| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Standard +0.3 This is a straightforward multi-part work-energy question requiring standard mechanics techniques: applying work-energy theorem with forces on an incline, using F=ma for acceleration, and calculating power. The steps are routine and clearly signposted, though it requires careful bookkeeping of multiple forces and energy terms across three parts. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gain in KE \(= \frac{1}{2}1250(8^2 - 5^2)\) | B1 | |
| Loss in PE \(= 1250g \times 400\sin4°\) | B1 | |
| M1 | For using WD by \(DF\) = Gain in KE \(-\) Loss in PE \(+\) WD by resistance | |
| \(400(DF) = \frac{1}{2}1250(8^2-5^2) - 1250g\times400\sin4° + 2000\times400\) | A1 | |
| Driving force is 1189 N or 1190 N | A1 | Total: 5 marks |
| SR for using Newton's second law (max 2/5): \(DF + 1250g\sin4° - 2000 = 1250a\) | B1 | |
| \(a = (8^2-5^2)/2\times400 \rightarrow DF = 1190\ \text{N}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using Newton's second law to find acceleration, or for finding \(v_C\) and using \(v^2 = u^2 + 2as\) to find acceleration | |
| \(1189\times2 - 2000 = 1250a\) or \(22.75^2 = 8^2 + 2a\times750\) | A1↓ | \(DF\) from part (i) |
| Acceleration is \(0.302\ \text{ms}^{-2}\) | A1 | Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v_c^2 = 64 + 2\times0.302\times750\) | B1↓ | acceleration from part (ii) |
| \(P/\ 22.75 - 2000 = 1250\times0.302\) | M1 | |
| Power is 54.1 kW or 54100 W | A1 | Total: 3 marks |
## Question 7(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gain in KE $= \frac{1}{2}1250(8^2 - 5^2)$ | B1 | |
| Loss in PE $= 1250g \times 400\sin4°$ | B1 | |
| | M1 | For using WD by $DF$ = Gain in KE $-$ Loss in PE $+$ WD by resistance |
| $400(DF) = \frac{1}{2}1250(8^2-5^2) - 1250g\times400\sin4° + 2000\times400$ | A1 | |
| Driving force is 1189 N or 1190 N | A1 | Total: 5 marks |
| **SR** for using Newton's second law (max 2/5): $DF + 1250g\sin4° - 2000 = 1250a$ | B1 | |
| $a = (8^2-5^2)/2\times400 \rightarrow DF = 1190\ \text{N}$ | B1 | |
## Question 7(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using Newton's second law to find acceleration, or for finding $v_C$ and using $v^2 = u^2 + 2as$ to find acceleration |
| $1189\times2 - 2000 = 1250a$ or $22.75^2 = 8^2 + 2a\times750$ | A1↓ | $DF$ from part (i) |
| Acceleration is $0.302\ \text{ms}^{-2}$ | A1 | Total: 3 marks |
## Question 7(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_c^2 = 64 + 2\times0.302\times750$ | B1↓ | acceleration from part (ii) |
| $P/\ 22.75 - 2000 = 1250\times0.302$ | M1 | |
| Power is 54.1 kW or 54100 W | A1 | Total: 3 marks |7 A straight hill $A B$ has length 400 m with $A$ at the top and $B$ at the bottom and is inclined at an angle of $4 ^ { \circ }$ to the horizontal. A straight horizontal road $B C$ has length 750 m . A car of mass 1250 kg has a speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $A$ when starting to move down the hill. While moving down the hill the resistance to the motion of the car is 2000 N and the driving force is constant. The speed of the car on reaching $B$ is $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) By using work and energy, find the driving force of the car.
On reaching $B$ the car moves along the road $B C$. The driving force is constant and twice that when the car was on the hill. The resistance to the motion of the car continues to be 2000 N . Find\\
(ii) the acceleration of the car while moving from $B$ to $C$,\\
(iii) the power of the car's engine as the car reaches $C$.
\hfill \mbox{\textit{CAIE M1 2015 Q7 [11]}}