| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2001 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Two particles over pulley, vertical strings |
| Difficulty | Moderate -0.3 This is a standard M1 pulley problem requiring application of Newton's second law to both particles and use of the constraint that accelerations are equal. The multi-part structure guides students through the solution methodically, and the techniques are routine for this topic. Slightly easier than average due to the scaffolding, though it requires careful algebraic manipulation. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For particle A: \(T - 3mg = 3ma\) | M1 | Note: \(T - mg = ma\) or \(T - m = ma\) etc scores M1 |
| \(T - 3mg = 3m\left(\frac{2}{5}g\right) \Rightarrow T = \frac{21}{5}mg\) | A1 → A1 | (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| String is inextensible | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For particle B: \(kmg - T = kma\) | M1 | |
| (or system): \(kmg - 3mg = (km + 3m)a\) | ||
| \(kg - \frac{21}{5}g = \frac{2}{5}kg\) (or equivalent equation in \(k\) only) | A1 f.t. | |
| Solving (DM1 dependent on first M1 in (c)) | DM1 | |
| \(k = 7\) | A1 cao | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Tension is of same magnitude throughout the string | B1 | (1) |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For particle A: $T - 3mg = 3ma$ | M1 | Note: $T - mg = ma$ or $T - m = ma$ etc scores M1 |
| $T - 3mg = 3m\left(\frac{2}{5}g\right) \Rightarrow T = \frac{21}{5}mg$ | A1 → A1 | (3 marks total) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| String is inextensible | B1 | (1) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For particle B: $kmg - T = kma$ | M1 | |
| (or system): $kmg - 3mg = (km + 3m)a$ | | |
| $kg - \frac{21}{5}g = \frac{2}{5}kg$ (or equivalent equation in $k$ only) | A1 f.t. | |
| Solving (DM1 dependent on first M1 in (c)) | DM1 | |
| $k = 7$ | A1 cao | (4) |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Tension is of same magnitude throughout the string | B1 | (1) |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{49169cc3-c353-430f-80ce-e14ae7fcd6ea-3_437_646_305_706}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
Two particles $A$ and $B$ have masses $3 m$ and $k m$ respectively, where $k > 3$. They are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with the string taut and the hanging parts of the string vertical, as shown in Fig. 3. While the particles are moving freely, $A$ has an acceleration of magnitude $\frac { 2 } { 5 } g$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$ and g , the tension in the string.
\item State why $B$ also has an acceleration of magnitude $\frac { 2 } { 5 } g$.
\item Find the value of $k$.
\item State how you have used the fact that the string is light.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2001 Q3 [9]}}