Introduction · quiz worked solutions

Tried the quiz first? Good. Here’s how to work through each answer. Follow the link beside a question to jump back to its original spot. Questions marked Extra exploration go beyond this AP Physics 1 course; Book controls questions explain the interface.

For Future Reference - Forces as Vectors · question 1

Back to the question · AP Topic 2.5

Question: A 50 kg box remains at rest on a horizontal floor. A man pulls upward on it with 10 N. With g = 9.81 m/s², what is the normal-force magnitude?

Answer: 480.5 N.

1. Write N + 10 N − mg = 0 with up positive. Why: The box stays at rest, so its vertical forces balance.

2. Find N = (50)(9.81) − 10 = 480.5 N. Why: The upward pull takes over part of the floor’s job.

For Future Reference - Forces as Vectors · question 2

Back to the question · AP Topic 2.5

Question: The shown 10 kg box is suspended above the ground. Only its weight (98.1 N downward) and a 5 N applied force at 40° above horizontal act on it. What is the box’s acceleration?

Answer: 9.497 m/s², 87.7° south of east.

1. Resolve the pull: Fₓ = 5 cos40° = 3.830 N and Fᵧ = 5 sin40° = 3.214 N. Why: The weight acts only vertically.

2. Find aₓ = 3.830/10 = 0.3830 and aᵧ = (3.214 − 98.1)/10 = −9.4886 m/s². Why: Only the stated two forces act on this suspended box.

3. Calculate the magnitude 9.496 m/s² and angle arctan(9.4886/0.3830) = 87.7° below +x. Why: The signs place the acceleration south of east.

For Future Reference - Forces as Vectors · question 3

Back to the question · AP Topic 2.7

Question: Refer to this Free Body Diagram. If the box is moving to the right, which force is incorrectly drawn?

Answer: Friction is drawn incorrectly.

1. Take the stated rightward sliding as motion relative to the floor. Why: Kinetic friction opposes relative sliding.

2. The friction arrow must point left. Why: An arrow to the right would aid that sliding instead of resisting it.

For Future Reference - Forces as Vectors · question 4

Back to the question · AP Topic 2.5

Question: Which of the following FBDs shows a box that will not experience any acceleration?

Answer: Option 2: balanced forces.

1. Add opposite horizontal forces, then opposite vertical forces. Why: Zero acceleration requires a zero vector sum, not an absence of forces.

2. In option 2, 50 cos45° ≈ 35.35 N balances the leftward force, and 50 sin45° + 14.65 N ≈ 50 N balances weight. Why: The displayed rounding gives zero net force to the figure’s precision, so a = 0.

For Future Reference - Forces as Vectors · question 5

Back to the question · AP Topic 2.5

Question: Four forces act on the same box. Each force has a magnitude of 40 N . If the box has a mass of 43 kg , what is the acceleration of the box?

Answer: 1.22 m/s² downward.

1. Cancel the two 40 N horizontal components. Why: Their equal magnitudes and opposite directions give zero horizontal resultant.

2. Take up positive: Fᵧ = 2(40 sin20°) − 40 − 40 = −52.638 N. Why: Both sloping forces point upward; both vertical forces point down.

3. Divide by 43 kg to get −1.224 m/s². Why: The negative sign means downward acceleration.

Newton’s Laws and FBDs · question 1

Back to the question · AP Topic 2.4

Question: Jake is at a lake on a boat. He wishes to see if he, like Jesus, can walk on water so he decides to stick one foot out of the boat. He then notices that the boat begins to move in the opposite direction. Which law is displayed?

Answer: Newton’s third law.

1. Choose Jake and the boat as separate objects. Why: Third-law partners act on different objects.

2. Jake pushes on the boat, and the boat pushes oppositely on Jake. Why: The interaction explains the boat’s opposite response; simply placing a foot over water without pushing would not propel it.

Newton’s Laws and FBDs · question 2

Back to the question · AP Topic 2.3

Question: A probe coasts through space with constant velocity while the net external force on it is negligible. Which of Newton’s laws most directly describes this coasting motion?

Answer: Newton’s first law.

