PHYSICS 5A Physics for Life Sciences Final – University Physics
Department – 2026/2027
DOMAIN 1: KINEMATICS AND NEWTON'S LAWS OF MOTION
Question 1
A sprinter accelerates uniformly from rest to 10.0 m/s in 3.5 s. What is the
sprinter's acceleration?
A) 2.86 m/s²
B) 3.50 m/s²
C) 1.43 m/s²
D) 5.71 m/s²
Correct Answer: A) 2.86 m/s²
Rationale: Acceleration equals change in velocity divided by time: a = Δv/Δt =
(10.0 m/s - 0)/3.5 s = 2.86 m/s². Option B incorrectly divides time by velocity;
Option C uses half the correct value; Option D uses an incorrect factor.
Question 2
A ball is thrown vertically upward with an initial velocity of 25 m/s. How long does
it take to reach its maximum height? (Assume g = 9.8 m/s²)
A) 1.28 s
B) 2.55 s
C) 3.82 s
D) 5.10 s
Correct Answer: B) 2.55 s
Rationale: At maximum height, v_f = 0. Using v_f = v_i - gt: 0 = 25 - 9.8t → t =
25/9.8 = 2.55 s. Option A uses g = 19.6; Option C uses 25/6.5; Option D uses
25/4.9 (half of g).
Question 3
,A car traveling at 30 m/s applies brakes and decelerates uniformly at 4.0 m/s².
What distance does the car travel before coming to rest?
A) 56.25 m
B) 112.5 m
C) 225 m
D) 450 m
Correct Answer: B) 112.5 m
Rationale: Using v_f² = v_i² + 2ad: 0 = (30)² + 2(-4.0)d → 0 = 900 - 8d → d =
112.5 m. Option A uses d = v²/(4a); Option C uses d = v²/a; Option D uses d =
2v²/a.
Question 4
SATA Which of the following statements about projectile motion are TRUE?
(Select all that apply)
A) The horizontal velocity is constant (neglecting air resistance)
B) The vertical acceleration is constant
C) The time to reach maximum height equals the time to return to the same
height
D) The trajectory is a parabola
E) The horizontal acceleration is -g
Correct Answers: A, B, C, D
Rationale: In projectile motion with no air resistance: horizontal velocity is
constant (A); vertical acceleration is constant at -g (B); symmetry gives equal rise
and fall times (C); the path is parabolic (D). Option E is false because horizontal
acceleration is 0, not -g.
Question 5
A 2.0 kg block is pushed across a frictionless horizontal surface with a force of 8.0
N. What is the block's acceleration?
,A) 2.0 m/s²
B) 4.0 m/s²
C) 6.0 m/s²
D) 16.0 m/s²
Correct Answer: B) 4.0 m/s²
Rationale: Newton's second law: F = ma → a = F/m = 8.0/2.0 = 4.0 m/s².
Option A uses m/2; Option C adds forces incorrectly; Option D multiplies F and m.
Question 6
A 10 kg box sits on a horizontal floor. The coefficient of static friction is 0.40 and
kinetic friction is 0.25. What horizontal force is required to start the box moving?
A) 24.5 N
B) 39.2 N
C) 98 N
D) 245 N
Correct Answer: B) 39.2 N
Rationale: To start moving, overcome static friction: f_s,max = μ_s N = μ_s mg
= 0.40 × 10 × 9.8 = 39.2 N. Option A uses μ_k; Option C uses mg only; Option D
uses μ_s × mg × 6.25.
Question 7
A 5.0 kg mass hangs from a rope. What is the tension in the rope if the mass is
accelerating upward at 2.0 m/s²?
A) 39 N
B) 49 N
C) 59 N
D) 69 N
Correct Answer: C) 59 N
, Rationale: T - mg = ma → T = m(g + a) = 5.0(9.8 + 2.0) = 59 N. Option A is m(g-
a) for downward acceleration; Option B is mg at rest; Option D uses incorrect
addition.
Question 8
SATA A 15 kg block is on an inclined plane at 30° above horizontal. Which of the
following correctly describe the forces? (Select all that apply)
A) The normal force is 127 N
B) The component of gravity parallel to the incline is 73.5 N
C) The component of gravity perpendicular to the incline is 127 N
D) The normal force equals mg cos θ
E) The net force down the incline is mg sin θ (frictionless)
Correct Answers: A, B, C, D, E
Rationale: N = mg cos θ = 15 × 9.8 × cos 30° = 127 N (A, D). F_parallel = mg sin
θ = 15 × 9.8 × 0.5 = 73.5 N (B). F_perpendicular = mg cos θ = 127 N (C). On a
frictionless incline, net force = mg sin θ (E).
Question 9
Two blocks of masses 3.0 kg and 5.0 kg are in contact on a frictionless surface. A
force of 24 N is applied to the 3.0 kg block. What is the force exerted by the 3.0 kg
block on the 5.0 kg block?
A) 9.0 N
B) 15.0 N
C) 24.0 N
D) 8.0 N
Correct Answer: B) 15.0 N
Rationale: Common acceleration a = F/(m₁+m₂) = 24/(3+5) = 3.0 m/s². Force
on 5.0 kg block: F = m₂a = 5.0 × 3.0 = 15.0 N. Option A uses m₁a; Option C is the
applied force; Option D uses incorrect mass ratio.
