| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | January |
| 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 a common two-phase motion scenario. Parts (a)-(d) involve routine application of Newton's second law, SUVAT equations, and energy considerations. Part (e) requires sketching a velocity-time graph with two distinct phases. While multi-part, each step follows predictable mechanics patterns with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Motion of \(P\): \(T - 3g = 3a\) | M1 | Equation of motion for \(P\) with \(T\) not substituted, condone sign errors |
| \(33.6 - 3g = 3a\) | A1 | Correct equation in \(a\) only (allow \(\pm a\)) |
| \(a = 1.4 \text{ (m s}^{-2}\text{)}\) Given Answer | A1 | For given answer (units not needed) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Motion of \(Q\): \(mg - T = ma\) | M1 | Equation of motion for \(Q\) with neither \(T\) nor \(a\) substituted, condone sign errors |
| \(mg - 33.6 = 1.4m\) | A1 | Correct equation in \(m\) only |
| \(m = 4\) | A1 | N.B. Whole system equation \(mg - 3g = a(m+3)\) may be used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(s = ut + \frac{1}{2}at^2\): \(10.5 = \frac{1}{2} \times 1.4 \times t^2\) | M1A1 | M1 for complete method to find \(T_1\) (M0 if \(g\) used); A1 for correct equation |
| \(T_1 = \sqrt{15} = 3.9\) or better | A1 | \(v = \sqrt{29.4}\) (5.4) may be found here but only gets credit if it appears in part (d) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(v^2 = u^2 + 2as\) to find speed when \(Q\) hits ground: \(v = \sqrt{2 \times 1.4 \times 10.5} \; (= \sqrt{29.4})\) | M1 | First M1 for complete method to find speed when \(Q\) hits ground (M0 if using \(g\)) |
| Use \(v = u + at\) to find additional time for \(P\) to come to rest: \(0 = \sqrt{29.4} - gt\) | DM1 | Second M1 dependent on first M1; complete method to find additional time for \(P\) to come to rest (must use \(g\)) |
| Total time: \(T_2 = \sqrt{15} + \dfrac{\sqrt{29.4}}{9.8} = 4.4\) or \(4.43\) | A1 | A1 for \(4.4\) or \(4.43\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape: increases to peak at \(v = 5.4\), decreases, then continues to negative values | B1 | Shape only; graph does not need to go down as far as it goes up; ignore gradients. B0 if outside range \(0 \leq t \leq T_3\) or if a continuous vertical line is included |
| Values \(5.4\), \(-5.4\), \(3.9\), \(4.4\) (or \(T_1\), \(T_2\)) correctly marked | DB1 | Dependent on first B1; ft on their \(\sqrt{29.4}\), \(T_1\) and \(T_2\). Allow \(T_1\) and \(T_2\) entered on graph rather than numerical values |
## Question 7:
---
### Part 7a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Motion of $P$: $T - 3g = 3a$ | M1 | Equation of motion for $P$ with $T$ not substituted, condone sign errors |
| $33.6 - 3g = 3a$ | A1 | Correct equation in $a$ only (allow $\pm a$) |
| $a = 1.4 \text{ (m s}^{-2}\text{)}$ **Given Answer** | A1 | For given answer (units not needed) |
---
### Part 7b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Motion of $Q$: $mg - T = ma$ | M1 | Equation of motion for $Q$ with neither $T$ nor $a$ substituted, condone sign errors |
| $mg - 33.6 = 1.4m$ | A1 | Correct equation in $m$ only |
| $m = 4$ | A1 | **N.B.** Whole system equation $mg - 3g = a(m+3)$ may be used |
---
### Part 7c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $s = ut + \frac{1}{2}at^2$: $10.5 = \frac{1}{2} \times 1.4 \times t^2$ | M1A1 | M1 for complete method to find $T_1$ (M0 if $g$ used); A1 for correct equation |
| $T_1 = \sqrt{15} = 3.9$ or better | A1 | $v = \sqrt{29.4}$ (5.4) may be found here but only gets credit if it appears in part (d) |
---
### Part 7d:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $v^2 = u^2 + 2as$ to find speed when $Q$ hits ground: $v = \sqrt{2 \times 1.4 \times 10.5} \; (= \sqrt{29.4})$ | M1 | First M1 for complete method to find speed when $Q$ hits ground (M0 if using $g$) |
| Use $v = u + at$ to find additional time for $P$ to come to rest: $0 = \sqrt{29.4} - gt$ | DM1 | Second M1 **dependent** on first M1; complete method to find additional time for $P$ to come to rest (must use $g$) |
| Total time: $T_2 = \sqrt{15} + \dfrac{\sqrt{29.4}}{9.8} = 4.4$ or $4.43$ | A1 | A1 for $4.4$ or $4.43$ |
---
### Part 7e:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape: increases to peak at $v = 5.4$, decreases, then continues to negative values | B1 | Shape only; graph does not need to go down as far as it goes up; ignore gradients. B0 if outside range $0 \leq t \leq T_3$ or if a continuous vertical line is included |
| Values $5.4$, $-5.4$, $3.9$, $4.4$ (or $T_1$, $T_2$) correctly marked | DB1 | **Dependent** on first B1; ft on their $\sqrt{29.4}$, $T_1$ and $T_2$. Allow $T_1$ and $T_2$ entered on graph rather than numerical values |
**Total: [14]**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ba698f74-a51c-409a-a9d9-e9080fc87be2-12_524_586_274_696}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Two particles $P$ and $Q$ have masses 3 kg and $m \mathrm {~kg}$ respectively ( $m > 3$ ). 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 and the hanging parts of the string vertical. The particle $Q$ is at a height of 10.5 m above the horizontal ground, as shown in Figure 5. The system is released from rest and $Q$ moves downwards. In the subsequent motion $P$ does not reach the pulley. After the system is released, the tension in the string is 33.6 N .
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the acceleration of $P$ is $1.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item Find the value of $m$.
The system is released from rest at time $t = 0$. At time $T _ { 1 }$ seconds after release, $Q$ strikes the ground and does not rebound. The string goes slack and $P$ continues to move upwards.
\item Find the value of $T _ { 1 }$
At time $T _ { 2 }$ seconds after release, $P$ comes to instantaneous rest.
\item Find the value of $T _ { 2 }$
At time $T _ { 3 }$ seconds after release ( $T _ { 3 } > T _ { 1 }$ ) the string becomes taut again.
\item Sketch a velocity-time graph for the motion of $P$ in the interval $0 \leqslant t \leqslant T _ { 3 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2017 Q7 [14]}}