| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| 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 requiring SUVAT equations, Newton's second law for connected particles, and energy/kinematics after string goes slack. All techniques are routine for this topic with straightforward numerical values, making it slightly easier than average but still requiring systematic application of multiple mechanics principles. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \([0.36 = \frac{1}{2}a(0.6)^2]\) | M1 | For using \(s = (ut) + \frac{1}{2}at^2\) |
| Acceleration is 2 ms⁻² | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \([0.45g - T = 0.45 \times 2]\) | M1 | For applying Newton's second law to A |
| Tension is 3.6 N | A1ft | [2] ft T = 0.45(10 - a) |
| Answer | Marks | Guidance |
|---|---|---|
| \([T - mg = 2m \text{ or}\) | M1 | For applying Newton's second law to B or |
| \(0.9 + 2m = 4.5 - 10m]\) | for using \((M + m)a = (M - m)g\) | |
| \((2 + g)m = 3.6\) (must have m terms combined) | ||
| Mass is 0.3 kg | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(u = 1.2\) | B1ft | |
| \([0 = 1.44 - 20s \to 0.072]\) | M1 | For using \(0 = u^2 - 2gs\) |
| Maximum height is 0.792 | A1ft | [3] ft 0.72 + 0.05\(u^2\) |
**(i)**
$[0.36 = \frac{1}{2}a(0.6)^2]$ | M1 | For using $s = (ut) + \frac{1}{2}at^2$
Acceleration is 2 ms⁻² | A1 | [2]
**(ii)**
$[0.45g - T = 0.45 \times 2]$ | M1 | For applying Newton's second law to A
Tension is 3.6 N | A1ft | [2] ft T = 0.45(10 - a)
**(iii)**
$[T - mg = 2m \text{ or}$ | M1 | For applying Newton's second law to B or
$0.9 + 2m = 4.5 - 10m]$ | | for using $(M + m)a = (M - m)g$
$(2 + g)m = 3.6$ (must have m terms combined) | |
Mass is 0.3 kg | A1 | [3]
**(iv)**
$u = 1.2$ | B1ft |
$[0 = 1.44 - 20s \to 0.072]$ | M1 | For using $0 = u^2 - 2gs$
Maximum height is 0.792 | A1ft | [3] ft 0.72 + 0.05$u^2$\includegraphics{figure_6}
Particles $A$ and $B$ are attached to the ends of a light inextensible string which passes over a smooth pulley. The system is held at rest with the string taut and its straight parts vertical. Both particles are at a height of 0.36 m above the floor (see diagram). The system is released and $A$ begins to fall, reaching the floor after 0.6 s.
\begin{enumerate}[label=(\roman*)]
\item Find the acceleration of $A$ as it falls. [2]
\end{enumerate}
The mass of $A$ is 0.45 kg. Find
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item the tension in the string while $A$ is falling, [2]
\item the mass of $B$, [3]
\item the maximum height above the floor reached by $B$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2009 Q6 [10]}}