| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough horizontal surface, particle hanging |
| Difficulty | Standard +0.3 This is a standard M1 pulley system question requiring routine application of Newton's second law to connected particles, followed by kinematics. The multi-part structure guides students through the solution methodically. While it requires careful bookkeeping of forces and two phases of motion, it involves no novel insights—just systematic application of standard mechanics techniques taught in M1. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
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| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For \(A\): \(T - F = 2ma\) | M1 A1 | First M1 for equation of motion for \(A\); First A1 correct (allow \(-T\) instead of \(T\)) |
| For \(B\): \(mg - T = ma\) | M1 A1 (4) | Second M1 for equation of motion for \(B\); Second A1 correct (allow consistent \(-T\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = 2mg\) | B1 | B1 for \(R = 2mg\) |
| \(mg(1 - 2\mu) = 3ma\) | M1 | M1 for using \(F = \mu R\) and eliminating to give equation in \(a\) and \(\mu\) only |
| \(\frac{g}{3}(1 - 2\mu) = a\) | A1 (3) | A1 for printed answer (must be identical to printed answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v^2 = \frac{2gh}{3}(1 - 2\mu)\) | M1 | M1 for using \(v^2 = u^2 + 2as\) or other complete method |
| \(v = \sqrt{\frac{2gh}{3}(1 - 2\mu)}\) | A1 (2) | A1 for correct answer in any form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-\mu R = 2ma'\) | M1 | First M1 for equation of motion for \(A\) with \(T=0\), \(F = \mu R\); e.g. \(\mu R = 2ma'\) (must be \(2m\)) |
| \(0^2 = \text{their } u^2 - 2a's\) | M1 | Second M1 for using \(v^2 = u^2 + 2as\) with \(u^2\) from (c), \(v=0\) and new \(a\) |
| \(0 = \frac{2gh}{3}\!\left(1 - \frac{2}{3}\right) - 2\!\left(\frac{1}{3}g\right)s\) (where \(s = d-h\)) | A1 (A1) | First A1 for correct equation in \(s\), \(g\), \(h\) with \(\mu = \frac{1}{3}\) |
| \(s = \frac{1}{3}h\) | A1 | Second A1 for \(s = \frac{1}{3}h\) |
| \(d = \frac{1}{3}h + h = \frac{4}{3}h\) | A1 (5) | Third A1 for \(d = \frac{4}{3}h\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mu R s = \frac{1}{2}(2m)u^2\) | M2 | M2 for work-energy; (M1 if they use \(m\)) |
| \(\frac{1}{3}(2m)gs = \frac{1}{2}(2m)\frac{2gh}{3}\!\left(1 - \frac{2}{3}\right)\) | A1 | First A1 |
| \(s = \frac{1}{3}h\) | A1 | Second A1 |
| \(d = \frac{4}{3}h\) | A1 | Third A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A\) (or \(B\)) would not move; OR \(A\) (or \(B\)) would remain in (limiting) equilibrium; OR the system would remain in (limiting) equilibrium | B1 (1) | B1 for any one of the listed alternatives |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $A$: $T - F = 2ma$ | M1 A1 | First M1 for equation of motion for $A$; First A1 correct (allow $-T$ instead of $T$) |
| For $B$: $mg - T = ma$ | M1 A1 (4) | Second M1 for equation of motion for $B$; Second A1 correct (allow consistent $-T$) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = 2mg$ | B1 | B1 for $R = 2mg$ |
| $mg(1 - 2\mu) = 3ma$ | M1 | M1 for using $F = \mu R$ and eliminating to give equation in $a$ and $\mu$ only |
| $\frac{g}{3}(1 - 2\mu) = a$ | A1 (3) | A1 for printed answer (must be identical to printed answer) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v^2 = \frac{2gh}{3}(1 - 2\mu)$ | M1 | M1 for using $v^2 = u^2 + 2as$ or other complete method |
| $v = \sqrt{\frac{2gh}{3}(1 - 2\mu)}$ | A1 (2) | A1 for correct answer in any form |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-\mu R = 2ma'$ | M1 | First M1 for equation of motion for $A$ with $T=0$, $F = \mu R$; e.g. $\mu R = 2ma'$ (must be $2m$) |
| $0^2 = \text{their } u^2 - 2a's$ | M1 | Second M1 for using $v^2 = u^2 + 2as$ with $u^2$ from (c), $v=0$ and new $a$ |
| $0 = \frac{2gh}{3}\!\left(1 - \frac{2}{3}\right) - 2\!\left(\frac{1}{3}g\right)s$ (where $s = d-h$) | A1 (A1) | First A1 for correct equation in $s$, $g$, $h$ with $\mu = \frac{1}{3}$ |
| $s = \frac{1}{3}h$ | A1 | Second A1 for $s = \frac{1}{3}h$ |
| $d = \frac{1}{3}h + h = \frac{4}{3}h$ | A1 (5) | Third A1 for $d = \frac{4}{3}h$ |
**Alternative (work-energy):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mu R s = \frac{1}{2}(2m)u^2$ | M2 | M2 for work-energy; (M1 if they use $m$) |
| $\frac{1}{3}(2m)gs = \frac{1}{2}(2m)\frac{2gh}{3}\!\left(1 - \frac{2}{3}\right)$ | A1 | First A1 |
| $s = \frac{1}{3}h$ | A1 | Second A1 |
| $d = \frac{4}{3}h$ | A1 | Third A1 |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A$ (or $B$) would not move; OR $A$ (or $B$) would remain in (limiting) equilibrium; OR the system would remain in (limiting) equilibrium | B1 (1) | B1 for any one of the listed alternatives |
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\caption{Figure 3}
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Two particles, $A$ and $B$, have masses $2 m$ and $m$ respectively. The particles are attached to the ends of a light inextensible string. Particle $A$ is held at rest on a fixed rough horizontal table at a distance $d$ from a small smooth light pulley which is fixed at the edge of the table at the point $P$. The coefficient of friction between $A$ and the table is $\mu$, where $\mu < \frac { 1 } { 2 }$. The string is parallel to the table from $A$ to $P$ and passes over the pulley. Particle $B$ hangs freely at rest vertically below $P$ with the string taut and at a height $h$, ( $h < d$ ), above a horizontal floor, as shown in Figure 3. Particle $A$ is released from rest with the string taut and slides along the table.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down an equation of motion for $A$.
\item Write down an equation of motion for $B$.
\end{enumerate}\item Hence show that, until $B$ hits the floor, the acceleration of $A$ is $\frac { g } { 3 } ( 1 - 2 \mu )$.
\item Find, in terms of $g , h$ and $\mu$, the speed of $A$ at the instant when $B$ hits the floor.
After $B$ hits the floor, $A$ continues to slide along the table. Given that $\mu = \frac { 1 } { 3 }$ and that $A$ comes to rest at $P$,
\item find $d$ in terms of $h$.
\item Describe what would happen if $\mu = \frac { 1 } { 2 }$
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\hfill \mbox{\textit{Edexcel M1 2017 Q8 [15]}}