PHYS 165 Module 4 Examination Physics: Two-
Dimensional Kinematics and Dynamics Portage
Learning – 2026/2027 Academic Year
Content Area Overview
Content Area Questions Key Topics Weight
Vector addition/subtraction,
Vectors and
components, unit vectors,
2D 1–15 25%
displacement, velocity,
Kinematics
acceleration in 2D
Horizontal and vertical motion,
Projectile
16–30 range, maximum height, flight 25%
Motion
time, projectile trajectory
Centripetal acceleration,
Circular
31–45 tangential velocity, uniform 25%
Motion
circular motion, banked curves
Relative Relative velocity, frames of
Motion & 2D 46–60 reference, forces in 2D, inclined 25%
Dynamics planes, pulleys
DOMAIN 1: VECTORS AND 2D KINEMATICS
Questions 1–15
,Q1. Vector A has magnitude 6.0 units and points in the +x direction.
Vector B has magnitude 8.0 units and points in the +y direction. What is
the magnitude of the resultant vector A + B?
A. 2.0 units
B. 10.0 units
C. 14.0 units
D. 48.0 units
Answer: B. 10.0 units
RATIONALE: Since the vectors are perpendicular (90° angle), the magnitude
of the resultant is found using the Pythagorean theorem: |R| = √(A² + B²) =
√(6.0² + 8.0²) = √(36 + 64) = √100 = 10.0 units. Option A (2.0 units)
incorrectly subtracts the magnitudes. Option C (14.0 units) incorrectly adds
the magnitudes directly. Option D (48.0 units) incorrectly multiplies the
magnitudes. This fundamental vector addition problem tests understanding of
vector components and the Pythagorean theorem, a cornerstone of two-
dimensional kinematics.
Q2. Vector A has components Aₓ = -3.0 m and Aᵧ = 4.0 m. What is the
direction of vector A measured counterclockwise from the positive x-
axis?
A. 53.1°
B. 126.9°
C. 233.1°
D. 306.9°
Answer: B. 126.9°
RATIONALE: The angle relative to the x-axis is θ = tan⁻¹(Aᵧ/Aₓ) = tan⁻¹(4.0/-
3.0) = tan⁻¹(-1.333) = -53.1°. Since Aₓ is negative and Aᵧ is positive, the vector
lies in Quadrant II. Adding 180° to the reference angle: θ = 180° - 53.1° =
126.9°. Option A (53.1°) is the reference angle in Quadrant I. Option C (233.1°)
is Quadrant III (both components negative). Option D (306.9°) is Quadrant IV
(Aₓ positive, Aᵧ negative). Proper vector direction determination is essential
for two-dimensional problem-solving.
, Q3. A displacement vector has components Δx = 12.0 m and Δy = -5.0 m.
What is the magnitude of the displacement?
A. 7.0 m
B. 13.0 m
C. 17.0 m
D. 60.0 m
Answer: B. 13.0 m
RATIONALE: Magnitude of displacement: |Δr| = √(Δx² + Δy²) = √(12.0² + (-
5.0)²) = √(144 + 25) = √169 = 13.0 m. Option A (7.0 m) incorrectly subtracts
components. Option C (17.0 m) incorrectly adds components. Option D (60.0
m) incorrectly multiplies components. The magnitude of a vector represents
the straight-line distance between two points, independent of path, a concept
central to displacement analysis.
Q4. Vector P = 5.0î + 3.0ĵ and vector Q = -2.0î + 4.0ĵ. What is the vector P -
Q?
A. 7.0î - 1.0ĵ
B. 3.0î + 7.0ĵ
C. 7.0î + 7.0ĵ
D. -7.0î - 1.0ĵ
Answer: A. 7.0î - 1.0ĵ
RATIONALE: P - Q = (5.0î + 3.0ĵ) - (-2.0î + 4.0ĵ) = (5.0 + 2.0)î + (3.0 - 4.0)ĵ =
7.0î - 1.0ĵ. Option B (3.0î + 7.0ĵ) represents P + Q. Option C (7.0î + 7.0ĵ)
incorrectly handles the y-component. Option D (-7.0î - 1.0ĵ) has the x-
component sign reversed. Vector subtraction is a critical operation in
determining relative displacements and velocities.
Q5. A particle moves from position (2.0 m, 5.0 m) to position (8.0 m, -1.0
m). What is the displacement vector?
