AQA Paper 2 2019 June — Question 12 1 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2019
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeForces in vector form: equilibrium (find unknowns)
DifficultyEasy -1.8 This is a 1-mark multiple choice question testing only Newton's First Law: constant velocity means zero resultant force. Students simply equate the i-components (400 + p = 0) to get p = -400. No calculation of acceleration, no consideration of mass, and the answer is given among options—this is significantly easier than average A-level mechanics.
Spec3.03b Newton's first law: equilibrium

A particle, under the action of two constant forces, is moving across a perfectly smooth horizontal surface at a constant speed of \(10 \text{ m s}^{-1}\) The first force acting on the particle is \((400\mathbf{i} + 180\mathbf{j})\) N. The second force acting on the particle is \((p\mathbf{i} - 180\mathbf{j})\) N. Find the value of \(p\). Circle your answer. [1 mark] \(-400\) \quad \(-390\) \quad \(390\) \quad \(400\)

Question 12:
AnswerMarks Guidance
12Circles correct answer 1.1b
Total1
QMarking Instructions AO
Question 12:
12 | Circles correct answer | 1.1b | B1 | -400
Total | 1
Q | Marking Instructions | AO | Marks | Typical Solution
A particle, under the action of two constant forces, is moving across a perfectly smooth horizontal surface at a constant speed of $10 \text{ m s}^{-1}$

The first force acting on the particle is $(400\mathbf{i} + 180\mathbf{j})$ N.

The second force acting on the particle is $(p\mathbf{i} - 180\mathbf{j})$ N.

Find the value of $p$.

Circle your answer.
[1 mark]

$-400$ \quad $-390$ \quad $390$ \quad $400$

\hfill \mbox{\textit{AQA Paper 2 2019 Q12 [1]}}