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US Common Core Common Core·🚀 US Physics

US Common Core High School Physics — Assessment

90 minutes📊 60 marks📄 Assessment
📚 Subject revision notes↩ All exam papers
ℹ️ About this paper: This is an exam-board-aligned practice paper written in the style of US Common Core Common Core — not an official past paper. Use it for timed practice, then check against the mark scheme included below. For official past papers, see the exam board's website.
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US Common Core High School Physics — Assessment

Total marks: 60 · Duration: 90 minutes

Instructions to students

• Answer all questions. • Show all your work; credit may be given for a correct method even if the final answer is incorrect. • Calculators may be used. • Useful equations: v = u + at; KE = ½mv²; W = Fd; F = ma; P = W/t; wave speed = fλ; Q = mcΔT. Use g = 9.8 m/s² unless told otherwise. • The total number of points for this assessment is 60. • All data you need is given within each question.

Assessment

Section A — Short-answer Questions (44 points)

1. Kinematics

(a) A car starts from rest and accelerates uniformly at 3.0 m/s² for 5.0 s. Calculate its final velocity. [3]

(b) Calculate the distance the car travels in this time. [3]

(c) Explain the difference between distance and displacement, and between speed and velocity. [3]

2. Forces

(a) A net force of 12 N acts on an object of mass 4.0 kg. Calculate its acceleration. [3]

(b) Explain the difference between the mass and the weight of an object. [2]

(c) State Newton's third law of motion, and give an example of a Newton's-third-law pair of forces. [4]

3. Energy and work

(a) A 2.0 kg object moves at 6.0 m/s. Calculate its kinetic energy. [3]

(b) A force of 20 N pushes a box 5.0 m across a floor. Calculate the work done by the force. [3]

(c) The push takes 4.0 s. Calculate the power developed, and state the law of conservation of energy. [3]

4. Waves and circular motion

(a) A wave has a frequency of 250 Hz and a wavelength of 1.4 m. Calculate its speed. [3]

(b) A ball on a string is swung in a horizontal circle at constant speed. Explain why the ball is accelerating even though its speed is constant, and state the direction of the force that keeps it moving in a circle. [3]

(c) The ball completes 5 revolutions in 2.0 s. Calculate its frequency and its period. [3]

5. Thermal physics

(a) Explain the difference between conduction, convection and radiation as mechanisms of heat transfer. [4]

(b) Calculate the energy needed to raise the temperature of 0.50 kg of water by 20 °C. (Specific heat capacity of water = 4200 J/kg°C) [4]

Section B — Extended Questions (16 points)

6. Electricity

(a) A resistor has a potential difference of 9.0 V across it and a current of 0.30 A. Calculate its resistance and the power it dissipates. [4]

(b) Explain what happens to the total resistance and the total current when a second identical resistor is added in parallel with the first (with the same supply voltage), and explain why. [6]

7. Conservation of energy

A roller-coaster car of mass 500 kg is released from rest at the top of a track 20 m high.

Using the principle of conservation of energy, calculate the speed of the car at the bottom of the track (ignoring friction), and explain how your answer would be different if friction were not ignored. [6]

Answer Key and Worked Solutions

Section A

1. (a) v = u + at = 0 + 3.0 × 5.0 (M1 M1) = 15 m/s (A1). [3] (b) distance = ut + ½at² = 0 + ½ × 3.0 × 5.0² (M1 M1) = ½ × 3.0 × 25 = 37.5 m (A1). (Or use average velocity × time = 7.5 × 5 = 37.5 m.) [3] (c) (up to 3) Distance is the total length of the path travelled (a scalar); displacement is the straight-line distance in a given direction from start to finish (a vector). Speed is how fast something moves (scalar); velocity is speed in a stated direction (vector). [3]

2. (a) a = F/m = 12 / 4.0 (M1 M1) = 3.0 m/s² (A1). [3] (b) (up to 2) Mass is the amount of matter in an object (kg) and is the same everywhere; weight is the force of gravity on the object (N) and depends on the gravitational field strength (so it changes with location). [2] (c) (up to 4) Newton's third law: when object A exerts a force on object B, B exerts an equal and opposite force on A (2). Example: a swimmer pushes the water backwards, and the water pushes the swimmer forwards (or rocket/gas, foot/ground) — a valid equal-and-opposite pair on two different objects (2). [4]

3. (a) KE = ½mv² = ½ × 2.0 × 6.0² (M1 M1) = ½ × 2.0 × 36 = 36 J (A1). [3] (b) W = Fd = 20 × 5.0 (M1 M1) = 100 J (A1). [3] (c) P = W/t = 100 / 4.0 = 25 W (M1 A1); conservation of energy: energy cannot be created or destroyed, only transferred/transformed from one form to another (A1). [3]

4. (a) v = fλ = 250 × 1.4 (M1 M1) = 350 m/s (A1). [3] (b) (up to 3) The ball's direction is constantly changing, so its velocity is changing, which means it is accelerating (even at constant speed); the force (centripetal force, provided by the tension in the string) acts towards the centre of the circle. [3] (c) frequency = 5 ÷ 2.0 = 2.5 Hz (M1 A1); period = 1/f = 1/2.5 = 0.4 s (A1). [3]

5. (a) (up to 4) Conduction — heat transfer through a solid by vibrating particles passing energy to neighbours (no net movement of the material); convection — heat transfer in fluids (liquids/gases) as warmer, less dense fluid rises and cooler fluid sinks, forming currents; radiation — heat transfer by infrared electromagnetic waves, needing no medium (can travel through a vacuum). Two marks per clear distinction, up to 4. [4] (b) Q = mcΔT = 0.50 × 4200 × 20 (M1 M1) = 42,000 J (M1 A1). [4]

Section B

6. (a) R = V/I = 9.0 / 0.30 = 30 Ω (M1 A1); P = VI = 9.0 × 0.30 = 2.7 W (M1 A1). [4] (b) (up to 6) Adding a second identical resistor in parallel decreases the total resistance (it halves, to 15 Ω), because there are now two paths for the current; since the supply voltage is unchanged, by I = V/R the total current increases (it doubles). Each branch still has the same voltage and current as before, but the supply now provides current to both branches. Award for: total resistance decreases/halves (2), reason of extra path (2), total current increases/doubles with reason (2). [6]

7. (up to 6) At the top, GPE = mgh = 500 × 9.8 × 20 = 98,000 J (M1). By conservation of energy, this converts to KE at the bottom: ½mv² = 98,000 (M1). v² = (2 × 98,000)/500 = 392, so v = √392 = 19.8 m/s (M1 A1). If friction were not ignored, some energy would be transferred to thermal energy (and sound) due to friction/air resistance, so less energy would become KE and the actual speed at the bottom would be lower than 19.8 m/s (M1 A1). [6]

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