| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Coefficient from constant speed |
| Difficulty | Moderate -0.8 This is a straightforward application of Newton's second law with friction. Given the deceleration, students simply equate the friction force (μR = μmg) to ma, leading directly to μ = a/g = 1/20. This requires only basic recall of friction formulas and one algebraic step, making it easier than average. |
| Spec | 3.03v Motion on rough surface: including inclined planes |
| Answer | Marks |
|---|---|
| 7(a) | Collision: ( k+1 ) mv=mu so v= u |
| k+1 | B1 |
| Answer | Marks |
|---|---|
| 2 4 | B1 |
| Answer | Marks |
|---|---|
| 4 | B1 |
| Answer | Marks |
|---|---|
| 2 4 ( k+1 )2 8 | M1 |
| Use value of u and solve | M1 |
| k =3 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(b) | 9g |
| Answer | Marks |
|---|---|
| 20 | M1 |
| Answer | Marks |
|---|---|
| a 5 | M1 |
| μR=4mg/5and compare with R=4mg | B1 |
| Answer | Marks |
|---|---|
| 5 | A1 |
Question 7:
--- 7(a) ---
7(a) | Collision: ( k+1 ) mv=mu so v= u
k+1 | B1
Loss in KE = 1 m ( k+1 ) v2 − 1 v2
2 4 | B1
2
a
Gain in EPE = ½ .4mg/a.
4 | B1
So, 1 . 3 .m ( k+1 ) u2 = mga
2 4 ( k+1 )2 8 | M1
Use value of u and solve | M1
k =3 | A1
6
Question | Answer | Marks
--- 7(b) ---
7(b) | 9g
T +F =4m.
20 | M1
a
9mg
4mg.4 +F =
a 5 | M1
μR=4mg/5and compare with R=4mg | B1
1
μ=
5 | A1
4At the point $C$ the horizontal surface becomes rough, with coefficient of friction $\mu$ between the combined particle and the surface. The deceleration of the combined particle at $C$ is $\frac{g}{20}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $\mu$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q7 [4]}}