| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Block on horizontal plane motion |
| Difficulty | Standard +0.3 This is a standard multi-part mechanics problem requiring resolution of forces, use of F=ma with constant acceleration kinematics, and friction calculations. While it involves several steps and careful bookkeeping of forces in two dimensions, all techniques are routine M1 content with no novel problem-solving insight required. The given information (distance, time, angle) makes the solution methodical rather than challenging. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks |
|---|---|
| 6(i) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | 1 |
| Answer | Marks |
|---|---|
| a=0.36 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 25 | M1 | Resolving horizontally. Allow use of θ=16.3 |
| F=4.68N | A1 |
| Answer | Marks |
|---|---|
| 6(ii) | 7 [ ] |
| Answer | Marks | Guidance |
|---|---|---|
| 25 | B1 | |
| 4.68=µ × 28.32 | M1 | Use of F=µR |
| µ= 0.165 (0.165254…) | A1 | 39 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Mark |
| Answer | Marks |
|---|---|
| 6(iii) | v=5×0.36 [= 1.8] |
| Answer | Marks | Guidance |
|---|---|---|
| or | B1FT | For velocity at t=5 ft on their a from 6(i) |
| 3a=–0.165 × 3g | M1 | Using Newton’s second law with new frictional force |
| 0=1.8–0.165gt (t = 1.09) | M1 | Using constant acceleration equations which would lead to a |
| Answer | Marks |
|---|---|
| Total time = 5 + 1.09 = 6.09s | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Mark |
Question 6:
--- 6(i) ---
6(i) | 1
4.5=0+ ×a×52
2 | M1 | 1
For use of s=ut+ at2 to find a
2
a=0.36 | A1
24
6× −F =3×0.36
25 | M1 | Resolving horizontally. Allow use of θ=16.3
F=4.68N | A1
4
--- 6(ii) ---
6(ii) | 7 [ ]
R=3g−6sin16.3=3g−6× =28.32
25 | B1
4.68=µ × 28.32 | M1 | Use of F=µR
µ= 0.165 (0.165254…) | A1 | 39
AG. Allow µ=
236
3
Question | Answer | Mark | Guidance
--- 6(iii) ---
6(iii) | v=5×0.36 [= 1.8]
v= ( 2×0.36×4.5 )[ =1.8 ]
or | B1FT | For velocity at t=5 ft on their a from 6(i)
3a=–0.165 × 3g | M1 | Using Newton’s second law with new frictional force
0=1.8–0.165gt (t = 1.09) | M1 | Using constant acceleration equations which would lead to a
positive value of t
Total time = 5 + 1.09 = 6.09s | A1
4
Question | Answer | Mark | GuidanceA block of mass 3 kg is initially at rest on a rough horizontal plane. A force of magnitude 6 N is applied to the block at an angle of $\theta$ above the horizontal, where $\cos \theta = \frac{24}{25}$. The force is applied for a period of 5 s, during which time the block moves a distance of 4.5 m.
\begin{enumerate}[label=(\roman*)]
\item Find the magnitude of the frictional force on the block. [4]
\item Show that the coefficient of friction between the block and the plane is 0.165, correct to 3 significant figures. [3]
\item When the block has moved a distance of 4.5 m, the force of magnitude 6 N is removed and the block then decelerates to rest. Find the total time for which the block is in motion. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q6 [11]}}