suggests equal pressures, confusing the fact that pressure acts in all directions with pressure magnitude. When comparing pressures in fluids, always consider the depth difference and remember that deeper points have higher pressure.