| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on smooth horizontal surface, particle hanging |
| Difficulty | Moderate -0.3 This is a standard two-stage pulley problem requiring basic application of Newton's second law and energy/kinematics. Part (i) tests understanding of inextensible strings (trivial). Part (ii) requires setting up equations for the connected system and using energy conservation or SUVAT, but follows a well-practiced template with straightforward arithmetic. Slightly easier than average due to the guided structure and routine nature of the mechanics. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 2 (i) | B1 | 1 |
| Answer | Marks |
|---|---|
| (ii) | Loss of PE = 0.15gh |
| Answer | Marks |
|---|---|
| h = 1.5 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 4 | For using loss of PE |
| Answer | Marks |
|---|---|
| (ii) | [0.15g – T = 0.15a and T = 0.35a |
| Answer | Marks |
|---|---|
| h = 1.5 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 4 | For applying Newton’s second law to A |
| Answer | Marks | Guidance |
|---|---|---|
| Page 5 | Mark Scheme | Syllabus |
| Cambridge International AS/A Level – May/June 2015 | 9709 | 43 |
| Answer | Marks |
|---|---|
| (ii) | [0.15g – T = 0.15a and T = 0.35a |
| Answer | Marks |
|---|---|
| h = 1.5 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 4 | For applying Newton’s second law to A |
Question 2:
--- 2 (i) ---
2 (i) | B1 | 1 | After B reaches the floor, A continues at
constant speed until it reaches the pulley
(no tension and the surface is smooth).
Thus A’s speed when B reaches the floor
–1
is the same as A’s speed (3 ms ) when it
reaches the pulley. Until the instant when
B reached the floor, A and B have the
same speed and hence B reaches the floor
-1
with speed 3 ms .
(ii) | Loss of PE = 0.15gh
1
2
Gain of KE = (0.35 + 0.15) × 3
2
1.5h = 0.25 × 9
h = 1.5 | B1
B1
M1
A1 | 4 | For using loss of PE
= gain of KE
Alternative Method for part (ii)
(ii) | [0.15g – T = 0.15a and T = 0.35a
or 0.15g = (0.35+0.15)a]
(cid:1) a = ....
–2
a = 3ms
2
[3 = 0 + 2 × 3h]
h = 1.5 | M1
A1
M1
A1 | 4 | For applying Newton’s second law to A
and to B or for using m Bg = (mA+mB)a to
find a
2 2
For using v = u + 2as
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 43
Alternative Method for part (ii)
(ii) | [0.15g – T = 0.15a and T = 0.35a
(cid:1) T = ....
T = 1.05N
1
0.15gh− ×0.15×32 =1.05h
2
or
1
×0.35×32 =1.05h
2
h = 1.5 | M1
A1
M1
A1 | 4 | For applying Newton’s second law to A
and to B to find T
For using PEB loss – KEB gain= WD
against T or
for using KEA gain = WD by T\includegraphics{figure_2}
Particles $A$ and $B$, of masses 0.35 kg and 0.15 kg respectively, are attached to the ends of a light inextensible string. $A$ is held at rest on a smooth horizontal surface with the string passing over a small smooth pulley fixed at the edge of the surface. $B$ hangs vertically below the pulley at a distance $h$ m above the floor (see diagram). $A$ is released and the particles move. $B$ reaches the floor and $A$ subsequently reaches the pulley with a speed of $3 \text{ m s}^{-1}$.
\begin{enumerate}[label=(\roman*)]
\item Explain briefly why the speed with which $B$ reaches the floor is $3 \text{ m s}^{-1}$. [1]
\item Find the value of $h$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2015 Q2 [5]}}