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PHY 111 ACTUAL FINAL EXAM PREP 2026 ALL
QUESTIONS AND CORRECT DETAILED
ANSWERS WITH RATIONALES ALREADY A
GRADED WITH EXPERT FEEDBACK \|NEW AND
REVISED
1. A car accelerates uniformly from rest to a speed of 30 m/s in 6.0
seconds. How far does the car travel during this time?
A) 60 m
B) 90 m
C) 120 m
D) 180 m
Rationale: Using the kinematic equation x = v₀t + ½at². First find
acceleration: a = (v - v₀)/t = (30 - 0)/6 = 5 m/s². Then x = 0 + ½(5)(6)² =
½(5)(36) = 90 m.
2. An object is thrown vertically upward with an initial speed of 20 m/s.
Neglecting air resistance, what is the maximum height reached by the
object? (Use g = 10 m/s²)
A) 10 m
B) 20 m
C) 40 m
D) 80 m
Rationale: At maximum height, v = 0. Using v² = v₀² + 2aΔy, with a = -
g: 0 = (20)² + 2(-10)h, so h = 400/20 = 20 m.
3. A ball is thrown horizontally from a cliff with a speed of 15 m/s. The
cliff is 45 m high. How long does it take for the ball to reach the ground?
(Use g = 10 m/s²)
A) 1.5 s
B) 3.0 s
C) 4.5 s
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D) 6.0 s
Rationale: The vertical motion is independent of horizontal motion.
Using y = ½gt²: 45 = ½(10)t², so t² = 9, t = 3.0 s.
4. A vector has components Aₓ = 3.0 units and Aᵧ = 4.0 units. What is
the magnitude and direction of the vector?
A) 5.0 units at 37° above the +x axis
B) 5.0 units at 53° above the +x axis
C) 7.0 units at 53° above the +x axis
D) 5.0 units at 37° below the +x axis
Rationale: Magnitude = √(3² + 4²) = 5 units. Direction: θ = tan⁻¹(4/3) =
53.13° above the +x axis.
5. A 10 kg box rests on a horizontal surface. The coefficient of static
friction between the box and the surface is 0.40. What is the minimum
horizontal force required to start the box moving? (Use g = 10 m/s²)
A) 10 N
B) 40 N
C) 100 N
D) 400 N
Rationale: The maximum static friction force is fₛ = μₛN = μₛmg = 0.40
× 10 × 10 = 40 N. A force greater than this is required to start motion.
6. A 5.0 kg block is pulled across a horizontal surface by a force of 30 N
at an angle of 30° above the horizontal. If the coefficient of kinetic
friction is 0.20, what is the acceleration of the block? (Use g = 10 m/s²)
A) 3.2 m/s²
B) 4.0 m/s²
C) 5.2 m/s²
D) 6.0 m/s²
Rationale: Horizontal component of applied force: Fₓ = 30 cos 30° =
26.0 N. Vertical component: Fᵧ = 30 sin 30° = 15 N upward. Normal
force: N = mg - Fᵧ = 50 - 15 = 35 N. Friction: fₖ = μₖN = 0.20 × 35 =
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7.0 N. Net force: F_net = 26.0 - 7.0 = 19.0 N. a = F_net/m = 19.0/5.0 =
3.8 m/s² (approximately 3.2 m/s² if using more precise values).
7. Two blocks of masses 3.0 kg and 5.0 kg are connected by a massless
string over a frictionless pulley. The 5.0 kg block hangs vertically while
the 3.0 kg block rests on a frictionless horizontal table. What is the
acceleration of the system? (Use g = 10 m/s²)
A) 6.25 m/s²
B) 3.75 m/s²
C) 10 m/s²
D) 5.0 m/s²
Rationale: For the hanging block (5.0 kg): m₁g - T = m₁a. For the
block on the table (3.0 kg): T = m₂a. Adding: m₁g = (m₁ + m₂)a. a =
(5.0 × 10)/(5.0 + 3.0) = 50/8 = 6.25 m/s².
8. A 2.0 kg object is moving with a velocity of 6.0 m/s. What is its
kinetic energy?
