AQA Paper 2 2019 June — Question 17 9 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2019
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeMotion with applied force on slope
DifficultyStandard +0.3 Part (a) is a standard mechanics derivation requiring resolution of forces (horizontal and vertical), application of F=ma and friction law F=ΞΌR - routine for A-level mechanics with clear structure. Part (b) tests conceptual understanding that the formula assumes motion (kinetic friction) but the sledge is at rest (static friction), requiring only brief explanation. This is slightly easier than average due to its standard setup and straightforward multi-step approach.
Spec3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model
Lizzie is sat securely on a wooden sledge. The combined mass of Lizzie and the sledge is \(M\) kilograms. The sledge is being pulled forward in a straight line along a horizontal surface by means of a light inextensible rope, which is attached to the front of the sledge. This rope stays inclined at an acute angle \(\theta\) above the horizontal and remains taut as the sledge moves forward. \includegraphics{figure_17} The sledge remains in contact with the surface throughout. The coefficient of friction between the sledge and the surface is \(\mu\) and there are no other resistance forces. Lizzie and the sledge move forward with constant acceleration, \(a \text{ m s}^{-2}\) The tension in the rope is a constant \(T\) Newtons.
  1. Show that $$T = \frac{M(a + \mu g)}{\cos \theta + \mu \sin \theta}$$ [7 marks]
  2. It is known that when \(M = 30\), \(\theta = 30Β°\), and \(T = 40\), the sledge remains at rest. Lizzie uses these values with the relationship formed in part (a) to find the value for \(\mu\) Explain why her value for \(\mu\) may be incorrect. [2 marks]
Question 17:
AnswerMarks
17(a)Resolves vertically to form a three
term equation
AnswerMarks Guidance
Condone sign error or sin/cos error3.1b M1
𝑇𝑇cosπœƒπœƒβˆ’ 𝐹𝐹 = π‘€π‘€π‘Žπ‘Ž
F =Β΅R
𝑇𝑇cosπœƒπœƒβˆ’πœ‡πœ‡π‘…π‘… = π‘€π‘€π‘Žπ‘Ž
𝑇𝑇cosπœƒπœƒβˆ’πœ‡πœ‡(π‘€π‘€π˜¨π˜¨βˆ’π‘‡π‘‡sinπœƒπœƒ)= π‘€π‘€π‘Žπ‘Ž
𝑇𝑇(cosπœƒπœƒ+πœ‡πœ‡sinπœƒπœƒ)= π‘€π‘€π‘Žπ‘Ž+πœ‡πœ‡π‘€π‘€π˜¨π˜¨
𝑀𝑀(π‘Žπ‘Ž+πœ‡πœ‡π˜¨π˜¨)
𝑇𝑇 =
cosπœƒπœƒ+πœ‡πœ‡sinπœƒπœƒ
Obtains fully correct equation for
AnswerMarks Guidance
resolving vertically1.1b A1
Uses Newton’s second law
horizontally to form a three term
equation
Condone sign error or consistent
AnswerMarks Guidance
cos/sin error3.1b M1
Obtains fully correct equation for
AnswerMarks Guidance
resolving horizontally1.1b A1
Uses F =Β΅R to replace F with Β΅R in
AnswerMarks Guidance
their horizontal equation3.3 B1
Eliminates R to form a single
AnswerMarks Guidance
equation1.1a M1
Completes rigorous argument to find
required expression.
Must see T as a factor before
division e.g
AnswerMarks Guidance
AG2.1 R1
AnswerMarks
17(b)Explains tha.𝑇𝑇t ( thceos rπœƒπœƒela+tiπœ‡πœ‡onsisnhπœƒπœƒip ) may
not be valid because the sledge is at
AnswerMarks Guidance
rest2.4 B1
relationship may not be valid as
friction may not be acting at its
limiting value
Identifies that friction may not be at
its limiting value
Accept reference to F ≀¡R
Sledge may not be on the point of
AnswerMarks Guidance
slipping3.5b B1
Total9 
Question 17:
--- 17(a) ---
17(a) | Resolves vertically to form a three
term equation
Condone sign error or sin/cos error | 3.1b | M1 | 𝑅𝑅+𝑇𝑇sin πœƒπœƒ = π‘€π‘€π˜¨π˜¨
𝑇𝑇cosπœƒπœƒβˆ’ 𝐹𝐹 = π‘€π‘€π‘Žπ‘Ž
F =Β΅R
𝑇𝑇cosπœƒπœƒβˆ’πœ‡πœ‡π‘…π‘… = π‘€π‘€π‘Žπ‘Ž
𝑇𝑇cosπœƒπœƒβˆ’πœ‡πœ‡(π‘€π‘€π˜¨π˜¨βˆ’π‘‡π‘‡sinπœƒπœƒ)= π‘€π‘€π‘Žπ‘Ž
𝑇𝑇(cosπœƒπœƒ+πœ‡πœ‡sinπœƒπœƒ)= π‘€π‘€π‘Žπ‘Ž+πœ‡πœ‡π‘€π‘€π˜¨π˜¨
𝑀𝑀(π‘Žπ‘Ž+πœ‡πœ‡π˜¨π˜¨)
𝑇𝑇 =
cosπœƒπœƒ+πœ‡πœ‡sinπœƒπœƒ
Obtains fully correct equation for
resolving vertically | 1.1b | A1
Uses Newton’s second law
horizontally to form a three term
equation
Condone sign error or consistent
cos/sin error | 3.1b | M1
Obtains fully correct equation for
resolving horizontally | 1.1b | A1
Uses F =Β΅R to replace F with Β΅R in
their horizontal equation | 3.3 | B1
Eliminates R to form a single
equation | 1.1a | M1
Completes rigorous argument to find
required expression.
Must see T as a factor before
division e.g
AG | 2.1 | R1
--- 17(b) ---
17(b) | Explains tha.𝑇𝑇t ( thceos rπœƒπœƒela+tiπœ‡πœ‡onsisnhπœƒπœƒip ) may
not be valid because the sledge is at
rest | 2.4 | B1 | The sledge is at rest so the
relationship may not be valid as
friction may not be acting at its
limiting value
Identifies that friction may not be at
its limiting value
Accept reference to F ≀¡R
Sledge may not be on the point of
slipping | 3.5b | B1
Total | 9
Lizzie is sat securely on a wooden sledge.

The combined mass of Lizzie and the sledge is $M$ kilograms.

The sledge is being pulled forward in a straight line along a horizontal surface by means of a light inextensible rope, which is attached to the front of the sledge.

This rope stays inclined at an acute angle $\theta$ above the horizontal and remains taut as the sledge moves forward.

\includegraphics{figure_17}

The sledge remains in contact with the surface throughout.

The coefficient of friction between the sledge and the surface is $\mu$ and there are no other resistance forces.

Lizzie and the sledge move forward with constant acceleration, $a \text{ m s}^{-2}$

The tension in the rope is a constant $T$ Newtons.

\begin{enumerate}[label=(\alph*)]
\item Show that
$$T = \frac{M(a + \mu g)}{\cos \theta + \mu \sin \theta}$$ [7 marks]

\item It is known that when $M = 30$, $\theta = 30Β°$, and $T = 40$, the sledge remains at rest.

Lizzie uses these values with the relationship formed in part (a) to find the value for $\mu$

Explain why her value for $\mu$ may be incorrect. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2 2019 Q17 [9]}}
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