| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| 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-particle pulley system problem requiring Newton's second law for connected particles and energy/kinematics after impact. Part (i) is routine application of F=ma to both masses with the given answer provided. Part (ii) requires recognizing that A stops but B continues upward, then using v²=u²+2as or energy methods. While it involves multiple stages of motion, the techniques are standard M1 content with no novel insight required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| 5(i) | A: 4 – T = 0.4a |
| Answer | Marks | Guidance |
|---|---|---|
| System: 4 – 2 = (0.4 + 0.2)a | M1 | Apply Newton’ second law to particle A (3 terms) |
| Answer | Marks |
|---|---|
| A1 | Two correct equations |
| M1 | Either solve the system equation for a |
| Answer | Marks | Guidance |
|---|---|---|
| 3 3 | A1 | Both correct AG |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | M1 | Apply v 2 = u 2 +2as to particle A or particle B with a = 10/3 |
| v2 = 0 + 2 × 10/3 × 0.5 | A1 | [v = 1.83 but not needed specifically] |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | M1 | Apply v2 = u 2 + 2as to particle B to find s, the distance |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | A1 | Maximum height = 1/2 + 1/2 + 1/6 = 7/6 = 1.17 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | A: 4 – T = 0.4a
B: T – 2 = 0.2a
System: 4 – 2 = (0.4 + 0.2)a | M1 | Apply Newton’ second law to particle A (3 terms)
or to particle B (3 terms)
or to the system (4 terms implied)
A1 | Two correct equations
M1 | Either solve the system equation for a
or solve two simultaneous equations for a or T
or verify the given value of a by finding the same T value in
both equations
10 8
a = , T =
3 3 | A1 | Both correct AG
4
--- 5(ii) ---
5(ii) | M1 | Apply v 2 = u 2 +2as to particle A or particle B with a = 10/3
v2 = 0 + 2 × 10/3 × 0.5 | A1 | [v = 1.83 but not needed specifically]
1
0 = 10/3 – 2 × 10 × s [s = ]
6 | M1 | Apply v2 = u 2 + 2as to particle B to find s, the distance
travelled by B after A has hit the ground
7
Maximum height = = 1.17 m
6 | A1 | Maximum height = 1/2 + 1/2 + 1/6 = 7/6 = 1.17
4
Question | Answer | Marks | Guidance\includegraphics{figure_5}
Two particles $A$ and $B$, of masses 0.4 kg and 0.2 kg respectively, are connected by a light inextensible string which passes over a fixed smooth pulley. Both $A$ and $B$ are 0.5 m above the ground. The particles hang vertically (see diagram). The particles are released from rest. In the subsequent motion $B$ does not reach the pulley and $A$ remains at rest after reaching the ground.
\begin{enumerate}[label=(\roman*)]
\item For the motion before $A$ reaches the ground, show that the magnitude of the acceleration of each particle is $\frac{10}{3}$ m s$^{-2}$ and find the tension in the string. [4]
\item Find the maximum height of $B$ above the ground. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q5 [8]}}