| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Three or more connected particles |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with a mass change mid-motion. Part (a) requires routine application of F=ma to connected particles (a 'show that' with given answer), parts (b-d) involve standard kinematics after the system changes. While multi-part with several steps, all techniques are textbook-standard for M1 with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4mg - T = 4ma\) | M1A1 | M1 resolving vertically for \(B+C\); A1 correct equation |
| \(T - 3mg = 3ma\) | M1A1 | M1 resolving vertically for \(A\); A1 correct equation |
| Reach given answer \(a = \dfrac{g}{7}\) correctly | A1 | Given answer; note \(1.4\) scores A0 |
| \(T = 3mg + 3\!\left(mg - \dfrac{T}{4}\right)\), or \(T = 3mg + 3m\dfrac{g}{7}\), or \(T = 4mg - 4m\dfrac{g}{7}\) | M1 | Form equation in \(T\) only |
| \(T = \dfrac{24}{7}mg\) or equivalent, \(33.6m\), \(34m\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v^2 = u^2 + 2as = 2 \times \dfrac{g}{7} \times 0.7 = 1.96\), \(v = 1.4\) ms\(^{-1}\) | M1A1 | M1 equation in \(v\) only; A1 for 1.4 ms\(^{-1}\), allow \(\sqrt{g/5}\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3mg - T = 3ma\) | M1A1 | M1 resolving for \(A\) or \(B\); M0 if tension from part (a) used; A1 correct equation for \(A\) |
| \(T - 2mg = 2ma\) | A1 | Correct equation for \(B\) |
| \(a = \dfrac{g}{5}\) | A1 | Allow \(g/5\) or 1.96 or 2.0 ms\(^{-2}\); allow negative answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0 = 1.96 - 2 \times \dfrac{g}{5} \times s\) | M1 | Equation in \(s\) only using their \(v\) from (b) and \(a\) from (c) |
| \(s = \dfrac{5 \times 1.96}{2g} = 0.5\) m | A1 | \(s = 0.5\) m correctly obtained |
| Total height \(= 0.7 + 0.5 = 1.2\) m | A1 ft | ft on their \(0.5 + 0.7\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3mgs - 2mgs = \dfrac{1}{2}3m(1.4)^2 + \dfrac{1}{2}2m(1.4)^2\) | M1 | Equation in \(s\) only, correct number of terms, using their \(v\) from (b) |
| \(s = \dfrac{2.5 \times 1.96^2}{g} = 0.5\) m | A1 | Correctly obtained |
| Total height \(= 0.7 + 0.5 = 1.2\) m | A1 ft |
# Question 7:
## Part 7a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4mg - T = 4ma$ | M1A1 | M1 resolving vertically for $B+C$; A1 correct equation |
| $T - 3mg = 3ma$ | M1A1 | M1 resolving vertically for $A$; A1 correct equation |
| Reach given answer $a = \dfrac{g}{7}$ correctly | A1 | Given answer; note $1.4$ scores A0 |
| $T = 3mg + 3\!\left(mg - \dfrac{T}{4}\right)$, or $T = 3mg + 3m\dfrac{g}{7}$, or $T = 4mg - 4m\dfrac{g}{7}$ | M1 | Form equation in $T$ only |
| $T = \dfrac{24}{7}mg$ or equivalent, $33.6m$, $34m$ | A1 | |
## Part 7b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v^2 = u^2 + 2as = 2 \times \dfrac{g}{7} \times 0.7 = 1.96$, $v = 1.4$ ms$^{-1}$ | M1A1 | M1 equation in $v$ only; A1 for 1.4 ms$^{-1}$, allow $\sqrt{g/5}$ oe |
## Part 7c:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3mg - T = 3ma$ | M1A1 | M1 resolving for $A$ or $B$; M0 if tension from part (a) used; A1 correct equation for $A$ |
| $T - 2mg = 2ma$ | A1 | Correct equation for $B$ |
| $a = \dfrac{g}{5}$ | A1 | Allow $g/5$ or 1.96 or 2.0 ms$^{-2}$; allow negative answer |
## Part 7d:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0 = 1.96 - 2 \times \dfrac{g}{5} \times s$ | M1 | Equation in $s$ only using their $v$ from (b) and $a$ from (c) |
| $s = \dfrac{5 \times 1.96}{2g} = 0.5$ m | A1 | $s = 0.5$ m correctly obtained |
| Total height $= 0.7 + 0.5 = 1.2$ m | A1 ft | ft on their $0.5 + 0.7$ |
**Alternative (energy):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3mgs - 2mgs = \dfrac{1}{2}3m(1.4)^2 + \dfrac{1}{2}2m(1.4)^2$ | M1 | Equation in $s$ only, correct number of terms, using their $v$ from (b) |
| $s = \dfrac{2.5 \times 1.96^2}{g} = 0.5$ m | A1 | Correctly obtained |
| Total height $= 0.7 + 0.5 = 1.2$ m | A1 ft | |
The images you've shared appear to be blank or nearly blank pages — they only show "PMT" watermarks in the corners and a Pearson Education Limited copyright notice at the bottom of the second page. There is no mark scheme content visible on these pages to extract.
Could you please share the actual mark scheme pages that contain the questions, answers, and mark allocations? I'd be happy to format those for you once I can see the content.
7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ed659098-c1cf-4ee1-a12a-bf8b6c42db95-13_490_316_267_815}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Three particles $A , B$ and $C$ have masses $3 m , 2 m$ and $2 m$ respectively. Particle $C$ is attached to particle $B$. Particles $A$ and $B$ are connected by a light inextensible string which passes over a smooth light fixed pulley. The system is held at rest with the string taut and the hanging parts of the string vertical, as shown in Figure 5. The system is released from rest and $A$ moves upwards.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the acceleration of $A$ is $\frac { g } { 7 }$
\item Find the tension in the string as $A$ ascends.
At the instant when $A$ is 0.7 m above its original position, $C$ separates from $B$ and falls away. In the subsequent motion, $A$ does not reach the pulley.
\end{enumerate}\item Find the speed of $A$ at the instant when it is 0.7 m above its original position.
\item Find the acceleration of $A$ at the instant after $C$ separates from $B$.
\item Find the greatest height reached by $A$ above its original position.\\
\includegraphics[max width=\textwidth, alt={}, center]{ed659098-c1cf-4ee1-a12a-bf8b6c42db95-14_115_161_2455_1784}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q7 [16]}}