| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | March |
| Marks | 12 |
| 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 question requiring application of Newton's second law to find acceleration and tension (routine for M1), followed by kinematics with a direction change when the heavier particle hits the floor. While part (ii) requires careful consideration of the motion in two phases, this is a well-practiced exam technique at this level, making it slightly easier than average overall. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 6(i) | M1 | Apply Newton’s law to either of the |
| Answer | Marks | Guidance |
|---|---|---|
| 12 – T = 1.2a and T – 8 = 0.8a | A1 | Both equations correct |
| M1 | Solve for a and T | |
| a = 2ms–2 and T = 9.6N | A1 | |
| Total: | 4 |
| Answer | Marks |
|---|---|
| 6(ii) | [0.64 = ½ × 2 × t 2] |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | M1 | Attempt to find time t taken for 1.2kg |
| Answer | Marks |
|---|---|
| 1 | A1 |
| v = 2 × 0.8 = 1.6 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | For attempting to find the time t |
| Answer | Marks |
|---|---|
| 2 | A1 |
| Answer | Marks |
|---|---|
| 2 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | B1 | Finding the distance s travelled |
Total distance travelled =
| Answer | Marks |
|---|---|
| 0.64 + 0.128 + 0.008 = 0.776m | B1 |
| Total: | 8 |
Question 6:
--- 6(i) ---
6(i) | M1 | Apply Newton’s law to either of the
particles
12 – T = 1.2a and T – 8 = 0.8a | A1 | Both equations correct
M1 | Solve for a and T
a = 2ms–2 and T = 9.6N | A1
Total: | 4
--- 6(ii) ---
6(ii) | [0.64 = ½ × 2 × t 2]
1
[v = 2t ]
1 | M1 | Attempt to find time t taken for 1.2kg
1
particle to reach ground and/or its
speed v at the ground
t = 0.8
1 | A1
v = 2 × 0.8 = 1.6 | A1
[0 = 1.6 – 10t ]
2
[1.62 = 2 × 10 × s ]
2 | M1 | For attempting to find the time t
2
and/or distance travelled s as 0.8kg
2
particle comes to rest
t = 0.16
2 | A1
s = 0.128
2 | A1
t = 1 – 0.8 – 0.16 = 0.04
3
s = ½ × 10 × 0.042
3 | B1 | Finding the distance s travelled
3
downwards in t seconds
3
Total distance travelled =
0.64 + 0.128 + 0.008 = 0.776m | B1
Total: | 8\includegraphics{figure_6}
Two particles of masses $1.2$ kg and $0.8$ kg are connected by a light inextensible string that passes over a fixed smooth pulley. The particles hang vertically. The system is released from rest with both particles $0.64$ m above the floor (see diagram). In the subsequent motion the $0.8$ kg particle does not reach the pulley.
\begin{enumerate}[label=(\roman*)]
\item Show that the acceleration of the particles is $2$ m s$^{-2}$ and find the tension in the string. [4]
\item Find the total distance travelled by the $0.8$ kg particle during the first second after the particles are released. [8]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2017 Q6 [12]}}