1. Check the net external force: it is negligible. Why: The coasting model has no appreciable acceleration.

2. The probe keeps its constant velocity. Why: The first law directly describes inertia in an inertial frame.

Newton’s Laws and FBDs · question 3

Back to the question · AP Topic 2.4

Question: Joe is trying to hit a target from far away. He uses a pistol to hit the target, but when he fires, he feels a recoil. Which of Newton’s laws is best displayed?

Answer: Newton’s third law.

1. Identify the gun–bullet interaction. Why: The gun pushes the bullet forward.

2. The bullet pushes the gun backward with equal force. Why: Those paired forces act on different masses, so equal force does not mean equal acceleration.

Newton’s Laws and FBDs · question 4

Back to the question · AP Topic 2.5

Question: Drew is trying to drive home to wish his family farewell before leaving on his trip. On the way home, he finds a huge boulder blocking his way. He moves the boulder by applying a force to it. Which of Newton’s Laws is best displayed?

Answer: Newton’s second law.

1. Choose the boulder and find its net force. Why: A push causes acceleration only when the other forces do not balance it.

2. Use ΣF = ma. Why: The second law connects the resultant force to the boulder’s change in motion.

Newton’s Laws and FBDs · question 5

Back to the question · AP Topic 2.3

Question: Which of the following laws describes the pull you feel when your car turns around a sharp bend?

Answer: Newton’s first law.

1. Imagine the passenger continuing along the old straight path. Why: Inertia resists a change in velocity direction.

2. The car and seat turn the passenger toward the curve. Why: The felt tendency to keep going straight illustrates the first law; the actual inward acceleration also obeys the second law.

Newton’s Laws and FBDs · question 1

Back to the question · AP Topic 2.2

Question: This man would prefer to not fall into the fire, so he gets a friend to pull him out. Match the forces with their respective arrows.

Answer: Targets 1–4: weight, friction, normal, applied force.

1. Identify the agents: Earth, the surface, and the pulling friend. Why: An FBD contains real external interactions on the man.

2. Match weight downward, normal perpendicular to the slope, friction along the contact opposing slipping, and the pull along the rope. Why: Each direction follows the relevant interaction rather than the man’s desired motion.

Newton’s Laws and FBDs · question 2

Back to the question · AP Topic 2.5

Question: In the previous problem, assume there is no frictional force acting on the man. Which force will then cause the man to fall into the fire?

Answer: Gravity’s component down the slope.

1. Resolve weight into perpendicular and parallel components. Why: The normal balances only the perpendicular component.

2. With no friction, the downhill component mg sinθ remains. Why: If the friend’s uphill pull is smaller than that component, the man accelerates downhill.

Newton’s Laws and FBDs · question 3

Back to the question · AP Topic 2.2

Question: A beach ball is falling through the air. Assuming no air resistance, what forces are acting on the ball?

Answer: Gravitational force only.

1. Choose the ball alone and list interacting objects. Why: The problem explicitly removes air resistance.

2. Earth exerts its weight downward; there is no supporting surface. Why: A normal force needs contact and cannot act on a freely falling ball.

Newton’s Laws and FBDs · question 4

Back to the question · AP Topic 2.7

Question: If both gravity and the normal force are acting on an object resting on an inclined plane, which force is preventing the object from moving down the plane

Answer: Static friction.

1. Resolve gravity down the incline and normal perpendicular to it. Why: The normal does not have a component along the surface.

2. Static friction points uphill and balances the downhill component. Why: For rest to be possible, the needed friction must not exceed μₛN.

Newton’s Laws and FBDs · question 5

Back to the question · AP Topic 2.5

Question: According to ΣF⃗ = ma⃗, which pair of quantities directly determines the net force required to give an object a specified acceleration?

Answer: Both mass and acceleration.

1. Start with the vector equation ΣF = ma. Why: This equation determines the required net force.

2. Multiply the specified acceleration by the object’s mass. Why: Gravity may contribute to the net force, but it is not an additional variable in this particular product.

Vector Math · question 1

Back to the question · AP Topic 1.5

Question: What is the magnitude of the sum of the following two vectors?​ a → = 41 N at 145 ∘ b → = 22 N at 107 ∘ ​

Answer: 59.9 N.

1. Resolve the vectors: Rₓ = 41 cos145° + 22 cos107° ≈ −40.02 N; Rᵧ = 41 sin145° + 22 sin107° ≈ 44.57 N. Why: Components add along the same axes.

