| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - horizontal motion with resistance (no driving force) |
| Difficulty | Moderate -0.3 This is a straightforward application of the work-energy principle with standard mechanics bookwork. Part (i) requires a single energy equation with given values, while part (ii) adds gravitational PE but remains routine. The calculations are direct with no conceptual challenges beyond standard M1 content. |
| Spec | 6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\frac{1}{2} \times 1.2 \times 7.5^2 - \frac{1}{2} \times 1.2 \times v^2 = 25]\) | M1 | For use of KE and 25 in a 3 term equation |
| \(v = 3.82 \text{ ms}^{-1}\) (3.81881...) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.2gd\sin30\) | B1 | Correct expression for PE |
| \([\frac{1}{2} \times 1.2 \times 7.5^2 - 25 + 1.2gd\sin30 = \frac{1}{2} \times 1.2 \times 9^2]\) | M1 | For 4 term work/energy equation |
| \(d = 6.64 \text{ m}\) (6.64166...) | A1 | |
| 3 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\frac{1}{2} \times 1.2 \times 7.5^2 - \frac{1}{2} \times 1.2 \times v^2 = 25]$ | M1 | For use of KE and 25 in a 3 term equation |
| $v = 3.82 \text{ ms}^{-1}$ (3.81881...) | A1 | |
| | [2] | |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.2gd\sin30$ | B1 | Correct expression for PE |
| $[\frac{1}{2} \times 1.2 \times 7.5^2 - 25 + 1.2gd\sin30 = \frac{1}{2} \times 1.2 \times 9^2]$ | M1 | For 4 term work/energy equation |
| $d = 6.64 \text{ m}$ (6.64166...) | A1 | |
| | 3 | |3 A particle of mass 1.2 kg moves in a straight line $A B$. It is projected with speed $7.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from $A$ towards $B$ and experiences a resistance force. The work done against this resistance force in moving from $A$ to $B$ is 25 J .\\
(i) Given that $A B$ is horizontal, find the speed of the particle at $B$.\\
(ii) It is given instead that $A B$ is inclined at $30 ^ { \circ }$ below the horizontal and that the speed of the particle at $B$ is $9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The work done against the resistance force remains the same. Find the distance $A B$.\\
\hfill \mbox{\textit{CAIE M1 2018 Q3 [5]}}