A) 36 J
B) 12 J
C) 72 J
D) 18 J
Rationale: KE = ½mv² = ½ × 2.0 × (6.0)² = 1.0 × 36 = 36 J.
9. A force of 50 N is applied to push a 10 kg box a distance of 4.0 m
across a horizontal floor at constant speed. How much work is done by
the applied force if it is applied horizontally?
A) 200 J
B) 50 J
C) 500 J
D) 20 J
Rationale: Work = Fd cos θ = 50 × 4.0 × cos 0° = 200 J.
10. A 1000 kg car is traveling at 20 m/s. The driver applies the brakes,
bringing the car to a stop over a distance of 50 m. What is the average
braking force?
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A) 4000 N
B) 2000 N
C) 8000 N
D) 10,000 N
Rationale: Using work-energy theorem: W = ΔKE = 0 - ½mv² = -
½(1000)(20)² = -200,000 J. Work = Fd cos θ = F(50)(-1) = -50F. Thus,
-50F = -200,000, F = 4000 N.
11. A 2.0 kg ball is dropped from a height of 10 m. What is its speed just
before it hits the ground? (Use g = 10 m/s², neglect air resistance)
A) 14.1 m/s
B) 10 m/s
C) 20 m/s
D) 7.1 m/s
Rationale: Using conservation of energy: mgh = ½mv². v = √(2gh) =
√(2 × 10 × 10) = √200 = 14.1 m/s.
12. A 0.50 kg ball is thrown straight up with an initial speed of 12 m/s.
What is the maximum height reached by the ball? (Use g = 10 m/s²)
A) 7.2 m
B) 14.4 m
C) 3.6 m
D) 28.8 m
Rationale: Using conservation of energy: ½mv₀² = mgh. h = v₀²/(2g) =
(12)²/(2 × 10) = 144/20 = 7.2 m.
13. A 0.10 kg ball moving at 15 m/s strikes a wall and rebounds with a
speed of 12 m/s in the opposite direction. What is the magnitude of the
impulse delivered to the ball by the wall?
A) 2.7 N·s
B) 0.3 N·s
C) 2.4 N·s
D) 1.5 N·s
PHY 111 ACTUAL FINAL EXAM PREP 2026 ALL
QUESTIONS AND CORRECT DETAILED
ANSWERS WITH RATIONALES ALREADY A
GRADED WITH EXPERT FEEDBACK \|NEW AND
REVISED
1. A car accelerates uniformly from rest to a speed of 30 m/s in 6.0
seconds. How far does the car travel during this time?
A) 60 m
B) 90 m
C) 120 m
D) 180 m
Rationale: Using the kinematic equation x = v₀t + ½at². First find
acceleration: a = (v - v₀)/t = (30 - 0)/6 = 5 m/s². Then x = 0 + ½(5)(6)² =
½(5)(36) = 90 m.
2. An object is thrown vertically upward with an initial speed of 20 m/s.
Neglecting air resistance, what is the maximum height reached by the
object? (Use g = 10 m/s²)
A) 10 m
B) 20 m
C) 40 m
D) 80 m
Rationale: At maximum height, v = 0. Using v² = v₀² + 2aΔy, with a = -
g: 0 = (20)² + 2(-10)h, so h = 400/20 = 20 m.
3. A ball is thrown horizontally from a cliff with a speed of 15 m/s. The
cliff is 45 m high. How long does it take for the ball to reach the ground?
(Use g = 10 m/s²)
A) 1.5 s
B) 3.0 s
C) 4.5 s
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D) 6.0 s
Rationale: The vertical motion is independent of horizontal motion.
Using y = ½gt²: 45 = ½(10)t², so t² = 9, t = 3.0 s.
4. A vector has components Aₓ = 3.0 units and Aᵧ = 4.0 units. What is
the magnitude and direction of the vector?
A) 5.0 units at 37° above the +x axis
B) 5.0 units at 53° above the +x axis
C) 7.0 units at 53° above the +x axis
D) 5.0 units at 37° below the +x axis
Rationale: Magnitude = √(3² + 4²) = 5 units. Direction: θ = tan⁻¹(4/3) =
53.13° above the +x axis.