2. Calculate √(Rₓ² + Rᵧ²) ≈ 59.90 N. Why: Perpendicular components form the legs of a right triangle.

Vector Math · question 2

Back to the question · AP Topic 1.5

Question: What is the magnitude of the sum of the following two vectors?​ a → = 4 m / s at 45 ∘ b → = 7 m / s at 23 ∘ ​

Answer: 10.81 m/s.

1. Find Rₓ = 4 cos45° + 7 cos23° ≈ 9.272 m/s and Rᵧ = 4 sin45° + 7 sin23° ≈ 5.564 m/s. Why: The angles are measured from +x.

2. Find √(Rₓ² + Rᵧ²) ≈ 10.81 m/s. Why: Adding magnitudes alone would ignore the angle between the vectors.

Vector Math · question 3

Back to the question · AP Topic 1.5

Question: In the following triangle, what is the value of β ?

Answer: More information is needed.

1. Check which side–angle pairs are actually supplied. Why: The sketch is not a scale measurement.

2. Do not infer β just from the drawn shape. Why: The given information does not fix a unique triangle and therefore does not fix β.

Vector Math · question 4

Back to the question · AP Topic 1.5

Question: In the following triangle, what is the value of A ?

Answer: 3.86 m.

1. Find the third angle: 180° − 110° − 30° = 40°. Why: Triangle angles add to 180°.

2. Use A/sin40° = 3 m/sin30° to get A = 3.857 m. Why: Each side is paired with its opposite angle in the sine rule.

Vector Math · question 5

Back to the question · AP Topic 1.5

Question: What is the magnitude of the sum of the following two vectors?​ a → = 10 m / s at 15 ∘ b → = 21 m / s at 65 ∘ ​

Answer: 28.5 m/s.

1. Find Rₓ = 10 cos15° + 21 cos65° ≈ 18.534 m/s and Rᵧ = 10 sin15° + 21 sin65° ≈ 21.621 m/s. Why: Cartesian components preserve direction when adding.

2. Find √(Rₓ² + Rᵧ²) ≈ 28.48 m/s. Why: This is the length of the resultant, not the sum of the original lengths.

Vector Math · question 1

Back to the question · AP Topic 1.5

Question: When lining up vectors for vector addition, what orientation is used?

Answer: Tip-to-tail.

1. Move the second vector without rotating it until its tail meets the first vector’s tip. Why: Translation preserves a free vector’s magnitude and direction.

2. Draw the resultant from the first tail to the last tip. Why: That arrow represents the total change.

Vector Math · question 2

Back to the question · AP Topic 1.5

Question: Which of the following situations shows the vectors correctly aligned for addition?

Answer: Option 2, the tip-to-tail arrangement.

1. Check that the second tail touches the first tip. Why: That joins successive vector changes.

2. Read the sum from the free tail to the free tip. Why: Tip-to-tip or tail-to-tail placement needs a different construction.

Vector Math · question 3

Back to the question · AP Topic 1.5

Question: What is the sum of the two vectors: a → = 4 i ˆ + 5 j ˆ b → = 5 i ˆ + 5 j ˆ

Answer: 9i + 10j.

1. Add x components: 4 + 5 = 9. Why: Both coefficients belong to the same i direction.

2. Add y components: 5 + 5 = 10. Why: The j direction is independent of i, so we keep both in the result.

Vector Math · question 4

Back to the question · AP Topic 1.5

Question: If two nonzero vectors point in the same direction, their sum points in that direction.

Answer: True.

1. Choose a unit vector in the common direction. Why: Each vector is a positive magnitude times that same unit vector.

2. Add the two positive magnitudes. Why: Their sum remains positive and points in the common direction.

Vector Math · question 5

Back to the question · AP Topic 1.5

Question: What is the magnitude of the displacement of a man if he walks 4 mi North, 3 mi East, and then 2 mi North?

Answer: 6.7 miles.

1. Combine the northward legs: 4 + 2 = 6 mi north; keep 3 mi east. Why: Displacement depends on the net change in each coordinate.

2. Find √(6² + 3²) = √45 ≈ 6.71 mi. Why: The east and north changes are perpendicular.

Vector Math · question 1

Back to the question · Extra exploration

Question: The cross product of two parallel vectors is always 0.

