| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Variable mass or unknown mass |
| Difficulty | Standard +0.3 This is a standard two-particle pulley problem requiring application of SUVAT equations (v=u+at gives a=2.5 m/s²), Newton's second law to find tension, and then solving simultaneous equations for the unknown mass. While it involves multiple steps and two applications of F=ma, the problem follows a well-established routine with no conceptual surprises—slightly easier than average for A-level mechanics. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| \([5 = 0 + 2a]\) | M1 | For using \(v = u + at\) with \(u = 0\) |
| Acceleration is \(2.5\text{ ms}^{-2}\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For applying Newton's second law to A (3 terms): (can be scored in (ii) by applying Newton's second law to B instead) | |
| \(0.5g - T = 0.5 \times 2.5\) | A1ft | |
| Tension is \(3.75\) N | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - mg = 2.5m\) | B1ft | ft from \(T - 0.5g = 0.5\times2.5\) in (i)(b) to allow \(mg - T = 2.5m\) or \(mg - 0.5g = 0.5\times2.5 + 2.5m\) |
| or \(0.5g - mg = 0.5\times2.5 + 2.5m\) | ||
| \([(10 + 2.5)m = 3.75]\) | M1 | For solving 3 term equation for m |
| \(m = 0.3\) | A1 | [3] |
# Question 5:
## Part (i)(a)
$[5 = 0 + 2a]$ | M1 | For using $v = u + at$ with $u = 0$
Acceleration is $2.5\text{ ms}^{-2}$ | A1 | [2]
## Part (i)(b)
| M1 | For applying Newton's second law to A (3 terms): (can be scored in **(ii)** by applying Newton's second law to B instead)
$0.5g - T = 0.5 \times 2.5$ | A1ft |
Tension is $3.75$ N | A1 | [3]
## Part (ii)
$T - mg = 2.5m$ | B1ft | ft from $T - 0.5g = 0.5\times2.5$ in (i)(b) to allow $mg - T = 2.5m$ or $mg - 0.5g = 0.5\times2.5 + 2.5m$
or $0.5g - mg = 0.5\times2.5 + 2.5m$ | |
$[(10 + 2.5)m = 3.75]$ | M1 | For solving 3 term equation for m
$m = 0.3$ | A1 | [3]
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{a4cb105b-55d2-4793-95d2-3d791990a1f6-3_643_481_274_831}
Particles $A$ and $B$, of masses 0.5 kg and $m \mathrm {~kg}$ respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. Particle $B$ is held at rest on the horizontal floor and particle $A$ hangs in equilibrium (see diagram). $B$ is released and each particle starts to move vertically. $A$ hits the floor 2 s after $B$ is released. The speed of each particle when $A$ hits the floor is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) For the motion while $A$ is moving downwards, find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of $A$,
\item the tension in the string.\\
(ii) Find the value of $m$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2008 Q5 [8]}}