| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work-energy over time interval |
| Difficulty | Moderate -0.8 This is a straightforward application of standard work-energy formulas with minimal problem-solving required. Part (i) uses W=Pt directly, part (ii) applies F=ma at two points then uses the work-energy principle, and part (iii) follows mechanically from energy conservation. All steps are routine textbook exercises requiring only recall and substitution of given values. |
| Spec | 6.02a Work done: concept and definition6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\text{WD} = 14000 \times 25]\) | M1 | For using \(P = \text{WD} \div \Delta t\) |
| Work done is \(350\text{ kJ}\) or \(350000\text{ J}\) | A1 | Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(\text{DF} = P/v\) and Newton's 2nd law to find the speed of the car at \(A\) or at \(B\) | |
| \(14000/v_A - 235 = 1600 \times 0.5 \rightarrow v_A = 13.53 \text{ ms}^{-1}\) | A1 | \(v_A = 2800/207\) |
| \(14000/v_B - 235 = 1600 \times 0.25 \rightarrow v_B = 22.05 \text{ ms}^{-1}\) | A1 | \(v_B = 2800/127\) |
| \([\text{KE gain} = \frac{1}{2} \times 1600(22.05^2 - 13.53^2)]\) | M1 | For using KE gain \(= \frac{1}{2}m(v_B^2 - v_A^2)\) |
| KE gain \(= 242.5\text{ kJ}\) or \(242500\text{ J}\) | A1 | Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using WD by DF \(=\) KE gain \(+\) resistance \(\times AB\) | |
| \(350000 = 242500 + 235 \times AB\) | A1\(^\checkmark\) | |
| Distance \(AB\) is \(457\text{ m}\) | A1 | Total: 3 marks |
# Question 7 (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\text{WD} = 14000 \times 25]$ | M1 | For using $P = \text{WD} \div \Delta t$ |
| Work done is $350\text{ kJ}$ or $350000\text{ J}$ | A1 | Total: 2 marks |
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# Question 7 (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $\text{DF} = P/v$ and Newton's 2nd law to find the speed of the car at $A$ **or** at $B$ |
| $14000/v_A - 235 = 1600 \times 0.5 \rightarrow v_A = 13.53 \text{ ms}^{-1}$ | A1 | $v_A = 2800/207$ |
| $14000/v_B - 235 = 1600 \times 0.25 \rightarrow v_B = 22.05 \text{ ms}^{-1}$ | A1 | $v_B = 2800/127$ |
| $[\text{KE gain} = \frac{1}{2} \times 1600(22.05^2 - 13.53^2)]$ | M1 | For using KE gain $= \frac{1}{2}m(v_B^2 - v_A^2)$ |
| KE gain $= 242.5\text{ kJ}$ or $242500\text{ J}$ | A1 | Total: 5 marks |
---
# Question 7 (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using WD by DF $=$ KE gain $+$ resistance $\times AB$ |
| $350000 = 242500 + 235 \times AB$ | A1$^\checkmark$ | |
| Distance $AB$ is $457\text{ m}$ | A1 | Total: 3 marks |7 A car of mass 1600 kg moves with constant power 14 kW as it travels along a straight horizontal road. The car takes 25 s to travel between two points $A$ and $B$ on the road.\\
(i) Find the work done by the car's engine while the car travels from $A$ to $B$.
The resistance to the car's motion is constant and equal to 235 N . The car has accelerations at $A$ and $B$ of $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ respectively. Find\\
(ii) the gain in kinetic energy by the car in moving from $A$ to $B$,\\
(iii) the distance $A B$.
{www.cie.org.uk} after the live examination series.
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\hfill \mbox{\textit{CAIE M1 2015 Q7 [10]}}