| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Coefficient of friction from motion |
| Difficulty | Standard +0.3 This is a standard two-part mechanics problem requiring force resolution on an inclined plane and then kinematics. Part (i) involves equilibrium (constant speed) with friction opposing motion, requiring resolution perpendicular and parallel to the plane to find μ. Part (ii) applies SUVAT equations after the force is removed. While it requires careful sign conventions and multiple steps (9 marks total), the techniques are routine for M1 students with no novel problem-solving insight needed—slightly easier than average A-level difficulty. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks |
|---|---|
| 6 (i) | 20 + 5gsin10° – F = 0 |
| Answer | Marks |
|---|---|
| Coefficient of friction is 0.582 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 5 | For resolving forces down the plane |
| Answer | Marks |
|---|---|
| (ii) | 5gsin10° – 0.582×49.24 =5a |
| Answer | Marks |
|---|---|
| Distance is 0.781 m | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 4 | For using Newton’s 2nd law |
| Answer | Marks |
|---|---|
| (ii) | ° |
| Answer | Marks |
|---|---|
| Distance is 0.781 m | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 4 | For using KE loss + PE loss = WD |
Question 6:
--- 6 (i) ---
6 (i) | 20 + 5gsin10° – F = 0
R = 5gcos10°
[µ = (20 + 8.6824)÷49.24]
Coefficient of friction is 0.582 | M1
A1
B1
M1
A1 | 5 | For resolving forces down the plane
For using µ = F ÷ R
(ii) | 5gsin10° – 0.582×49.24 =5a
[ ]
0=2.52 −2×4s
Distance is 0.781 m | M1
A1
M1
A1 | 4 | For using Newton’s 2nd law
ft µ from (i) (µ > 0)
2 2
For using v = u + 2as
Alternative Method for part (ii)
(ii) | °
PE loss = 5gdsin10
1
2 ° °
×5×2.5 +5gdsin10 = 0.582×5gdcos10
2
Distance is 0.781 m | B1
M1
A1
A1 | 4 | For using KE loss + PE loss = WD
against friction
ft µ (µ >0)A small box of mass 5 kg is pulled at a constant speed of $2.5 \text{ m s}^{-1}$ down a line of greatest slope of a rough plane inclined at $10°$ to the horizontal. The pulling force has magnitude 20 N and acts downwards parallel to a line of greatest slope of the plane.
\begin{enumerate}[label=(\roman*)]
\item Find the coefficient of friction between the box and the plane. [5]
\end{enumerate}
The pulling force is removed while the box is moving at $2.5 \text{ m s}^{-1}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the distance moved by the box after the instant at which the pulling force is removed. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2015 Q6 [9]}}