| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | String at angle to slope |
| Difficulty | Standard +0.3 This is a standard mechanics problem requiring resolution of forces in two directions on an inclined plane with friction. Part (a) involves systematic application of Newton's second law with multiple force components (7 marks suggests routine working), while part (b) is a straightforward kinematics calculation using constant acceleration equations. The problem requires no novel insight—just careful bookkeeping of force components and standard techniques taught in M1. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03e Resolve forces: two dimensions3.03v Motion on rough surface: including inclined planes |
| Answer | Marks |
|---|---|
| 4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a) | Attempt at N2L parallel to the plane | *M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Tcos26−8gsin18−F =80.2 | A1 | Allow with their F . |
| Attempt at resolving perpendicular to the plane | *M1 | 3 terms |
| Answer | Marks | Guidance |
|---|---|---|
| R+Tsin26=8gcos18 | A1 | |
| Use of F =0.65R to get an equation in T only | DM1 | R is a linear combination of a component of T |
| Answer | Marks | Guidance |
|---|---|---|
| Solve for T | M1 | Dependent on all 3 previous M marks. |
| T =64 ( .0 ) N | A1 |
| Answer | Marks |
|---|---|
| 4(b) | Complete method to find s using constant acceleration formula(e) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 2 2 | M1 | Finding distance moved between t =3 and |
| Answer | Marks | Guidance |
|---|---|---|
| Distance = 0.7 m | A1 | If 0 marks scored then |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | Attempt at N2L parallel to the plane | *M1 | 4 terms.
Allow sign errors, sin/cos mix, allow g
missing.
Tcos26−8gsin18−F =80.2 | A1 | Allow with their F .
Attempt at resolving perpendicular to the plane | *M1 | 3 terms
Allow sign errors, sin/cos mix, allow g
missing.
R+Tsin26=8gcos18 | A1
Use of F =0.65R to get an equation in T only | DM1 | R is a linear combination of a component of T
and a component of weight.
Using equations with no missing terms.
Solve for T | M1 | Dependent on all 3 previous M marks.
T =64 ( .0 ) N | A1
7
--- 4(b) ---
4(b) | Complete method to find s using constant acceleration formula(e)
1 1 1 1
s= 0.242− 0.232 OR s= ( 0+0.24 )4− ( 0+0.23 )3
2 2 2 2 | M1 | Finding distance moved between t =3 and
t =4, must be using a=0.2
Distance = 0.7 m | A1 | If 0 marks scored then
1
SCB1 for s = 0.242 =1.6
2
2
Question | Answer | Marks | Guidance
\includegraphics{figure_4}
A block of mass 8 kg is placed on a rough plane which is inclined at an angle of 18° to the horizontal. The block is pulled up the plane by a light string that makes an angle of 26° above a line of greatest slope. The tension in the string is $T$ N (see diagram). The coefficient of friction between the block and plane is 0.65.
\begin{enumerate}[label=(\alph*)]
\item The acceleration of the block is 0.2 m s$^{-2}$.
Find $T$. [7]
\item The block is initially at rest.
Find the distance travelled by the block during the fourth second of motion. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q4 [9]}}