| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - vertical strings |
| Difficulty | Moderate -0.3 This is a standard two-particle pulley system problem requiring Newton's second law to find acceleration and tension (routine application of F=ma to connected particles), followed by energy/kinematics to find maximum height after one particle stops. The multi-step nature and need to consider motion in two phases elevates it slightly above pure recall, but it follows a well-practiced template with no novel insight required. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| 5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a) | 0.8g – T = 0.8a, T – 0.2g = 0.2a, | |
| For system: 0.8g – 0.2g = (0.8 + 0.2)a | M1 | Apply Newton’s 2nd law to either particle or to the system |
| A1 | Any 2 correct equations | |
| Attempt to solve for either a or T | M1 | |
| a = 6 ms–2 and T = 3.2 N | A1 | AG. Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| 5(b) | v2 = 2 × 6 × 0.5 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0 = 6 – 20s | M1 | Attempt to find the extra height reached by 0.2 kg particle |
| Answer | Marks |
|---|---|
| Greatest height = 0.5 + 0.5 + 0.3 = 1.3 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | 0.8g – T = 0.8a, T – 0.2g = 0.2a,
For system: 0.8g – 0.2g = (0.8 + 0.2)a | M1 | Apply Newton’s 2nd law to either particle or to the system
A1 | Any 2 correct equations
Attempt to solve for either a or T | M1
a = 6 ms–2 and T = 3.2 N | A1 | AG. Both correct
4
--- 5(b) ---
5(b) | v2 = 2 × 6 × 0.5 | M1 | Attempt to find v or v2 as 0.8 kg particle reaches the ground using
a from 5(a)
0 = 6 – 20s | M1 | Attempt to find the extra height reached by 0.2 kg particle
using v2 from previous M1 mark
Greatest height = 0.5 + 0.5 + 0.3 = 1.3 m | A1
3
Question | Answer | Marks | Guidance\includegraphics{figure_5}
Two particles of masses 0.8 kg and 0.2 kg are connected by a light inextensible string that passes over a fixed smooth pulley. The system is released from rest with both particles 0.5 m above a horizontal floor (see diagram). In the subsequent motion the 0.2 kg particle does not reach the pulley.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the acceleration of the particles is $6 \text{ m s}^{-2}$ and find the tension in the string. [4]
\item When the 0.8 kg particle reaches the floor it comes to rest.
Find the greatest height of the 0.2 kg particle above the floor. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q5 [7]}}