Newton’s laws of motion — OCR A-Level Physics
Test yourself on Newton’s laws of motion with OCR A-Level practice questions.
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Newton’s laws of motion explained
Newton's three laws describe how forces change motion.
Read the full explanation
The first law: a body remains at rest or moves with constant velocity unless acted on by a resultant force, so zero net force means zero acceleration. The second law: the resultant force equals the rate of change of momentum, which for constant mass reduces to F = ma, so acceleration is proportional to net force and inversely proportional to mass. The third law: if body A exerts a force on body B, then B exerts an equal-magnitude force on A in the opposite direction, acting on a different body. For example, a rocket pushes exhaust gases backwards while the gases push the rocket forwards.
(b) linear momentum; p = mv ; vector nature of momentum
Linear momentum p is defined as the product of mass and velocity: p = mv. Its SI unit is kg m s⁻¹, equivalently N s. Momentum is a vector, so it has both magnitude and direction; the direction of p is the direction of the velocity. When solving problems, choose a positive direction first, then assign signs to velocities and momenta. For example, a 2.0 kg ball moving at 3.0 m s⁻¹ to the right has momentum 6.0 kg m s⁻¹ to the right; if it moves left at the same speed, its momentum is −6.0 kg m s⁻¹. Total momentum of a system is found by vector addition of individual momenta.
(c) net force = rate of change of momentum; F = Δp/Δt
Newton's second law states that the net (resultant) force on a body equals the rate of change of its momentum: F = Δp/Δt, where Δp is the change in momentum and Δt is the time interval over which it occurs. For constant mass, Δp = mΔv, so F = mΔv/Δt = ma. The force and the change in momentum are in the same direction. For example, a 0.15 kg ball changing velocity from 4.0 m s⁻¹ to −4.0 m s⁻¹ in 0.20 s experiences Δp = 0.15 × (−8.0) = −1.2 kg m s⁻¹, giving F = −1.2 / 0.20 = −6.0 N, i.e. 6.0 N opposite to the initial direction.
(d) impulse of a force; impulse = F t T
The impulse of a force is the product of the force and the time for which it acts: impulse = FΔt. Since F = Δp/Δt, impulse equals the change in momentum, so impulse = Δp = mΔv. Impulse is a vector with the same direction as the force, and its SI unit is N s, equivalent to kg m s⁻¹. For example, a force of 20 N acting for 0.50 s gives an impulse of 10 N s, changing the momentum of the body by 10 kg m s⁻¹ in the force's direction. Increasing the contact time for a given momentum change reduces the force, which is why crumple zones and airbags reduce injury.
(e) impulse is equal to the area under a force–time graph.
Impulse is the product of a force and the time for which it acts, and it equals the change in momentum. When the force varies, you cannot simply multiply one force value by the time. Instead, plot force on the y-axis against time on the x-axis; the area between the graph line and the time axis equals the impulse. Split the area into rectangles, triangles or trapezia, or count squares, then state the unit: N s, equivalent to kg m s⁻¹. For example, a constant 20 N for 3 s gives a rectangular area of 60 N s, so the momentum change is 60 kg m s⁻¹. A triangular spike of peak 40 N lasting 0.5 s gives area ½ × 40 × 0.5 = 10 N s.
Your focus
- State each of Newton's three laws of motion accurately.
- Identify which law applies to a described physical situation.
- Distinguish third-law pairs from balanced forces on one body.
Show all 15 objectives
- Define linear momentum and state its SI unit.
- Calculate momentum using p = mv with correct signs.
- Add momenta as vectors in one-dimensional problems.
- State Newton's second law as F = Δp/Δt.
- Calculate the net force from a change in momentum and time interval.
- Show that F = Δp/Δt reduces to F = ma for constant mass.
- Define impulse as FΔt and state its unit.
- Relate impulse to change in momentum using impulse = Δp.
- Explain how increasing contact time reduces the force for a given momentum change.
- Interpret the area under a force–time graph as impulse.
- Calculate impulse from rectangular, triangular or composite areas.
- Relate impulse to the change in momentum of a body.
Newton’s laws of motion exam tips
Marking Points
- First law: a body stays at rest or at constant velocity unless a resultant force acts on it.
- Second law: resultant force equals rate of change of momentum; for constant mass, F = ma.
