| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Particle on incline, hanging counterpart |
| Difficulty | Standard +0.3 This is a standard connected particles problem requiring routine application of Newton's second law to both masses, resolving forces on the incline, and calculating friction. Part (a) is a 'show that' requiring straightforward simultaneous equations. Part (b) tests understanding of limiting friction vs component down the plane. The given tan α simplifies to sin α = 5/13, cos α = 12/13 which makes calculations clean. Part (c) is a standard modelling question. Slightly easier than average due to the structured nature and clean arithmetic. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(R = 2mg\cos\alpha\) | B1 | Normal reaction between \(A\) and plane seen or implied; \(\cos\alpha\) does not need to be substituted |
| \(F = \frac{2}{3}R\) | B1 | Seen or implied anywhere including part (b) |
| Equation of motion for \(A\): \(T - F - 2mg\sin\alpha = 2ma\) | M1, A1 | Must include all relevant terms, correct mass, condone consistent missing \(m\)'s, condone sign errors and sin/cos confusion. N.B. If \(T-2mg=2ma\) seen with no working, M0A0 unless both B1 marks scored |
| Equation of motion for \(B\): \(3mg - T = 3ma\) | M1, A1 | Correct mass on RHS; condone consistent missing \(m\)'s, sign errors and sin/cos confusion |
| Complete strategy to find equation in \(T\), \(m\) and \(g\) only | M1 | Independent mark; must have two simultaneous equations in \(T\) and \(a\). N.B. Allow whole system equation \(3mg - F - 2mg\sin\alpha = 5ma\) to replace equation for \(A\) or \(B\) |
| \(T = \frac{12mg}{5}\) * | A1* | Obtain given answer from correct working using EXACT trig ratios (not available if using decimal angle) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \((F_{\max} =)\ \frac{16mg}{13} > \frac{10mg}{13}\) | M1 | Comparison of their \(F_{\max}\) \((\frac{2}{3}R)\) and their component of weight down the slope; must compare numerical values. Allow comparison of \(\mu\) and \(\tan\alpha\) with numerical values |
| \(\therefore A\) will not move | A1 | Correctly justified conclusion and no errors seen. N.B. If they equate difference to an '\(ma\)' term then A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Any two correct from: Extensible string; Weight of string; Friction at pulley (e.g. rough pulley); Allow for dimensions of blocks (e.g. "Do not model blocks as particles"; "include air resistance"; "include rotational effects of forces on blocks i.e. spin") | B1, B1 | Deduct 1 mark for each extra (more than 2) incorrect answer up to max 2 incorrect. Ignore extra correct answers. Ignore incorrect reasons or consequences. Ignore any mention of wind or general reference to friction |
# Question 3:
## Part 3(a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $R = 2mg\cos\alpha$ | B1 | Normal reaction between $A$ and plane seen or implied; $\cos\alpha$ does not need to be substituted |
| $F = \frac{2}{3}R$ | B1 | Seen or implied anywhere including part (b) |
| Equation of motion for $A$: $T - F - 2mg\sin\alpha = 2ma$ | M1, A1 | Must include all relevant terms, correct mass, condone consistent missing $m$'s, condone sign errors and sin/cos confusion. N.B. If $T-2mg=2ma$ seen with no working, M0A0 unless both B1 marks scored |
| Equation of motion for $B$: $3mg - T = 3ma$ | M1, A1 | Correct mass on RHS; condone consistent missing $m$'s, sign errors and sin/cos confusion |
| Complete strategy to find equation in $T$, $m$ and $g$ only | M1 | Independent mark; must have two simultaneous equations in $T$ and $a$. N.B. Allow whole system equation $3mg - F - 2mg\sin\alpha = 5ma$ to replace equation for $A$ or $B$ |
| $T = \frac{12mg}{5}$ * | A1* | Obtain given answer from correct working using EXACT trig ratios (not available if using decimal angle) |
## Part 3(b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $(F_{\max} =)\ \frac{16mg}{13} > \frac{10mg}{13}$ | M1 | Comparison of their $F_{\max}$ $(\frac{2}{3}R)$ and their component of weight down the slope; must compare numerical values. Allow comparison of $\mu$ and $\tan\alpha$ with numerical values |
| $\therefore A$ will not move | A1 | Correctly justified conclusion and no errors seen. N.B. If they equate difference to an '$ma$' term then A0 |
## Part 3(c):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Any two correct from: Extensible string; Weight of string; Friction at pulley (e.g. rough pulley); Allow for dimensions of blocks (e.g. "Do not model blocks as particles"; "include air resistance"; "include rotational effects of forces on blocks i.e. spin") | B1, B1 | Deduct 1 mark for each extra (more than 2) incorrect answer up to max 2 incorrect. Ignore extra correct answers. Ignore incorrect reasons or consequences. Ignore any mention of wind or general reference to friction |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8399dae8-1b9d-4564-a95b-7ab857368b86-06_339_812_242_628}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Two blocks, $A$ and $B$, of masses $2 m$ and $3 m$ respectively, are attached to the ends of a light string.
Initially $A$ is held at rest on a fixed rough plane.\\
The plane is inclined at angle $\alpha$ to the horizontal ground, where $\tan \alpha = \frac { 5 } { 12 }$\\
The string passes over a small smooth pulley, $P$, fixed at the top of the plane.\\
The part of the string from $A$ to $P$ is parallel to a line of greatest slope of the plane. Block $B$ hangs freely below $P$, as shown in Figure 1.
The coefficient of friction between $A$ and the plane is $\frac { 2 } { 3 }$\\
The blocks are released from rest with the string taut and $A$ moves up the plane.\\
The tension in the string immediately after the blocks are released is $T$.\\
The blocks are modelled as particles and the string is modelled as being inextensible.
\begin{enumerate}[label=(\alph*)]
\item Show that $T = \frac { 12 m g } { 5 }$
After $B$ reaches the ground, $A$ continues to move up the plane until it comes to rest before reaching $P$.
\item Determine whether $A$ will remain at rest, carefully justifying your answer.
\item Suggest two refinements to the model that would make it more realistic.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 2019 Q3 [12]}}