Skip to topic
    ← Back to course topics

    Topic 2 – Motion and forces — Edexcel GCSE Physics

    Test yourself on Topic 2 – Motion and forces with PEARSON EDEXCEL GCSE practice questions.

    Start free

    7 days Premium · Then the free plan continues · No card, no charge

    Your focus

    1. 2.1 Explain that a scalar quantity has magnitude (size) but no specific direction

    Topic 2 – Motion and forces exam tips

    Quick Revision Summary (Key Takeaway)

    Pearson Edexcel GCSE Physics Topic 2 covers motion, vectors, Newton's laws of motion, and stopping distances. Mastering this topic requires applying key kinematic equations, interpreting velocity-time graphs, and calculating resultant forces to explain acceleration and momentum.

    Topic Overview

    Topic 2: Motion and Forces forms the mechanical backbone of Pearson Edexcel GCSE Physics, linking kinematic descriptions of how objects move with dynamic explanations of why they move. It covers linear motion, graph analysis, Newton's three laws, momentum, and practical road safety factors such as stopping distances.

    A deep grasp of this topic provides essential foundations for higher-tier mechanics, energy transformations, and astrophysical systems. Exam success demands fluent mathematical manipulation of motion formulas alongside clear conceptual descriptions of vector interactions and safety mechanisms.

    Key Concepts
    • →Distinction between scalar quantities (magnitude only) and vector quantities (magnitude and direction).
    • →Graphical interpretation: gradient of a distance-time graph represents speed; gradient of a velocity-time graph represents acceleration; area under a velocity-time graph represents displacement.
    • →Newton's laws of motion: inertia (First Law), F = ma (Second Law), and equal/opposite action-reaction pairs on interacting bodies (Third Law).
    • →Stopping distance equals thinking distance (affected by reaction time, alcohol, drugs, fatigue) plus braking distance (affected by speed, road conditions, tyre/brake wear, mass).
    • →Conservation of momentum in closed systems: total initial momentum equals total final momentum during collisions and explosions.
    Examiner Tips
    • 💡When calculating area under a velocity-time graph to find distance, split complex shapes accurately into standard geometric rectangles and triangles, and annotate them clearly.
    • 💡State the appropriate formula in standard algebraic terms before substituting numerical values to secure method marks even if a calculator entry error occurs.
    • 💡Remember to convert units before calculating: speeds in km/h must be converted to m/s (divide by 3.6), and masses in grams must be converted to kilograms (divide by 1000).
    Common Mistakes
    • Assuming an object must experience a net force to stay in motion at a constant speed, forgetting Newton's first law where zero resultant force maintains constant velocity.
    • Thinking that thinking distance increases if road conditions are icy; icy roads affect friction and braking distance, while thinking distance depends purely on reaction time and vehicle speed.
    • Confusing negative acceleration (deceleration) with moving backwards on a velocity-time graph, rather than looking at whether velocity values lie above or below the zero axis.
    Revision Plan
    1. 1Day 1-2: Master vectors vs scalars, distance-time graphs, and velocity-time graphs; practice calculating gradients and areas.
    2. 2Day 3-5: Memorise and apply the core equations: v = s/t, a = (v - u)/t, and v^2 - u^2 = 2as in multi-step problems.
    3. 3Day 6-8: Study Newton's three laws, free-body force diagrams, and terminal velocity step-by-step sequences.
    4. 4Day 9-10: Revise momentum (conservation and F = (mv - mu)/t) and road safety factors including thinking and braking distances.
    5. 5Day 11-14: Complete Pearson Edexcel past-paper questions under timed conditions and review against official mark schemes.
    Exam Question Types
    • 📋Graph interpretation questions: requiring calculation of acceleration via gradient or distance travelled via area under a line.
    • 📋Multi-step calculation questions: combining kinematic equations with F = ma or momentum conservation.
    • 📋6-mark extended response questions: explaining terminal velocity in terms of balanced forces or comparing vehicle safety features using rate of change of momentum.
    Command Word Expectations (PEARSON EDEXCEL)
    Explain

    Identify a physical effect and provide a linked scientific reason or mechanism using terms like 'because', 'therefore', or 'as a result'.

    Describe

    State the characteristics of a process, trend, or graph pattern without needing to explain the underlying theoretical cause.

    Determine

    Obtain an answer by calculating, measuring directly, or taking readings directly from a supplied graph or data set.

