| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Topic | Pulley systems |
| Type | Particle on rough incline, particle hanging |
| Difficulty | Standard +0.3 This is a standard two-particle mechanics problem involving friction on a slope. Part (i) requires drawing force diagrams and resolving forces in equilibrium (limiting friction case), while part (ii) involves applying F=ma with the system in motion. The given sin θ = 3/5 simplifies calculations. All techniques are routine for A-level mechanics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03o Advanced connected particles: and pulleys3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| At \(A\): weight, normal contact, tension and friction shown correctly. | B1 | Friction must be up the slope. |
| At \(B\): weight and tension shown correctly. | B1 | Tension should be the same as at \(A\). |
| Answer | Marks | Guidance |
|---|---|---|
| At \(A\): \(F + T = m_1g\sin\theta = 6m_1\) | B1 | Resolve up the slope. |
| At \(A\): \(N = m_1g\cos\theta = 8m_1\) | B1 | Resolve perpendicular to the slope. Can be awarded in Part (iii) if not here. |
| At \(A\): \(F = \mu N = \mu \times 8m_1\) | B1 | Limiting friction. Can be awarded in Part (iii) if not here. |
| At \(B\): \(T = m_2g = 5m_1\) | B1 | Resolve vertically. |
| \(\mu \times 8m_1 + 5m_1 = 6m_1\) | M1 | Eliminate \(F\), \(N\) and \(T\). |
| \(\therefore \mu = \frac{1}{8}\) | A1 | c.a.o. SR: Allow max 5/6 when \(T\) is eliminated as follows: \(T + F - m_1g\sin\theta = T - m_2g\) or equivalent, i.e setting '0 = 0'. |
| Answer | Marks | Guidance |
|---|---|---|
| At \(A\): \(m_1a = T - F - m_1g\sin\theta = T - 6m_1\) | B1 | N2 applied up the slope. Friction must now be down the slope. |
| At \(B\): \(m_2a = m_2g - T\) or \(m_1a = 10m_1 - T\) | B1 | N2 applied downwards. |
| \(2m_1a = 10m_1 - \frac{1}{8} \times 8m_1 - 6m_1 = 3m_1\) | M1 | Eliminate \(T\), \(N\) and \(F\) using expressions from above. |
| \(\therefore a = 1.5\) (ms\(^{-2}\)) | A1 | Ft c's \(\mu\), provided \(a > 0(0 < \mu < \frac{1}{2})\). |
### (i)
#### (a)
At $A$: weight, normal contact, tension and friction shown correctly. | B1 | Friction must be up the slope.
At $B$: weight and tension shown correctly. | B1 | Tension should be the same as at $A$. | [2]
#### (b)
Values for $g$, $\sin\theta$ and $\cos\theta$, and $m_2 = \frac{1}{2}m_1$ can be substituted at any stage.
At $A$: $F + T = m_1g\sin\theta = 6m_1$ | B1 | Resolve up the slope.
At $A$: $N = m_1g\cos\theta = 8m_1$ | B1 | Resolve perpendicular to the slope. Can be awarded in Part (iii) if not here.
At $A$: $F = \mu N = \mu \times 8m_1$ | B1 | Limiting friction. Can be awarded in Part (iii) if not here.
At $B$: $T = m_2g = 5m_1$ | B1 | Resolve vertically.
$\mu \times 8m_1 + 5m_1 = 6m_1$ | M1 | Eliminate $F$, $N$ and $T$.
$\therefore \mu = \frac{1}{8}$ | A1 | c.a.o. SR: Allow max 5/6 when $T$ is eliminated as follows: $T + F - m_1g\sin\theta = T - m_2g$ or equivalent, i.e setting '0 = 0'. | [6]
### (ii)
Values for $g$, $\sin\theta$ and $\cos\theta$, and $m_2 = m_1$ can be substituted at any stage.
At $A$: $m_1a = T - F - m_1g\sin\theta = T - 6m_1$ | B1 | N2 applied up the slope. Friction must now be down the slope.
At $B$: $m_2a = m_2g - T$ or $m_1a = 10m_1 - T$ | B1 | N2 applied downwards.
$2m_1a = 10m_1 - \frac{1}{8} \times 8m_1 - 6m_1 = 3m_1$ | M1 | Eliminate $T$, $N$ and $F$ using expressions from above.
$\therefore a = 1.5$ (ms$^{-2}$) | A1 | Ft c's $\mu$, provided $a > 0(0 < \mu < \frac{1}{2})$. | [4]
---\includegraphics{figure_11}
The diagram shows a particle, $A$, of mass $m_1$ at rest on a rough slope at an angle $\theta$ to the horizontal, where $\sin \theta = \frac{3}{5}$. Particle $A$ is connected by a light inextensible string to another particle, $B$, of mass $m_2$. The string passes over a smooth peg at the top of the slope and particle $B$ is hanging freely.
\begin{enumerate}[label=(\roman*)]
\item In the case when $m_2 = \frac{1}{4}m_1$, particle $A$ is on the point of sliding down the slope.
\begin{enumerate}[label=(\alph*)]
\item Draw a fully labelled diagram to show all the forces acting on the particles. [2]
\item Find the coefficient of friction between $A$ and the slope. [6]
\end{enumerate}
\item In the case when $m_2 = m_1$, find the acceleration of the particles. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q11 [12]}}