Answer: True.

1. Use the magnitude rule for a cross product: ab sinθ. Why: A cross product measures the perpendicular combination.

2. For parallel or antiparallel vectors, θ is 0° or 180°, and sinθ = 0. Why: Both alignments therefore give zero cross product.

Vector Math · question 2

Back to the question · Extra exploration

Question: If vector a → was directed northwards, and vector b → was directed eastwards, where would: the vector a → × b → be directed? the vector b → × a → be directed?

Answer: a × b points into the ground; b × a points toward the sky.

1. Take east as +x and north as +y. Why: This sets a right-handed frame with +z upward.

2. Use j × i = −k, then reverse the order. Why: Swapping the two vectors reverses a cross product’s direction.

Vector Math · question 3

Back to the question · Extra exploration

Question: Is the following coordinate system right-handed? That is, is the k ˆ vector in the direction of the cross product i ˆ × j ˆ ?

Answer: Yes.

1. Curl the right-hand fingers from the pictured i direction toward j. Why: The thumb gives the cross-product direction.

2. The thumb agrees with the pictured k direction. Why: That is the definition of a right-handed basis.

Vector Math · question 4

Back to the question · Extra exploration

Question: Two nonzero vectors have dot product zero. What is the angle between them?

Answer: 90°.

1. Use a·b = ab cosθ. Why: Both vectors are nonzero, so ab is nonzero.

2. Set cosθ = 0 to get θ = 90° for the angle between vectors. Why: Zero dot product identifies perpendicular vectors in this case.

Vector Math · question 5

Back to the question · Extra exploration

Question: What is the dot product of the following vectors? a → = 5 mi at 45 ∘ North of West b → = 10 mi at 45 ∘ North of East

Answer: 0 mi².

1. The two directions are 90° apart. Why: One lies 45° west of north and the other 45° east of north.

2. Calculate (5 mi)(10 mi)cos90° = 0 mi². Why: The scalar product vanishes for perpendicular vectors.

Vectors · question 1

Back to the question · AP Topic 1.5

Question: What are the two properties every vector has?

Answer: Magnitude and direction.

1. Ask how long the vector is. Why: That gives its magnitude.

2. Ask which way it points. Why: Magnitude alone describes a scalar, while a vector also needs direction.

Vectors · question 2

Back to the question · AP Topic 1.5

Question: Which of the following is a correct way of identifying a vector? Pick all applicable.

Answer: Options 1, 3 and 4 specify vectors.

1. Check whether each notation supplies direction as well as magnitude. Why: Components and coordinate pairs encode direction.

2. Accept the two component forms and the magnitude-plus-angle form; reject 5 m alone. Why: The angle form is a valid vector description, although 5 m at 45° is not the same vector as (3,4) m.

Vectors · question 3

Back to the question · AP Topic 1.5

Question: Given the following vector, provide the vector in terms of its x and y components in m / s :

Answer: v = 8.66i + 5j m/s.

1. Read 10 m/s at 30° above +x in the figure. Why: The horizontal component is adjacent to the shown angle.

2. Calculate vₓ = 10 cos30° = 8.66 m/s and vᵧ = 10 sin30° = 5.00 m/s. Why: Both components are positive in the first quadrant.

Vectors · question 4

Back to the question · AP Topic 1.5

Question: If a man walks 4 miles north and 3 miles east, what is his displacement?

Answer: 5 mi, 53.1° north of east.

1. Find the magnitude √(3² + 4²) = 5 mi. Why: East and north are perpendicular components.

2. Find the angle arctan(4/3) = 53.1° above east. Why: The angle must be measured from the direction named in the answer.

Vectors · question 5

Back to the question · AP Topic 1.1

Question: If a person walks a full circular path of radius 5 miles , what is that person’s displacement and distance traveled, respectively?

Answer: Displacement 0 mi; distance 31.4 mi.

1. Compare the start and finish positions after one full loop. Why: They coincide, giving zero displacement.

2. Calculate the path length 2π(5 mi) ≈ 31.4 mi. Why: Distance counts the entire path even when displacement vanishes.

Model references

These original worked steps use the stated question models. Physics references: Vectors, Newton’s laws. AP topic labels follow the supplied-edition College Board CED, Effective Fall 2024, ©2026, pinned in the course audit.