| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force up incline, find KE/PE changes as sub-parts |
| Difficulty | Moderate -0.3 This is a straightforward application of standard energy formulas (KE = ½mv², PE = mgh) with clearly given values, followed by a work-energy principle calculation. All steps are routine with no problem-solving insight required, making it slightly easier than average but not trivial due to the multi-step nature and need to apply the work-energy theorem correctly. |
| Spec | 6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| M1 | Attempt KE and/or PE with correct dimensions | |
| KE gain \(= \frac{1}{2} \times 800 \times (14^2 - 8^2) = 52800\) J | A1 | |
| PE gain \(= 800 \times 10 \times 120 \times 0.15 = 144000\) J | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| WD by engine \(= 32000 \times 12\) | B1 | |
| \(32000 \times 12 = 144000 + 52800 +\) WD against \(F\) | M1 | Work/Energy equation, 4 terms |
| WD against \(F = 187200\) J | A1 | WD \(= 187000\) to 3sf |
| Total: 3 |
## Question 4(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| | M1 | Attempt KE and/or PE with correct dimensions |
| KE gain $= \frac{1}{2} \times 800 \times (14^2 - 8^2) = 52800$ J | A1 | |
| PE gain $= 800 \times 10 \times 120 \times 0.15 = 144000$ J | A1 | |
| **Total: 3** | | |
---
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| WD by engine $= 32000 \times 12$ | B1 | |
| $32000 \times 12 = 144000 + 52800 +$ WD against $F$ | M1 | Work/Energy equation, 4 terms |
| WD against $F = 187200$ J | A1 | WD $= 187000$ to 3sf |
| **Total: 3** | | |
---4 A car of mass 800 kg is moving up a hill inclined at $\theta ^ { \circ }$ to the horizontal, where $\sin \theta = 0.15$. The initial speed of the car is $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Twelve seconds later the car has travelled 120 m up the hill and has speed $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find the change in the kinetic energy and the change in gravitational potential energy of the car.\\
(ii) The engine of the car is working at a constant rate of 32 kW . Find the total work done against the resistive forces during the twelve seconds.\\
\hfill \mbox{\textit{CAIE M1 2017 Q4 [6]}}