| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - inclined plane involved |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with connected particles on an inclined plane. Part (i) requires routine application of Newton's second law to both particles (5 marks for standard equations). Parts (ii) and (iii) involve straightforward kinematics after the system is set up. The slack string extension requires understanding that P continues up the plane then returns, but this is a common textbook scenario. Slightly above average due to the three-part structure and the slack string consideration, but all techniques are standard M1 material with no novel insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | M1 | Use of Newton’s second law for P or Q or the system |
| Answer | Marks | Guidance |
|---|---|---|
| or 0.2g – 0.3g sin 36.9 = (0.2 + 0.3)a | A1 | Two correct equations |
| Answer | Marks | Guidance |
|---|---|---|
| [0.2g–0.18g = 0.5a] | M1 | For solving either the system for a or for solving a pair of |
| Answer | Marks |
|---|---|
| a=0.4ms–2 | A1 |
| T=1.92N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Mark |
| Answer | Marks |
|---|---|
| 7(ii) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | For use of the constant acceleration equations with their a from |
| Answer | Marks |
|---|---|
| t = 2 s | A1 |
| Answer | Marks |
|---|---|
| 7(iii) | Speed when Q hits the floor=2×0.4(=0.8) |
| Answer | Marks | Guidance |
|---|---|---|
| or | B1FT | Using v=u+at with u=0 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | M1 | Using Newton’s second law for P to find a ≠ ± g |
| Answer | Marks | Guidance |
|---|---|---|
| 15 15 | M1 | Use of the constant acceleration equation(s) to find the time |
Total time = 2 + 0.266... = 2 + = 2.27 = s
| Answer | Marks |
|---|---|
| 15 15 | A1 |
Question 7:
--- 7(i) ---
7(i) | M1 | Use of Newton’s second law for P or Q or the system
3
For P: T −0.3g× =T −0.3gsin36.9=0.3a
5
For Q: 0.2g – T = 0.2a
System: 0.2g−0.3g× 3 =( 0.2+0.3 ) a
5
or 0.2g – 0.3g sin 36.9 = (0.2 + 0.3)a | A1 | Two correct equations
Allow use of θ = 36.9
[0.2g–0.18g = 0.5a] | M1 | For solving either the system for a or for solving a pair of
simultaneous equations for a or T
a=0.4ms–2 | A1
T=1.92N | A1
5
Question | Answer | Mark | Guidance
--- 7(ii) ---
7(ii) | 1
0.8=0+ ×0.4×t2a
2 | M1 | For use of the constant acceleration equations with their a from
7(i) and a ≠ ± g for a complete method to find t
t = 2 s | A1
2
--- 7(iii) ---
7(iii) | Speed when Q hits the floor=2×0.4(=0.8)
v= ( 2×0.4×0.8 )[ =0.8 ]
or | B1FT | Using v=u+at with u=0
Allow FT for their unsimplified v = at or v2 = 2as with
a from (i), t from (ii) and s = 0.8
3
−0.3g× =−0.3gsin36.9=0.3a [a = –6]
5 | M1 | Using Newton’s second law for P to find a ≠ ± g
0=0.8t+ 1 ×( −6 ) t2 ( t=0.2666... )
2
or
0 = 0.8 – 6T
2 4
(T = 0.13333 = and t = 2T = 0.26666 = )
15 15 | M1 | Use of the constant acceleration equation(s) to find the time
taken for P to return to the position where the string first became
slack.
4 34
Total time = 2 + 0.266... = 2 + = 2.27 = s
15 15 | A1
4\includegraphics{figure_7}
Two particles $P$ and $Q$, of masses 0.3 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to the edge of a smooth plane. The plane is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac{3}{5}$. $P$ lies on the plane and $Q$ hangs vertically below the pulley at a height of 0.8 m above the floor (see diagram). The string between $P$ and the pulley is parallel to a line of greatest slope of the plane. $P$ is released from rest and $Q$ moves vertically downwards.
\begin{enumerate}[label=(\roman*)]
\item Find the tension in the string and the magnitude of the acceleration of the particles. [5]
\end{enumerate}
$Q$ hits the floor and does not bounce. It is given that $P$ does not reach the pulley in the subsequent motion.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the time, from the instant at which $P$ is released, for $Q$ to reach the floor. [2]
\item When $Q$ hits the floor the string becomes slack. Find the time, from the instant at which $P$ is released, for the string to become taut again. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q7 [11]}}