Department – 2026/2027
DOMAIN 1: KINEMATICS AND NEWTON'S LAWS OF MOTION
Question 1
A sprinter accelerates uniformly from rest to 10.0 m/s in 3.5 s. What is the
sprinter's acceleration?
A) 2.86 m/s²
B) 3.50 m/s²
C) 1.43 m/s²
D) 5.71 m/s²
Correct Answer: A) 2.86 m/s²
Rationale: Acceleration equals change in velocity divided by time: a = Δv/Δt =
(10.0 m/s - 0)/3.5 s = 2.86 m/s². Option B incorrectly divides time by velocity;
Option C uses half the correct value; Option D uses an incorrect factor.
Question 2
A ball is thrown vertically upward with an initial velocity of 25 m/s. How long does
it take to reach its maximum height? (Assume g = 9.8 m/s²)
A) 1.28 s
B) 2.55 s
C) 3.82 s
D) 5.10 s
Correct Answer: B) 2.55 s
Rationale: At maximum height, v_f = 0. Using v_f = v_i - gt: 0 = 25 - 9.8t → t =
25/9.8 = 2.55 s. Option A uses g = 19.6; Option C uses 25/6.5; Option D uses
25/4.9 (half of g).
Question 3
,A car traveling at 30 m/s applies brakes and decelerates uniformly at 4.0 m/s².
What distance does the car travel before coming to rest?
A) 56.25 m
B) 112.5 m
C) 225 m
D) 450 m
Correct Answer: B) 112.5 m
Rationale: Using v_f² = v_i² + 2ad: 0 = (30)² + 2(-4.0)d → 0 = 900 - 8d → d =
112.5 m. Option A uses d = v²/(4a); Option C uses d = v²/a; Option D uses d =
2v²/a.
Question 4
SATA Which of the following statements about projectile motion are TRUE?
(Select all that apply)
A) The horizontal velocity is constant (neglecting air resistance)
B) The vertical acceleration is constant
C) The time to reach maximum height equals the time to return to the same
height
D) The trajectory is a parabola
E) The horizontal acceleration is -g
Correct Answers: A, B, C, D
Rationale: In projectile motion with no air resistance: horizontal velocity is
constant (A); vertical acceleration is constant at -g (B); symmetry gives equal rise
and fall times (C); the path is parabolic (D). Option E is false because horizontal
acceleration is 0, not -g.
Question 5
A 2.0 kg block is pushed across a frictionless horizontal surface with a force of 8.0
N. What is the block's acceleration?
,A) 2.0 m/s²
B) 4.0 m/s²
C) 6.0 m/s²
D) 16.0 m/s²
Correct Answer: B) 4.0 m/s²
Rationale: Newton's second law: F = ma → a = F/m = 8.0/2.0 = 4.0 m/s².
Option A uses m/2; Option C adds forces incorrectly; Option D multiplies F and m.
Question 6
A 10 kg box sits on a horizontal floor. The coefficient of static friction is 0.40 and
kinetic friction is 0.25. What horizontal force is required to start the box moving?
A) 24.5 N
B) 39.2 N
C) 98 N
D) 245 N
Correct Answer: B) 39.2 N
Rationale: To start moving, overcome static friction: f_s,max = μ_s N = μ_s mg
= 0.40 × 10 × 9.8 = 39.2 N. Option A uses μ_k; Option C uses mg only; Option D
uses μ_s × mg × 6.25.
Question 7
A 5.0 kg mass hangs from a rope. What is the tension in the rope if the mass is
accelerating upward at 2.0 m/s²?
A) 39 N
B) 49 N
C) 59 N
D) 69 N
Correct Answer: C) 59 N
, Rationale: T - mg = ma → T = m(g + a) = 5.0(9.8 + 2.0) = 59 N. Option A is m(g-
a) for downward acceleration; Option B is mg at rest; Option D uses incorrect
addition.
Question 8
SATA A 15 kg block is on an inclined plane at 30° above horizontal. Which of the
following correctly describe the forces? (Select all that apply)
A) The normal force is 127 N
B) The component of gravity parallel to the incline is 73.5 N
C) The component of gravity perpendicular to the incline is 127 N
D) The normal force equals mg cos θ
E) The net force down the incline is mg sin θ (frictionless)
Correct Answers: A, B, C, D, E
Rationale: N = mg cos θ = 15 × 9.8 × cos 30° = 127 N (A, D). F_parallel = mg sin
θ = 15 × 9.8 × 0.5 = 73.5 N (B). F_perpendicular = mg cos θ = 127 N (C). On a
frictionless incline, net force = mg sin θ (E).
Question 9
Two blocks of masses 3.0 kg and 5.0 kg are in contact on a frictionless surface. A
force of 24 N is applied to the 3.0 kg block. What is the force exerted by the 3.0 kg
block on the 5.0 kg block?
A) 9.0 N
B) 15.0 N
C) 24.0 N
D) 8.0 N
Correct Answer: B) 15.0 N
Rationale: Common acceleration a = F/(m₁+m₂) = 24/(3+5) = 3.0 m/s². Force
on 5.0 kg block: F = m₂a = 5.0 × 3.0 = 15.0 N. Option A uses m₁a; Option C is the
applied force; Option D uses incorrect mass ratio.
