| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Moderate -0.5 This is a standard SUVAT kinematics problem involving a velocity-time graph with constant acceleration phases. While the OCR corruption makes the exact details unclear, typical M1 questions of this type require reading values from a graph, applying basic SUVAT equations (v=u+at, s=area under graph), and calculating distances or accelerations. These are routine mechanics exercises slightly easier than the average A-level question across all topics, as they involve direct application of memorized formulas with minimal problem-solving insight required. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.03a Force: vector nature and diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4^2 = 0^2 + 2a \times 12.5 \Rightarrow a = 0.64\) | B1 | |
| \([35 \times 0.96 - 3g \times 0.6 - F = 3 \times 0.64]\) | M1 | For using Newton's 2nd law to find \(F\) |
| \(F = 13.68\) | A1 | |
| WD against \(F = 13.68 \times 12.5 = 171\) J | B1 | [4 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R_{\text{from O to A}} = 3g \times 0.8 - 35 \times 0.28\) | B1 | |
| \([\mu = 13.68 \div 14.2\ (= 0.96338)]\) | M1 | For using \(\mu = F \div R\) |
| Coefficient is \(0.963\) (accept \(0.96\)) | A1 | [3 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([-3g \times 0.6 - 0.96338 \times (3g \times 0.8) = 3a]\) | M1 | For applying Newton's 2nd law to block to find \(a\) |
| Acceleration is \(-13.7 \text{ ms}^{-2}\) | A1 | |
| \([0 = 16 + 2(-13.7)s]\) | M1 | For using \(v^2 = u^2 + 2as\) to find \(s\) |
| Distance travelled is \(0.584\) m | A1 | [4 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gain in \(KE = \frac{1}{2} \times 3 \times 4^2\ (= 24 \text{ J})\) | B1 | |
| Gain in \(PE = 3g \times 12.5 \times 0.6\ (= 225 \text{ J})\) | B1 | |
| \([WD = 35 \times 12.5 \times 0.96 - \frac{1}{2} \times 3 \times 4^2 - 3g \times 12.5 \times 0.6]\) | M1 | For using \(WD\) against \(F\) = WD by applied force \(-\) KE gain \(-\) PE gain |
| WD against \(F\) is \(171\) J | A1 | [4 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| WD against \(F = 0.96(338...) \times 3g \times 0.8s\) | B1 | |
| M1 | For using KE loss \(=\) PE gain \(+\) WD against friction | |
| \(\frac{1}{2} \times 3 \times 4^2 = 3gs(0.6) + 0.96(338...) \times 3g \times 0.8s\) | A1 | |
| Distance travelled is \(0.584\) m | A1 | [4 marks] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4^2 = 0^2 + 2a \times 12.5 \Rightarrow a = 0.64$ | B1 | |
| $[35 \times 0.96 - 3g \times 0.6 - F = 3 \times 0.64]$ | M1 | For using Newton's 2nd law to find $F$ |
| $F = 13.68$ | A1 | |
| WD against $F = 13.68 \times 12.5 = 171$ J | B1 | **[4 marks]** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R_{\text{from O to A}} = 3g \times 0.8 - 35 \times 0.28$ | B1 | |
| $[\mu = 13.68 \div 14.2\ (= 0.96338)]$ | M1 | For using $\mu = F \div R$ |
| Coefficient is $0.963$ (accept $0.96$) | A1 | **[3 marks]** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[-3g \times 0.6 - 0.96338 \times (3g \times 0.8) = 3a]$ | M1 | For applying Newton's 2nd law to block to find $a$ |
| Acceleration is $-13.7 \text{ ms}^{-2}$ | A1 | |
| $[0 = 16 + 2(-13.7)s]$ | M1 | For using $v^2 = u^2 + 2as$ to find $s$ |
| Distance travelled is $0.584$ m | A1 | **[4 marks]** |
### Part (i) — Alternative:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gain in $KE = \frac{1}{2} \times 3 \times 4^2\ (= 24 \text{ J})$ | B1 | |
| Gain in $PE = 3g \times 12.5 \times 0.6\ (= 225 \text{ J})$ | B1 | |
| $[WD = 35 \times 12.5 \times 0.96 - \frac{1}{2} \times 3 \times 4^2 - 3g \times 12.5 \times 0.6]$ | M1 | For using $WD$ against $F$ = WD by applied force $-$ KE gain $-$ PE gain |
| WD against $F$ is $171$ J | A1 | **[4 marks]** |
### Part (iii) — Alternative:
| Answer/Working | Mark | Guidance |
|---|---|---|
| WD against $F = 0.96(338...) \times 3g \times 0.8s$ | B1 | |
| | M1 | For using KE loss $=$ PE gain $+$ WD against friction |
| $\frac{1}{2} \times 3 \times 4^2 = 3gs(0.6) + 0.96(338...) \times 3g \times 0.8s$ | A1 | |
| Distance travelled is $0.584$ m | A1 | **[4 marks]** |7\\
\includegraphics[max width=\textwidth, alt={}, center]{9c7e8624-c4cd-4a8e-83d9-f92d0bd6f95b-4_668_848_260_653}
A small block of mass 3 kg is initially at rest at the bottom $O$ of a rough plane inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = 0.6$ and $\cos \alpha = 0.8$. A force of magnitude 35 N acts on the block at an angle $\beta$ above the plane, where $\sin \beta = 0.28$ and $\cos \beta = 0.96$. The block starts to move up a line of greatest slope of the plane and passes through a point $A$ with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The distance $O A$ is 12.5 m (see diagram).\\
(i) For the motion of the block from $O$ to $A$, find the work done against the frictional force acting on the block.\\
(ii) Find the coefficient of friction between the block and the plane.
At the instant that the block passes through $A$ the force of magnitude 35 N ceases to act.\\
(iii) Find the distance the block travels up the plane after passing through $A$.
\end{document}
\hfill \mbox{\textit{CAIE M1 2014 Q7}}