    How Students Lose Marks (Examiner Pitfalls)
    Pitfall: Confusing scalar and vector quantities, particularly speed versus velocity and distance versus displacement.
    ❌ Weak Answer (Loses Marks):The car has a constant velocity because it travels around the roundabout at 30 mph.
    Example improved answer:Although the car travels at a constant speed of 30 mph, its direction is continuously changing as it moves along a curved path. Because velocity is a vector quantity defined by both magnitude and direction, the car's velocity is constantly changing, meaning it is accelerating.
    Examiner Tip: Always state that velocity depends on direction. Mentioning circular motion must immediately prompt you to discuss changing velocity and resultant centripetal force.
    Pitfall: Muddling Newton's third law pairs with balanced forces acting on a single object (Newton's first law).
    ❌ Weak Answer (Loses Marks):A book on a desk does not move because gravity pulls it down and the normal reaction pushes up, which is Newton's third law.
    Example improved answer:The weight of the book and the upward normal contact force act on the same object, illustrating Newton's first law (resultant force equals zero). According to Newton's third law, the interaction pair for the Earth's gravitational pull on the book is the equal and opposite gravitational pull of the book on the Earth.
    Examiner Tip: Newton's third law action-reaction pairs must act on two different bodies and be of the exact same type of force (e.g., both gravitational or both contact forces).
    Step-by-Step Worked Solutions

    Question: A car of mass 1200 kg accelerates uniformly from rest to a velocity of 25 m/s over a distance of 125 m. Calculate the resultant force acting on the car.

    1. 1.Step 1: Identify given quantities: initial velocity u = 0 m/s, final velocity v = 25 m/s, displacement s = 125 m, mass m = 1200 kg.
    2. 2.Step 2: Choose and rearrange the kinematic formula v^2 - u^2 = 2as to find acceleration (a): a = (v^2 - u^2) / (2s).
    3. 3.Step 3: Calculate acceleration: a = (25^2 - 0^2) / (2 * 125) = 625 / 250 = 2.5 m/s^2.
    4. 4.Step 4: Use Newton's second law, F = m * a, to determine resultant force: F = 1200 kg * 2.5 m/s^2 = 3000 N.
    Final Answer: Resultant force = 3000 N

    Question: A lorry with a mass of 8000 kg travelling at 15 m/s collides with a stationary van of mass 2000 kg. After the collision, the two vehicles lock together. Calculate their common velocity directly after the collision.

    1. 1.Step 1: State the principle of conservation of momentum: total momentum before collision = total momentum after collision.
    2. 2.Step 2: Calculate initial momentum: p_initial = (m1 * u1) + (m2 * u2) = (8000 * 15) + (2000 * 0) = 120000 kg m/s.
    3. 3.Step 3: Set up equation for combined final mass and unknown velocity (v): (m1 + m2) * v = 120000.
    4. 4.Step 4: Solve for v: (8000 + 2000) * v = 120000 -> 10000 * v = 120000 -> v = 12 m/s.
    Final Answer: Common velocity = 12 m/s in the initial direction of the lorry
    Active Recall Memory Test
    What does the gradient of a velocity-time graph represent?
    Key Fact: Acceleration.
    State the formula linking initial velocity (u), final velocity (v), acceleration (a), and distance (s).
    Key Fact: v^2 - u^2 = 2as
    How does vehicle safety equipment like crumple zones reduce passenger injury during a crash?
    Key Fact: They increase the collision impact time, which decreases the rate of change of momentum and therefore reduces the force exerted on passengers (F = delta p / t).
    What two separate components sum together to give total stopping distance?
    Key Fact: Thinking distance and braking distance.
    Frequently Asked Questions
    How do you calculate distance from a velocity-time graph if the line is curved?
    When the line on a velocity-time graph is curved, you cannot use simple triangle or rectangle formulas. Instead, estimate the displacement by counting the total number of grid squares underneath the curve and multiplying by the area value of one individual square. Alternatively, you can split the area into approximate vertical trapeziums.
    What is the difference between thinking distance and braking distance?
    Thinking distance is the distance travelled by the vehicle during the driver's reaction time, before the brakes are physically applied. Braking distance is the distance travelled after the brakes are applied until the vehicle comes to a complete standstill. Together, thinking distance plus braking distance equals the total stopping distance.
    Why is circular motion at constant speed considered acceleration?
    Acceleration is defined as the rate of change of velocity. Velocity is a vector quantity that has both magnitude (speed) and direction. When an object moves in a circle, its direction is continually changing, which means its velocity is also continually changing. Because the velocity changes over time, the object is accelerating towards the centre of the circle.
    Which equation should I use when time is not given in a kinematics question?
    If an exam question asks for acceleration, distance, or final velocity and does not state or ask for time (t), use the formula v^2 - u^2 = 2as. Rearrange this equation to solve for the missing variable: a = (v^2 - u^2) / (2s) or s = (v^2 - u^2) / (2a). This is given on the Edexcel physics equation sheet.