- Third law: forces come in pairs, equal in magnitude, opposite in direction, acting on two different bodies.
- Newton's third-law pairs act on different objects, so they never cancel on a single body.
- Applying F = ma requires the resultant force, not any single applied force.
- Momentum is defined as p = mv, the product of mass and velocity.
- The SI unit of momentum is kg m s⁻¹, which is equivalent to N s.
- Momentum is a vector quantity with the same direction as the velocity.
- Signs must be assigned consistently when adding momenta in one dimension.
- Total momentum of a system is the vector sum of the individual momenta.
- Net force equals the rate of change of momentum: F = Δp/Δt.
- Δp is the change in momentum, calculated as final momentum minus initial momentum.
- For constant mass, F = Δp/Δt reduces to F = ma.
- The direction of the net force is the same as the direction of the change in momentum.
- The time interval Δt must be the duration over which the momentum change occurs.
- Impulse is defined as the product of force and the time for which it acts: impulse = FΔt.
- Impulse equals the change in momentum: impulse = Δp = mΔv.
- The SI unit of impulse is N s, equivalent to kg m s⁻¹.
- Impulse is a vector in the direction of the applied force.
- For a fixed momentum change, a longer contact time gives a smaller average force.
- Impulse equals force × time for a constant force, and equals the area under a force–time graph when the force varies.
- Impulse equals the change in momentum, so the area has units N s or kg m s⁻¹.
- Area below the time axis represents negative impulse, reducing momentum in the chosen positive direction.
- For a varying force, split the area into standard shapes or count squares rather than using a single force value.
Examiner Tips
- 💡Underline the word resultant in the question before choosing an answer.
- 💡Check whether the question describes one body or two interacting bodies.
- 💡For constant-velocity situations, immediately set resultant force to zero.
- 💡Always state your chosen positive direction before substituting values.
- 💡Check units: kg m s⁻¹ and N s are both acceptable for momentum.
- 💡When two objects move in opposite directions, give one a negative velocity.
- 💡Write down p_initial and p_final separately before finding Δp.
- 💡Keep signs consistent with your chosen positive direction throughout.
- 💡Check that your answer's direction matches the direction of Δp.
- 💡Use impulse = Δp when the force is not constant or is unknown.
- 💡Check that time is in seconds before multiplying by force.
- 💡Link safety features to increased contact time reducing peak force.
- 💡Sketch the graph and shade the area before calculating, so the shape and limits are clear.
- 💡Convert milliseconds to seconds and grams to kilograms before finding any area.
- 💡State the unit of impulse as N s or kg m s⁻¹ and link the numerical answer to the momentum change.
Common Mistakes
- Thinking a moving object always needs a forward force; correction: constant velocity needs zero resultant force.
- Treating Newton's third-law pairs as balanced forces on one body; correction: the two forces act on different bodies.
- Using any single force in F = ma instead of the resultant force; correction: resolve all forces first, then divide the resultant by mass.
- Treating momentum as a scalar and ignoring direction; correction: assign a positive direction and use signs.
- Using speed instead of velocity in p = mv; correction: velocity includes direction, so use the signed value.
- Forgetting that momentum depends on mass as well as velocity; correction: calculate p = mv with both quantities.
- Using initial momentum instead of the change in momentum; correction: calculate Δp = p_final − p_initial.
- Ignoring direction when momentum reverses; correction: treat reversal as a larger change, e.g. from +4.0 to −4.0 gives Δv = −8.0 m s⁻¹.
- Confusing F = Δp/Δt with F = p/t; correction: divide the change in momentum by the time interval, not the momentum itself.
- Confusing impulse with force; correction: impulse is force multiplied by time, measured in N s.
- Forgetting that impulse equals change in momentum; correction: use impulse = Δp to link force and motion.
- Assuming a longer contact time increases the force; correction: for a fixed Δp, longer time means smaller average force.
- Multiplying the peak force by the total time when the force varies: the correct method is to find the area under the graph, which is smaller than the peak-force rectangle.
- Reading the gradient of a force–time graph as impulse: the gradient gives the rate of change of force, while impulse is the area.
- Ignoring the sign of areas below the time axis: the correct treatment is to subtract those areas because they represent impulse in the opposite direction.