"to":"speed","from":"speed_substituted","color":"orange","label":"calculate 0 + 0.4 × 15"}]},"order":2,"title":"Find the final speed","description":"Use the constant-acceleration equation v = u + at. Substituting the starting speed, acceleration and time earns the second method mark (M1). This correct value earns the accuracy mark (A1)."}]}],"generatedAt":"2026-07-22T08:15:32.373Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"6c6ecc21-eddf-4d09-be54-a05fe149bad0","revisionId":"5713036e-c9ab-4c33-81c6-3326672f1be2","questionNumber":null,"questionType":"LA","specification":"A toy car of mass $1.5 \\text{ kg}$ is accelerated from rest by a constant resultant force of $4.5 \\text{ N}$.","difficulty":5,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"0674e035-895a-40c7-a585-019c77396b68","label":null,"content":"Calculate the speed of the toy car after $8.0 \\text{ s}$.","markscheme":"- Uses $F = ma$ to find the acceleration: $a = \\frac{4.5}{1.5} = 3.0 \\text{ m/s}^2$ M1\n- Uses $v = u + at$ with $u = 0$ and $t = 8.0$ to find the final speed: $v = 0 + (3.0 \\times 8.0)$ M1\n- Obtains $v = 24 \\text{ m/s}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"newtons_laws_then_velocity","label":"Newton's laws then velocity","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"a=\\frac{F}{m}"},{"latex":"a=\\frac{4.5\\ \\mathrm{N}}{1.5\\ \\mathrm{kg}}"},{"latex":"a=3.0\\ \\mathrm{m\\,s^{-2}}"}]},"order":2,"title":"Calculate the acceleration","description":"Substitute the force and mass, then divide. This earns M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"v=u+at"},{"latex":"v=0+\\left(3.0\\ \\mathrm{m\\,s^{-2}}\\times8.0\\ \\mathrm{s}\\right)"},{"latex":"v=24\\ \\mathrm{m\\,s^{-1}}"}]},"order":3,"title":"Find the final speed","description":"From rest means u = 0 m/s. Substitute u = 0 m/s, a = 3.0 m/s² and t = 8.0 s into v = u + at; this earns the second M1. This gives v = 24 m/s, earning A1."}]}],"generatedAt":"2026-07-22T12:01:15.021Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"6d497900-811a-4699-b09c-53c9d3325a16","revisionId":"e1609277-7737-4c82-ad74-ea2bedd8f356","questionNumber":null,"questionType":"LA","specification":"Newton's third law, when applied to the collision between two bumper cars, A and B, as shown in the diagram, can be stated as follows:\n\n*The force exerted by bumper car A on bumper car B is equal in magnitude and opposite in direction to the force exerted by bumper car B on bumper car A.*\n\n![Diagram showing two bumper cars, Car A and Car B, in a head-on collision. Arrows indicate that Car A exerts a force to the right on Car B, and Car B exerts an equal and opposite force to the left on Car A during the impact.](https://assets.mathsgenie.co.uk/aqa-qb/7d11ae7b-94af-4262-8593-0add6e42f16d.png)","difficulty":8,"difficultyBand":"medium","totalMarks":4,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"9b16df9e-cc7e-4fc4-ad59-1e771e8246ea","label":null,"content":"Explain how this statement links to the conservation of momentum in the collision.","markscheme":"- States that according to Newton's third law, the forces on each car are equal in magnitude and opposite in direction ($F_A = -F_B$) B1\n- States that the contact time ($t$) during which the collision occurs is identical for both bumper cars B1\n- Relates force and momentum change using the equation $F = \\frac{\\Delta p}{t}$ or impulse $\\Delta p = F t$ M1\n- Explains that because the forces are equal and opposite and act for the same time, the momentum gained by one car is equal and opposite to the momentum lost by the other, meaning the total change in momentum is zero (momentum is conserved) A1","marks":4,"order":0,"markingRubric":null,"ex

Newton's laws and momentum

Question 18

Newton's laws and momentum Questions

Newton's laws and momentum

Question 18

Newton's laws and momentum Questions