| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Particle on inclined plane motion |
| Difficulty | Standard +0.3 This is a standard mechanics problem requiring resolution of forces on an inclined plane and application of F=ma, followed by a straightforward kinematics calculation using s=ut+½at². While it involves multiple steps, the techniques are routine for A-level mechanics students with no novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States \(F = \mu R\) | B1 (3.3) | Seen anywhere; PI by use of \(\mu R\) in four-term equation or on diagram |
| \(mg\sin 25\) (or better) from resolving parallel to slope | B1 (1.1b) | Resolves weight parallel to slope |
| \(R = mg\cos 25\) (or better) from resolving perpendicular | B1 (1.1b) | Resolves perpendicular to slope |
| \(T - \text{weight} - \text{Friction} = ma\), i.e. \(T - 20g\sin 25 - F = m \times 1.2\) | M1 (3.3) | Uses \(F = ma\) for four-term equation with consistent signs; condone omission of \(g\) in weight and friction |
| Substitutes \(T = 230\) and \(F = \mu mg\cos 25\) into four-term equation | M1 (1.1a) | Condone '\(mga\)' in \(F = ma\) |
| \(230 - 196\sin 25 - 196\cos 25\,\mu = 24\) | A1 (1.1b) | Single correct equation with all numerical values substituted |
| \(\mu = 0.69\) | A1 (3.2a) | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = ut + \frac{1}{2}at^2\) with \(u=0\), \(a=1.2\), \(t=3.8\) | M1 (1.1a) | Or uses appropriate constant acceleration equations forming complete method to obtain \(s\) |
| \(s = \frac{1}{2} \times 1.2 \times 3.8^2 = 8.664\); AWRT 8.7 | A1 (1.1b) | |
| \(OA = 10 - 8.664 = 1.3\text{ m}\); AWRT 1.3 | A1F (1.1b) | FT their 8.7 provided less than 10; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The crate is a particle | E1 (3.5b) | States the crate has been modelled as a particle OE |
## Question 19(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States $F = \mu R$ | B1 (3.3) | Seen anywhere; PI by use of $\mu R$ in four-term equation or on diagram |
| $mg\sin 25$ (or better) from resolving parallel to slope | B1 (1.1b) | Resolves weight parallel to slope |
| $R = mg\cos 25$ (or better) from resolving perpendicular | B1 (1.1b) | Resolves perpendicular to slope |
| $T - \text{weight} - \text{Friction} = ma$, i.e. $T - 20g\sin 25 - F = m \times 1.2$ | M1 (3.3) | Uses $F = ma$ for four-term equation with consistent signs; condone omission of $g$ in weight and friction |
| Substitutes $T = 230$ and $F = \mu mg\cos 25$ into four-term equation | M1 (1.1a) | Condone '$mga$' in $F = ma$ |
| $230 - 196\sin 25 - 196\cos 25\,\mu = 24$ | A1 (1.1b) | Single correct equation with all numerical values substituted |
| $\mu = 0.69$ | A1 (3.2a) | CAO |
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## Question 19(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = ut + \frac{1}{2}at^2$ with $u=0$, $a=1.2$, $t=3.8$ | M1 (1.1a) | Or uses appropriate constant acceleration equations forming complete method to obtain $s$ |
| $s = \frac{1}{2} \times 1.2 \times 3.8^2 = 8.664$; AWRT 8.7 | A1 (1.1b) | |
| $OA = 10 - 8.664 = 1.3\text{ m}$; AWRT 1.3 | A1F (1.1b) | FT their 8.7 provided less than 10; condone missing units |
---
## Question 19(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The crate is a particle | E1 (3.5b) | States the crate has been modelled as a particle OE |19
\begin{enumerate}[label=(\alph*)]
\item The tension in the rope is 230 N\\
The crate accelerates up the ramp at $1.2 \mathrm {~ms} ^ { - 2 }$\\
Find the coefficient of friction between the crate and the ramp.\\
19
\item (i) The crate takes 3.8 seconds to reach the top of the ramp.\\
Find the distance $O A$.\\[0pt]
[3 marks]\\
19 (b) (ii) Other than air resistance, state one assumption you have made about the crate in answering part (b)(i).\\
\includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-32_2492_1721_217_150}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2022 Q19 [11]}}