| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Moderate -0.3 This is a straightforward work-energy problem requiring standard formulas (work = force × distance × cos θ, and kinetic energy calculations). Part (i) is direct substitution, part (ii) requires applying the work-energy principle with given values. No novel insight needed, just careful bookkeeping of the energy transfers across two stages. |
| Spec | 6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(WD = 25 \times 40\cos30°\) | M1 | For using \(WD = Fd\cos\theta\) |
| Work done is 866 J | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(50 \times 40\cos30° = 866 + \text{KE gain}\) | M1 | For using \(WD\) by \(P = WD\) against resistance \(+\) KE gain |
| KE gain is 866 J | A1ft | ft incorrect ans (i) |
| \(\frac{1}{2} \times 35(v^2 - 1.2^2) = 866\) | M1 | For using KE gain \(= \frac{1}{2}m(v^2 - u^2)\) |
| A1ft | ft incorrect KE | |
| Speed is \(7.14\text{ ms}^{-1}\) | A1 | |
| [5] | ||
| SR (max 2/3 for last three marks): \(50\cos30° - 25\cos30° = 35a\); \(v^2 = 1.2^2 + 2\times40a\) M1 \(\Rightarrow\) speed is \(7.14\text{ ms}^{-1}\) A1 | Using Newton's second law and constant acceleration |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $WD = 25 \times 40\cos30°$ | M1 | For using $WD = Fd\cos\theta$ |
| Work done is 866 J | A1 | |
| **[2]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $50 \times 40\cos30° = 866 + \text{KE gain}$ | M1 | For using $WD$ by $P = WD$ against resistance $+$ KE gain |
| KE gain is 866 J | A1ft | ft incorrect ans (i) |
| $\frac{1}{2} \times 35(v^2 - 1.2^2) = 866$ | M1 | For using KE gain $= \frac{1}{2}m(v^2 - u^2)$ |
| | A1ft | ft incorrect KE |
| Speed is $7.14\text{ ms}^{-1}$ | A1 | |
| **[5]** | | |
| SR (max 2/3 for last three marks): $50\cos30° - 25\cos30° = 35a$; $v^2 = 1.2^2 + 2\times40a$ M1 $\Rightarrow$ speed is $7.14\text{ ms}^{-1}$ A1 | | Using Newton's second law and constant acceleration |
---3 A load is pulled along a horizontal straight track, from $A$ to $B$, by a force of magnitude $P \mathrm {~N}$ which acts at an angle of $30 ^ { \circ }$ upwards from the horizontal. The distance $A B$ is 80 m . The speed of the load is constant and equal to $1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ as it moves from $A$ to the mid-point $M$ of $A B$.\\
(i) For the motion from $A$ to $M$ the value of $P$ is 25 . Calculate the work done by the force as the load moves from $A$ to $M$.
The speed of the load increases from $1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ as it moves from $M$ towards $B$. For the motion from $M$ to $B$ the value of $P$ is 50 and the work done against resistance is the same as that for the motion from $A$ to $M$. The mass of the load is 35 kg .\\
(ii) Find the gain in kinetic energy of the load as it moves from $M$ to $B$ and hence find the speed with which it reaches $B$.
\hfill \mbox{\textit{CAIE M1 2010 Q3 [7]}}