Dimensional Kinematics and Dynamics Portage
Learning – 2026/2027 Academic Year
Content Area Overview
Content Area Questions Key Topics Weight
Vector addition/subtraction,
Vectors and
components, unit vectors,
2D 1–15 25%
displacement, velocity,
Kinematics
acceleration in 2D
Horizontal and vertical motion,
Projectile
16–30 range, maximum height, flight 25%
Motion
time, projectile trajectory
Centripetal acceleration,
Circular
31–45 tangential velocity, uniform 25%
Motion
circular motion, banked curves
Relative Relative velocity, frames of
Motion & 2D 46–60 reference, forces in 2D, inclined 25%
Dynamics planes, pulleys
DOMAIN 1: VECTORS AND 2D KINEMATICS
Questions 1–15
,Q1. Vector A has magnitude 6.0 units and points in the +x direction.
Vector B has magnitude 8.0 units and points in the +y direction. What is
the magnitude of the resultant vector A + B?
A. 2.0 units
B. 10.0 units
C. 14.0 units
D. 48.0 units
Answer: B. 10.0 units
RATIONALE: Since the vectors are perpendicular (90° angle), the magnitude
of the resultant is found using the Pythagorean theorem: |R| = √(A² + B²) =
√(6.0² + 8.0²) = √(36 + 64) = √100 = 10.0 units. Option A (2.0 units)
incorrectly subtracts the magnitudes. Option C (14.0 units) incorrectly adds
the magnitudes directly. Option D (48.0 units) incorrectly multiplies the
magnitudes. This fundamental vector addition problem tests understanding of
vector components and the Pythagorean theorem, a cornerstone of two-
dimensional kinematics.
Q2. Vector A has components Aₓ = -3.0 m and Aᵧ = 4.0 m. What is the
direction of vector A measured counterclockwise from the positive x-
axis?
A. 53.1°
B. 126.9°
C. 233.1°
D. 306.9°
Answer: B. 126.9°
RATIONALE: The angle relative to the x-axis is θ = tan⁻¹(Aᵧ/Aₓ) = tan⁻¹(4.0/-
3.0) = tan⁻¹(-1.333) = -53.1°. Since Aₓ is negative and Aᵧ is positive, the vector
lies in Quadrant II. Adding 180° to the reference angle: θ = 180° - 53.1° =
126.9°. Option A (53.1°) is the reference angle in Quadrant I. Option C (233.1°)
is Quadrant III (both components negative). Option D (306.9°) is Quadrant IV
(Aₓ positive, Aᵧ negative). Proper vector direction determination is essential
for two-dimensional problem-solving.
, Q3. A displacement vector has components Δx = 12.0 m and Δy = -5.0 m.
What is the magnitude of the displacement?
A. 7.0 m
B. 13.0 m
C. 17.0 m
D. 60.0 m
Answer: B. 13.0 m
RATIONALE: Magnitude of displacement: |Δr| = √(Δx² + Δy²) = √(12.0² + (-
5.0)²) = √(144 + 25) = √169 = 13.0 m. Option A (7.0 m) incorrectly subtracts
components. Option C (17.0 m) incorrectly adds components. Option D (60.0
m) incorrectly multiplies components. The magnitude of a vector represents
the straight-line distance between two points, independent of path, a concept
central to displacement analysis.
Q4. Vector P = 5.0î + 3.0ĵ and vector Q = -2.0î + 4.0ĵ. What is the vector P -
Q?
A. 7.0î - 1.0ĵ
B. 3.0î + 7.0ĵ
C. 7.0î + 7.0ĵ
D. -7.0î - 1.0ĵ
Answer: A. 7.0î - 1.0ĵ
RATIONALE: P - Q = (5.0î + 3.0ĵ) - (-2.0î + 4.0ĵ) = (5.0 + 2.0)î + (3.0 - 4.0)ĵ =
7.0î - 1.0ĵ. Option B (3.0î + 7.0ĵ) represents P + Q. Option C (7.0î + 7.0ĵ)
incorrectly handles the y-component. Option D (-7.0î - 1.0ĵ) has the x-
component sign reversed. Vector subtraction is a critical operation in
determining relative displacements and velocities.
Q5. A particle moves from position (2.0 m, 5.0 m) to position (8.0 m, -1.0
m). What is the displacement vector?
