| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | November |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - inclined plane involved |
| Difficulty | Standard +0.3 This is a standard M1 pulley system question with straightforward application of Newton's second law to connected particles. Parts (a)-(c) involve routine equation setup and algebraic manipulation. Parts (e)-(f) use basic kinematics (SUVAT equations). While multi-part with several steps, each component follows textbook methods without requiring novel insight or complex problem-solving, making it slightly easier than the average A-level question. |
| 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 |
|---|---|---|
| \(0.4g - T = 0.4 \times \frac{1}{5}g\) | M1 | A1 |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = \frac{8}{25}g\) or \(3.14\) or \(3.1\) N | M1 | A1 |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - mg \sin 30° = m \times \frac{1}{5}g\) | M1 | A1 |
| \(\rightarrow m = \underline{\frac{16}{35}}\) ★ | M1 | A1 |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Same \(T\) for A & B | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = 2 \times \frac{1}{5}g \times 1\) | M1 | |
| \(v = \sqrt{\frac{2g}{5}} \approx 1.98\) or \(2\) m s\(^{-1}\) | A1 | |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| A: \(-\frac{1}{2}mg = ma \Rightarrow a = -\frac{1}{2}g\) | M1 | A1 |
| \(v^2 = \frac{2g}{5} - 2 \times \frac{1}{2}g \times 0.4\) | M1 | A1 ✓ |
| \(\Rightarrow v = 0\) | A1 | |
| (5) |
## (a)
![Diagram showing forces]
| |
$0.4g - T = 0.4 \times \frac{1}{5}g$ | M1 | A1
| | (2)
## (b)
$T = \frac{8}{25}g$ or $3.14$ or $3.1$ N | M1 | A1
| | (2)
## (c)
![Diagram showing inclined plane with tension and weight]
$T - mg \sin 30° = m \times \frac{1}{5}g$ | M1 | A1
$\rightarrow m = \underline{\frac{16}{35}}$ ★ | M1 | A1
| | (4)
## (d)
Same $T$ for A & B | B1 | (1)
## (e)
$v^2 = 2 \times \frac{1}{5}g \times 1$ | M1 |
$v = \sqrt{\frac{2g}{5}} \approx 1.98$ or $2$ m s$^{-1}$ | | A1
| | (2)
## (f)
A: $-\frac{1}{2}mg = ma \Rightarrow a = -\frac{1}{2}g$ | M1 | A1
$v^2 = \frac{2g}{5} - 2 \times \frac{1}{2}g \times 0.4$ | M1 | A1 ✓
$\Rightarrow v = 0$ | | A1
| | (5)
**Total: 16 marks**\includegraphics{figure_3}
Figure 3 shows two particles $A$ and $B$, of mass $m$ kg and 0.4 kg respectively, connected by a light inextensible string. Initially $A$ is held at rest on a fixed smooth plane inclined at 30° to the horizontal. The string passes over a small light smooth pulley $P$ fixed at the top of the plane. The section of the string from $A$ to $P$ is parallel to a line of greatest slope of the plane. The particle $B$ hangs freely below $P$. The system is released from rest with the string taut and $B$ descends with acceleration $\frac{1}{8}g$.
\begin{enumerate}[label=(\alph*)]
\item Write down an equation of motion for $B$. [2]
\item Find the tension in the string. [2]
\item Prove that $m = \frac{16}{35}$. [4]
\item State where in the calculations you have used the information that $P$ is a light smooth pulley. [1]
\end{enumerate}
On release, $B$ is at a height of one metre above the ground and $AP = 1.4$ m. The particle $B$ strikes the ground and does not rebound.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Calculate the speed of $B$ as it reaches the ground. [2]
\item Show that $A$ comes to rest as it reaches $P$. [5]
\end{enumerate}
END
\hfill \mbox{\textit{Edexcel M1 2003 Q7 [16]}}