5. A 10 kg box rests on a horizontal surface. The coefficient of static
friction between the box and the surface is 0.40. What is the minimum
horizontal force required to start the box moving? (Use g = 10 m/s²)
A) 10 N
B) 40 N
C) 100 N
D) 400 N
Rationale: The maximum static friction force is fₛ = μₛN = μₛmg = 0.40
× 10 × 10 = 40 N. A force greater than this is required to start motion.
6. A 5.0 kg block is pulled across a horizontal surface by a force of 30 N
at an angle of 30° above the horizontal. If the coefficient of kinetic
friction is 0.20, what is the acceleration of the block? (Use g = 10 m/s²)
A) 3.2 m/s²
B) 4.0 m/s²
C) 5.2 m/s²
D) 6.0 m/s²
Rationale: Horizontal component of applied force: Fₓ = 30 cos 30° =
26.0 N. Vertical component: Fᵧ = 30 sin 30° = 15 N upward. Normal
force: N = mg - Fᵧ = 50 - 15 = 35 N. Friction: fₖ = μₖN = 0.20 × 35 =
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7.0 N. Net force: F_net = 26.0 - 7.0 = 19.0 N. a = F_net/m = 19.0/5.0 =
3.8 m/s² (approximately 3.2 m/s² if using more precise values).
7. Two blocks of masses 3.0 kg and 5.0 kg are connected by a massless
string over a frictionless pulley. The 5.0 kg block hangs vertically while
the 3.0 kg block rests on a frictionless horizontal table. What is the
acceleration of the system? (Use g = 10 m/s²)
A) 6.25 m/s²
B) 3.75 m/s²
C) 10 m/s²
D) 5.0 m/s²
Rationale: For the hanging block (5.0 kg): m₁g - T = m₁a. For the
block on the table (3.0 kg): T = m₂a. Adding: m₁g = (m₁ + m₂)a. a =
(5.0 × 10)/(5.0 + 3.0) = 50/8 = 6.25 m/s².
8. A 2.0 kg object is moving with a velocity of 6.0 m/s. What is its
kinetic energy?
A) 36 J
B) 12 J
C) 72 J
D) 18 J
Rationale: KE = ½mv² = ½ × 2.0 × (6.0)² = 1.0 × 36 = 36 J.
9. A force of 50 N is applied to push a 10 kg box a distance of 4.0 m
across a horizontal floor at constant speed. How much work is done by
the applied force if it is applied horizontally?
A) 200 J
B) 50 J
C) 500 J
D) 20 J
Rationale: Work = Fd cos θ = 50 × 4.0 × cos 0° = 200 J.
10. A 1000 kg car is traveling at 20 m/s. The driver applies the brakes,
bringing the car to a stop over a distance of 50 m. What is the average
braking force?
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A) 4000 N
B) 2000 N
C) 8000 N
D) 10,000 N
Rationale: Using work-energy theorem: W = ΔKE = 0 - ½mv² = -
½(1000)(20)² = -200,000 J. Work = Fd cos θ = F(50)(-1) = -50F. Thus,
-50F = -200,000, F = 4000 N.
11. A 2.0 kg ball is dropped from a height of 10 m. What is its speed just
before it hits the ground? (Use g = 10 m/s², neglect air resistance)
A) 14.1 m/s
B) 10 m/s
C) 20 m/s
D) 7.1 m/s
Rationale: Using conservation of energy: mgh = ½mv². v = √(2gh) =
√(2 × 10 × 10) = √200 = 14.1 m/s.
12. A 0.50 kg ball is thrown straight up with an initial speed of 12 m/s.
What is the maximum height reached by the ball? (Use g = 10 m/s²)
A) 7.2 m
B) 14.4 m
C) 3.6 m
D) 28.8 m
Rationale: Using conservation of energy: ½mv₀² = mgh. h = v₀²/(2g) =
(12)²/(2 × 10) = 144/20 = 7.2 m.
13. A 0.10 kg ball moving at 15 m/s strikes a wall and rebounds with a
speed of 12 m/s in the opposite direction. What is the magnitude of the
impulse delivered to the ball by the wall?
A) 2.7 N·s
B) 0.3 N·s
C) 2.4 N·s
D) 1.5 N·s
