AQA AS Paper 1 2018 June — Question 12 1 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2018
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton's laws and connected particles
TypeSingle particle, Newton's second law – scalar (1D, horizontal or inclined)
DifficultyEasy -1.8 This is a straightforward application of Newton's second law (F=ma) requiring only one step: 5 × 0.6 = 3, so F - R = 3. It's a 1-mark multiple choice question testing basic recall of a fundamental formula with no problem-solving or conceptual challenge required.
Spec3.03c Newton's second law: F=ma one dimension

An object of mass \(5\,\mathrm{kg}\) is moving in a straight line. As a result of experiencing a forward force of \(F\) newtons and a resistant force of \(R\) newtons it accelerates at \(0.6\,\mathrm{m}\,\mathrm{s}^{-2}\) Which one of the following equations is correct? Circle your answer. [1 mark] \(F - R = 0\) \quad \(F - R = 5\) \quad \(F - R = 3\) \quad \(F - R = 0.6\)

Question 12:
AnswerMarks Guidance
12Circles correct answer AO2.2a
Total1
QMarking Instructions AO
Question 12:
12 | Circles correct answer | AO2.2a | B1 | F - R = 3
Total | 1
Q | Marking Instructions | AO | Marks | Typical Solution
An object of mass $5\,\mathrm{kg}$ is moving in a straight line.

As a result of experiencing a forward force of $F$ newtons and a resistant force of $R$ newtons it accelerates at $0.6\,\mathrm{m}\,\mathrm{s}^{-2}$

Which one of the following equations is correct?

Circle your answer.
[1 mark]

$F - R = 0$ \quad $F - R = 5$ \quad $F - R = 3$ \quad $F - R = 0.6$

\hfill \mbox{\textit{AQA AS Paper 1 2018 Q12 [1]}}