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Its thruster is activated, exerting a constant forward force of $1.8 \\text{ N}$ on the rover for a duration of $15 \\text{ s}$.","difficulty":4,"difficultyBand":"medium","tier":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"b55cecf1-e65c-4944-a6be-fb80519d4294","label":null,"content":"Calculate the speed of the rover after $15 \\text{ s}$.","markscheme":"- Calculates acceleration using Newton's Second Law: $a = \\frac{1.8}{4.5} = 0.4 \\text{ m/s}^2$ OR calculates impulse: $I = 1.8 \\times 15 = 27 \\text{ N s}$ M1\n- Uses $v = u + at$ with $u = 0$, $a = 0.4$ and $t = 15$ OR equates impulse to change in momentum: $27 = 4.5 \\times v$ M1\n- Obtains a final speed of $6 \\text{ m/s}$ (or $6.0 \\text{ m/s}$) A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"impulse_momentum","label":"Impulse and momentum","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"I=F\\Delta t"},{"latex":"\\htmlData{anchor=impulse_substituted}{I=1.8\\ \\mathrm{N}\\times15\\ \\mathrm{s}}"},{"latex":"\\htmlData{anchor=impulse}{I=27\\ \\mathrm{N\\,s}}"}],"highlights":[{"color":"green","style":"box","anchor":"impulse"}],"connections":[{"to":"impulse","from":"impulse_substituted","color":"teal","label":"calculate 1.8 × 15"}]},"order":1,"title":"Calculate the impulse","description":"Multiply the constant force by the time for which it acts. This earns the first method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"I=\\Delta p"},{"latex":"27\\ \\mathrm{N\\,s}=4.5\\ \\mathrm{kg}\\times(v-0\\ \\mathrm{m/s})"},{"latex":"\\htmlData{anchor=momentum_equation}{27=4.5v}"},{"latex":"\\htmlData{anchor=momentum_speed}{v=6\\ \\mathrm{m/s}}"}],"highlights":[{"color":"green","style":"box","anchor":"momentum_speed"}],"connections":[{"to":"momentum_speed","from":"momentum_equation","color":"orange","label":"divide both sides by 4.5"}]},"order":2,"title":"Equate momentum change","description":"Set the impulse equal to the rover's change in momentum. Since it starts at rest, its initial momentum is zero. This earns the second method mark (M1), and this correct final value earns the accuracy mark (A1)."}]},{"id":"newtons_second_law","label":"Newton's second law","steps":[{"type":"text","order":0,"title":"Identify the starting speed","content":"The rover is initially stationary, so its initial speed is $u=0\\ \\mathrm{m/s}$. The surface is frictionless, so the thruster provides the forward resultant force of $1.8\\ \\mathrm{N}$.","highlights":[{"text":"initially stationary","location":"specification"},{"text":"horizontal, frictionless surface","location":"specification"},{"text":"constant forward force of $1.8 \\text{ N}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"a=\\frac{F}{m}"},{"latex":"\\htmlData{anchor=acceleration_substituted}{a=\\frac{1.8\\ \\mathrm{N}}{4.5\\ \\mathrm{kg}}}"},{"latex":"\\htmlData{anchor=acceleration}{a=0.4\\ \\mathrm{m/s^2}}"}],"highlights":[{"color":"green","style":"box","anchor":"acceleration"}],"connections":[{"to":"acceleration","from":"acceleration_substituted","color":"sky","label":"calculate 1.8 ÷ 4.5"}]},"order":1,"title":"Calculate the acceleration","description":"Newton's second law gives acceleration as resultant force divided by mass. This calculation earns the first method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"v=u+at"},{"latex":"\\htmlData{anchor=speed_substituted}{v=0\\ \\mathrm{m/s}+(0.4\\ \\mathrm{m/s^2})\\times15\\ \\mathrm{s}}"},{"latex":"\\htmlData{anchor=speed}{v=6\\ \\mathrm{m/s}}"}],"highlights":[{"color":"green","style":"box","anchor":"speed"}],"connections":[{"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,"walkthroughVideo":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","tier":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"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,"tableImages":[],"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,"walkthroughVideo":null},{"id":"7def9745-ea62-44c3-8782-91db27719a87","revisionId":"53b25c58-af4d-4522-b546-47058e01d290","questionNumber":null,"questionType":"LA","specification":"A speedboat accelerates at a constant rate of $1.65 \\text{ m/s}^2$ across a calm lake.\n\nThe forward thrust force acting on the speedboat is $2.450 \\text{ kN}$.","difficulty":4,"difficultyBand":"medium","tier":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"9089993b-17ae-44ca-9042-592d77312ce3","label":null,"content":"Calculate the mass of the speedboat.\n\nGive your answer to three significant figures.","markscheme":"- Converts thrust force to $2450 \\text{ N}$ B1\n- Rearranges $F = ma$ and substitutes values to get $m = \\frac{2450}{1.65}$ M1\n- Obtains $1480 \\text{ kg}$ (to 3 s.f.) A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"newtons_second_law","label":"Use Newton's second law","steps":[{"type":"text","order":0,"title":"Convert the thrust","content":"The thrust is given in kilonewtons, but the force unit in $F=ma$ is newtons.\n\nSince $1\\text{ kN}=1000\\text{ N}$,\n\n$$2.450\\text{ kN}=2450\\text{ N}.$$\n\nThis unit conversion earns the **B1** mark.","highlights":[{"text":"$$2.450 \\text{ kN}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=law}{F=ma}"},{"latex":"m=\\frac{\\htmlData{anchor=force-symbol}{F}}{\\htmlData{anchor=acceleration-symbol}{a}}"},{"latex":"m=\\frac{\\htmlData{anchor=force-value}{2450}}{\\htmlData{anchor=acceleration-value}{1.65}}"}],"connections":[{"to":"force-symbol","from":"law","color":"sky","label":"divide by a"},{"to":"force-value","from":"force-symbol","color":"orange","label":"substitute force"},{"to":"acceleration-value","from":"acceleration-symbol","color":"orange","label":"substitute acceleration"}]},"order":1,"title":"Rearrange the equation","description":"Use Newton's second law: F = ma. Divide both sides by the acceleration to make mass the subject. This rearrangement and substitution earn the M1 mark."},{"type":"text","order":2,"title":"Round the mass","content":"$$$m=1484.848\\ldots\\text{ kg}$$\n\nTo three significant figures, this is\n\n$$\\boxed{1480\\text{ kg}}.$$\n\nThe zero is needed to show the answer is given to three significant figures. This earns the **A1** mark.","highlights":[{"text":"Give your answer to three significant 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Forces and Newton's laws","topicId":"1ab93471-f52f-5707-8456-d4bc733a362c","topicSidebar":"$L3c","topicSlug":"newton-s-laws-and-momentum","topicTitle":"2.3 Forces and Newton's laws"}]}],"$L3d","$L3e"]}]

2.3 Forces and Newton's laws

Question 8

2.3 Forces and Newton's laws Questions

2.3 Forces and Newton's laws

Question 8

2.3 Forces and Newton's laws Questions