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2026 Statistics
Key Facts: VWO Natuurkunde Exam
3 hours
CE Exam Duration
CvTE 2026 Syllabus
1–10
Dutch Grading Scale
Examenblad.nl
≥ 5.5
Passing Subject Mark
CvTE Regulations
Binas 7e
Allowed Data Book
CvTE Allowed Aids
100 MCQs
Practice Adaptations
OpenExamPrep
50% CE / 50% SE
Grade Weighting
Dutch Secondary Ed
9 Subdomains
CvTE Core Topics
Examenblad Syllabus
Graphic Calc
Permitted Aid
CvTE Calculator Policy
Official VWO Natuurkunde 2026 is CE (3-hour written exam) plus school SE. These 100 English MCQs provide practice aligned to CvTE VWO physics CE subdomains with numerical calculation questions.
Sample VWO Natuurkunde Practice Questions
Try these sample questions to test your VWO Natuurkunde exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A car accelerates uniformly from rest to a speed of 25.0 m/s in 5.0 seconds. What is the magnitude of the car's acceleration?
A.2.5 m/s²
B.5.0 m/s²
C.10.0 m/s²
D.125 m/s²
Explanation: Uniform acceleration is defined as a = Δv / Δt. Here, Δv = 25.0 m/s - 0 m/s = 25.0 m/s and Δt = 5.0 s. Therefore, a = 25.0 / 5.0 = 5.0 m/s².
2A cyclist travels a distance of 18.0 km in 45.0 minutes. What is the average speed of the cyclist in meters per second (m/s)?
A.4.00 m/s
B.6.67 m/s
C.10.0 m/s
D.24.0 m/s
Explanation: Convert distance to meters: 18.0 km = 18,000 m. Convert time to seconds: 45.0 min = 45.0 × 60 = 2700 s. Average speed v = s / t = 18,000 / 2700 = 6.67 m/s.
3Two perpendicular forces act on an object: F1 = 30.0 N pointing East and F2 = 40.0 N pointing North. What is the magnitude of the resultant net force?
A.10.0 N
B.50.0 N
C.70.0 N
D.1200 N
Explanation: Because the forces are perpendicular, the magnitude of the resultant vector is calculated using the Pythagorean theorem: F_res = √(F1² + F2²) = √(30² + 40²) = √(900 + 1600) = √2500 = 50.0 N.
4An object is dropped from rest near the Earth's surface. Neglecting air resistance, what vertical distance does it cover during the first 3.0 seconds of free fall? (Use g = 9.81 m/s²)
A.14.7 m
B.29.4 m
C.44.1 m
D.88.3 m
Explanation: For free fall from rest (v0 = 0), distance s = 0.5 × g × t². Plugging in g = 9.81 m/s² and t = 3.0 s: s = 0.5 × 9.81 × (3.0)² = 0.5 × 9.81 × 9.0 = 44.1 m.
5A 1200 kg vehicle moves along a straight horizontal road at a constant velocity of 15.0 m/s. What is the linear momentum of the vehicle?
A.80.0 kg·m/s
B.9.00 × 10³ kg·m/s
C.1.80 × 10⁴ kg·m/s
D.1.35 × 10⁵ kg·m/s
Explanation: Linear momentum p = m × v = 1200 kg × 15.0 m/s = 18,000 kg·m/s = 1.80 × 10⁴ kg·m/s.
6A car traveling at 20.0 m/s applies its brakes, causing a uniform deceleration of 4.00 m/s² until it comes to a complete stop. What braking distance is required?
A.25.0 m
B.50.0 m
C.80.0 m
D.100 m
Explanation: Using kinematic formula v² = u² + 2as, where final velocity v = 0 m/s, initial velocity u = 20.0 m/s, acceleration a = -4.00 m/s²: 0 = (20.0)² + 2(-4.00)s => 8.00 s = 400 => s = 50.0 m.
7A 0.50 kg stone tied to a string is rotated in a horizontal circle of radius 1.20 m at a constant speed of 6.00 m/s. What is the centripetal force acting on the stone?
A.2.50 N
B.15.0 N
C.18.0 N
D.30.0 N
Explanation: Centripetal force Fc = m × v² / r. Substituting m = 0.50 kg, v = 6.00 m/s, and r = 1.20 m: Fc = 0.50 × (6.00)² / 1.20 = 0.50 × 36.0 / 1.20 = 15.0 N.
8A ball is launched horizontally from a cliff of height 45.0 m with an initial horizontal velocity of 12.0 m/s. How far from the cliff's base does the ball land? (g = 9.81 m/s²)
A.18.0 m
B.36.3 m
C.45.0 m
D.109 m
Explanation: Time of flight t = √(2h / g) = √(2 × 45.0 / 9.81) = √9.174 ≈ 3.029 s. Horizontal distance x = vx × t = 12.0 m/s × 3.029 s ≈ 36.3 m.
9A 1000 kg car moving East at 10.0 m/s collides inelastically with a stationary 1500 kg truck and sticks to it. What is the velocity of the combined wreck immediately after impact?
A.2.00 m/s
B.4.00 m/s
C.6.00 m/s
D.10.0 m/s
Explanation: Conservation of momentum: m1 × v1 + m2 × v2 = (m1 + m2) × vf. (1000 × 10.0) + 0 = (1000 + 1500) × vf => 10,000 = 2500 vf => vf = 4.00 m/s.
10A velocity-time graph shows a linear speed increase from 0 to 30.0 m/s over 6.0 s, followed by a constant speed of 30.0 m/s for 4.0 s. What is the total displacement?
A.150 m
B.180 m
C.210 m
D.300 m
Explanation: Displacement equals the area under the v-t graph. Acceleration phase area (triangle) = 0.5 × 6.0 s × 30.0 m/s = 90 m. Constant speed area (rectangle) = 4.0 s × 30.0 m/s = 120 m. Total displacement = 90 + 120 = 210 m.
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