| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Particle on incline, hanging counterpart |
| Difficulty | Standard +0.3 This is a standard connected particles problem requiring resolution of forces parallel to the plane, application of F=ma to the system, and basic kinematics. While it involves multiple forces (applied force at an angle, weight components, tension, resistance), the solution follows a routine procedure taught in M1 with no novel insight required. The multi-part structure and 7 total marks indicate slightly above-average length, but the techniques are entirely standard for this topic. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| 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 |
|---|---|
| 5(a) | T – 40g sin20 – 50 = 40a T −136.8−50=40a |
| Answer | Marks | Guidance |
|---|---|---|
| 482.96−273.61−136.8−50=120a | M1 | Attempt at Newton’s second law for at least one case. |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | Any 2 equations. | |
| For attempt to solve for T or a | M1 | From equation(s) with no missing/extra terms. Allow g |
| Answer | Marks | Guidance |
|---|---|---|
| Acceleration = 0.188 ms−2 | A1 | Allow AWRT 0.19. |
| Tension = 194 N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(b) | [1.2 = 0 + 0.188t] | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Allow if a is negative in part (a) and use | a | here. |
| Time = 6.39 s | A1 | Allow 6.38 |
| 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) | T – 40g sin20 – 50 = 40a T −136.8−50=40a
500 cos15 – 80g sin20 – T = 80a 482.96−273.61−T =80a
500 cos15 – 80g sin20 – 40g sin20−50 = (80+40)a
482.96−273.61−136.8−50=120a | M1 | Attempt at Newton’s second law for at least one case.
Allow sign errors. Do not allow g missing. Correct
number of terms. Allow sin/cos mix.
A1 | Any 2 equations.
For attempt to solve for T or a | M1 | From equation(s) with no missing/extra terms. Allow g
missing. Must get to ‘T =’ or ‘α =’.
Acceleration = 0.188 ms−2 | A1 | Allow AWRT 0.19.
Tension = 194 N | A1
5
--- 5(b) ---
5(b) | [1.2 = 0 + 0.188t] | M1 | For use of constant acceleration formula(e) and solving
for t with their positive a, leading to a positive value of
t
a10,ag
Allow if a is negative in part (a) and use |a| here.
Time = 6.39 s | A1 | Allow 6.38
Allow 6.32 from a = 0.19
2
Question | Answer | Marks | Guidance\includegraphics{figure_5}
A block $A$ of mass 80 kg is connected by a light, inextensible rope to a block $B$ of mass 40 kg. The rope joining the two blocks is taut and is parallel to a line of greatest slope of a plane which is inclined at an angle of $20°$ to the horizontal. A force of magnitude 500 N inclined at an angle of $15°$ above the same line of greatest slope acts on $A$ (see diagram). The blocks move up the plane and there is a resistance force of 50 N on $B$, but no resistance force on $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the blocks and the tension in the rope. [5]
\end{enumerate}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the time that it takes for the blocks to reach a speed of $1.2 \text{ m s}^{-1}$ from rest. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q5 [7]}}