| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| Marks | 9 |
| 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 two-stage pulley problem requiring Newton's second law for connected particles, then projectile motion after one particle stops. The setup is routine for M1, with straightforward application of F=ma, kinematics equations (suvat), and recognizing the string goes slack. Slightly above average due to the two-stage nature and careful bookkeeping, but all techniques are standard M1 fare with no novel insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([T - 0.3g = 0.3a\) or \(0.5g - T = 0.5a]\) | M1 | Use of Newton's second law for \(P\) or \(Q\) or use of \(a = (m_Q - m_P)g/(m_P + m_Q)\) |
| \(T - 0.3g = 0.3a\) and \(0.5g - T = 0.5a\) or \(a = (0.5g - 0.3g)/(0.5 + 0.3)\) | A1 | |
| \([0.5g - 0.3g = 0.8a]\) | M1 | Solve for \(a\) |
| \(a = 2.5\) | A1 | |
| \([h = 0 + \frac{1}{2} \times 2.5 \times 0.6^2]\) | M1 | For use of \(s = ut + \frac{1}{2}at^2\) |
| \(h = 0.45\) | A1 | |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Velocity of \(P\) when \(Q\) reaches floor \(= 0 + 0.6 \times 2.5 = 1.5 \text{ ms}^{-1}\) | B1ft | ft \(a\) from (i) \(\times 0.6\) |
| \([0 = 1.5 - gt \rightarrow t = \ldots]\) \((t = 0.15)\) | M1 | Use of suvat to find time to highest point |
| Total time \(= 2 \times 0.15 + 0.6 = 0.9\) s | A1 | |
| 3 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[T - 0.3g = 0.3a$ or $0.5g - T = 0.5a]$ | M1 | Use of Newton's second law for $P$ or $Q$ or use of $a = (m_Q - m_P)g/(m_P + m_Q)$ |
| $T - 0.3g = 0.3a$ and $0.5g - T = 0.5a$ or $a = (0.5g - 0.3g)/(0.5 + 0.3)$ | A1 | |
| $[0.5g - 0.3g = 0.8a]$ | M1 | Solve for $a$ |
| $a = 2.5$ | A1 | |
| $[h = 0 + \frac{1}{2} \times 2.5 \times 0.6^2]$ | M1 | For use of $s = ut + \frac{1}{2}at^2$ |
| $h = 0.45$ | A1 | |
| | 6 | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Velocity of $P$ when $Q$ reaches floor $= 0 + 0.6 \times 2.5 = 1.5 \text{ ms}^{-1}$ | B1ft | ft $a$ from (i) $\times 0.6$ |
| $[0 = 1.5 - gt \rightarrow t = \ldots]$ $(t = 0.15)$ | M1 | Use of suvat to find time to highest point |
| Total time $= 2 \times 0.15 + 0.6 = 0.9$ s | A1 | |
| | 3 | |5\\
\includegraphics[max width=\textwidth, alt={}, center]{007ccd92-79ba-409a-97e8-a4cf1f0a6cc5-08_538_414_260_868}
Two particles $P$ and $Q$, of masses 0.3 kg and 0.5 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley with the particles hanging freely below it. $Q$ is held at rest with the string taut at a height of $h \mathrm {~m}$ above a horizontal floor (see diagram). $Q$ is now released and both particles start to move. The pulley is sufficiently high so that $P$ does not reach it at any stage. The time taken for $Q$ to reach the floor is 0.6 s .\\
(i) Find the acceleration of $Q$ before it reaches the floor and hence find the value of $h$.\\
$Q$ remains at rest when it reaches the floor, and $P$ continues to move upwards.\\
(ii) Find the velocity of $P$ at the instant when $Q$ reaches the floor and the total time taken from the instant at which $Q$ is released until the string becomes taut again.\\
\hfill \mbox{\textit{CAIE M1 2018 Q5 [9]}}