| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - vertical strings |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with three routine parts: (a) finding tension using Newton's second law for connected particles (show-that question), (b) applying energy/kinematics after one particle stops, and (c) calculating impulse from momentum change. All techniques are textbook exercises requiring no novel insight, though part (b) requires careful tracking of the motion phases. Slightly easier than average due to the guided structure and show-that in part (a). |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(3mg - T = 3ma\) | M1A1 | |
| \(T - 2mg = 2ma\) | M1A1 | |
| \(T = 2mg + 2\left(mg - \frac{T}{3}\right)\) | DM1 | Dependent on both previous M marks |
| \(T = \frac{12}{5}mg\) (Given Answer) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(a = \frac{g}{5}\) | B1 | |
| At time of impact: \(v^2 = u^2 + 2as = 2\times\frac{g}{5}\times1.5 = 0.6g\) | M1A1 | |
| Vertical motion under gravity: \(0 = 0.6g - 2gs\) | M1 | |
| \(s = 0.3\ \text{(m)}\) | ||
| Total distance \(= 2\times0.3 = 0.6\ \text{(m)}\) | DM1 A1 | Dependent on previous M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Impulse \(= 3m(v-u) = -3mu\) | M1 | |
| Magnitude \(= 3m\sqrt{0.6g} = 3.6\ \text{(Ns)}\) (3.64) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Resolving vertically for \(B\) (up or down) | M1 | Correct no. of terms; allow if \(m\) omitted but have the 3 |
| Correct equation for \(B\) | A1 | |
| Resolving vertically for \(A\) (up or down) | M1 | Correct no. of terms; allow if \(m\) omitted but have the 2 |
| Correct equation for \(A\) | A1 | |
| Eliminating \(a\) | DM1 | Dependent on first two M marks |
| \(T = \frac{12mg}{5}\) | A1 | Given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(a = \frac{g}{5}\) found and used | B1 | Possibly found in part (a) |
| Using \(suvat\) with their \(a\) from (a) to find speed \(v\) (or \(v^2\)) of \(B\) at impact | M1 | |
| \(\sqrt{0.6g}\), \(2.4\) or better | A1 | Found correctly; may be implied |
| Using \(suvat\) with \(a = \pm g\), equation in \(s\) only, using their \(v\) (or \(v^2\)) with final velocity \(= 0\) | M1 | |
| Doubling their \(s\) value | DM1 | Dependent on second M1 |
| \(0.6\) (m) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\pm 3m \times\) (their \(v\)) or \(\pm 1.5 \times\) (their \(v\)) or \(\pm m \times\) (their \(v\)) or \(\pm 0.5 \times\) (their \(v\)) | M1 | M0 if \(3m\) missing or extra \(g\) |
| \(3.6\) or \(3.64\) (Ns) | A1 |
# Question 5:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $3mg - T = 3ma$ | M1A1 | |
| $T - 2mg = 2ma$ | M1A1 | |
| $T = 2mg + 2\left(mg - \frac{T}{3}\right)$ | DM1 | Dependent on both previous M marks |
| $T = \frac{12}{5}mg$ **(Given Answer)** | A1 | |
## Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $a = \frac{g}{5}$ | B1 | |
| At time of impact: $v^2 = u^2 + 2as = 2\times\frac{g}{5}\times1.5 = 0.6g$ | M1A1 | |
| Vertical motion under gravity: $0 = 0.6g - 2gs$ | M1 | |
| $s = 0.3\ \text{(m)}$ | | |
| Total distance $= 2\times0.3 = 0.6\ \text{(m)}$ | DM1 A1 | Dependent on previous M1 |
## Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| Impulse $= 3m(v-u) = -3mu$ | M1 | |
| Magnitude $= 3m\sqrt{0.6g} = 3.6\ \text{(Ns)}$ (3.64) | A1 | |
## Question 5:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Resolving vertically for $B$ (up or down) | M1 | Correct no. of terms; allow if $m$ omitted but have the 3 |
| Correct equation for $B$ | A1 | |
| Resolving vertically for $A$ (up or down) | M1 | Correct no. of terms; allow if $m$ omitted but have the 2 |
| Correct equation for $A$ | A1 | |
| Eliminating $a$ | DM1 | Dependent on first two M marks |
| $T = \frac{12mg}{5}$ | A1 | **Given answer** |
**N.B.** Either equation can be replaced by whole system equation: $3mg - 2mg = 5ma$ scores M1A1; any error loses both marks.
**N.B.** If $m$ omitted in (a) giving dimensionally incorrect $a$: max B0M1A0M1M1A0 in (b) and M1A0 in (c).
---
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $a = \frac{g}{5}$ found and used | B1 | Possibly found in part (a) |
| Using $suvat$ with their $a$ from (a) to find speed $v$ (or $v^2$) of $B$ at impact | M1 | |
| $\sqrt{0.6g}$, $2.4$ or better | A1 | Found correctly; may be implied |
| Using $suvat$ with $a = \pm g$, equation in $s$ only, using their $v$ (or $v^2$) with final velocity $= 0$ | M1 | |
| Doubling their $s$ value | DM1 | Dependent on second M1 |
| $0.6$ (m) | A1 | |
---
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\pm 3m \times$ (their $v$) or $\pm 1.5 \times$ (their $v$) or $\pm m \times$ (their $v$) or $\pm 0.5 \times$ (their $v$) | M1 | M0 if $3m$ missing or extra $g$ |
| $3.6$ or $3.64$ (Ns) | A1 | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b896c631-00a0-46c5-bce9-16d65f6e3095-09_364_422_269_753}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Two particles $A$ and $B$ have masses $2 m$ and $3 m$ respectively. The particles 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. The hanging parts of the string are vertical and $A$ and $B$ are above a horizontal plane, as shown in Figure 2. The system is released from rest.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the string immediately after the particles are released is $\frac { 12 } { 5 } m g$.
After descending $1.5 \mathrm {~m} , B$ strikes the plane and is immediately brought to rest. In the subsequent motion, $A$ does not reach the pulley.
\item Find the distance travelled by $A$ between the instant when $B$ strikes the plane and the instant when the string next becomes taut.
Given that $m = 0.5 \mathrm {~kg}$,
\item find the magnitude of the impulse on $B$ due to the impact with the plane.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q5 [14]}}