| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion up rough slope |
| Difficulty | Moderate -0.3 This is a standard two-part mechanics question on limiting equilibrium and motion on an inclined plane. Part (a) is routine application of F=μR with given answer to show, requiring resolving forces perpendicular and parallel to the plane. Part (b) applies Newton's second law with friction now opposing motion, then uses SUVAT equations. While it requires careful handling of forces and multiple steps (7 marks total), all techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| 2(a) | R = 0.4g cos 30 =2 3 or F or R=0.4gsin30 =2 | |
| | B1 | Use of m instead of 0.4 condoned. |
| 0.4g sin 30 – µ 0.4g cos 30 = 0 | M1 | For using F = µR. Allow sin/cos mix. |
| Answer | Marks | Guidance |
|---|---|---|
| 4cos30 3 3 | A1 | AG (exact answer only) |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 2(b) | 7.2 – 0.4g sin 30 –F = 0.4a | M1 |
| Answer | Marks |
|---|---|
| a = 8 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | For use of constant acceleration formula(e) and solving |
| Answer | Marks | Guidance |
|---|---|---|
| Allow if a is negative in part (a) and use | a | here. |
| Time = 0.5 s | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 2:
--- 2(a) ---
2(a) | R = 0.4g cos 30 =2 3 or F or R=0.4gsin30 =2
| B1 | Use of m instead of 0.4 condoned.
0.4g sin 30 – µ 0.4g cos 30 = 0 | M1 | For using F = µR. Allow sin/cos mix.
Both must be different components of their weight
only, not a 2 term R. Allow sign errors. Allow g
omitted.
4sin30 1 3
= = 3 or .
4cos30 3 3 | A1 | AG (exact answer only)
If zero scored then SC B1 for [Angle of friction = 30°
1
so] µ = tan 30 = 3.
3
Allow full marks if using m in place of 0.4 or W in
place of mg or 0.4g
1
A0 for =0.577= 3, but A1(ISW) for
3
1
= 3=0.577
3
3
Question | Answer | Marks | Guidance
--- 2(b) ---
2(b) | 7.2 – 0.4g sin 30 –F = 0.4a | M1 | Newton’s second law. Four terms. Second term must
be a component of their weight. F ≠ 0 and F .
Allow sin/cos mix. Allow sign errors.
F must be a numerical expression
May use their F from part (a).
a = 8 | A1
1
1=0+ (their positive 8)t2
2 | M1 | For use of constant acceleration formula(e) and solving
for t. a10,ag.
Allow if a is negative in part (a) and use |a| here.
Time = 0.5 s | A1
4
Question | Answer | Marks | Guidance
A particle $P$ of mass 0.4 kg is in limiting equilibrium on a plane inclined at $30°$ to the horizontal.
\begin{enumerate}[label=(\alph*)]
\item Show that the coefficient of friction between the particle and the plane is $\frac{1}{3}\sqrt{3}$. [3]
\end{enumerate}
A force of magnitude 7.2 N is now applied to $P$ directly up a line of greatest slope of the plane.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Given that $P$ starts from rest, find the time that it takes for $P$ to move 1 m up the plane. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q2 [7]}}