| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Standard +0.3 This is a standard M1 work-energy problem requiring application of P=Fv, F=ma, and work-energy principle across two sections. While multi-step with several calculations, it follows routine mechanics procedures without requiring novel insight—slightly above average due to the two-part structure and careful bookkeeping of energy terms. |
| Spec | 3.03d Newton's second law: 2D vectors6.02d Mechanical energy: KE and PE concepts6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([144000/v - 4800 = 12500a]\) | M1 | For using \(DF = P/v\) and Newton's 2nd law at A or at B |
| Acceleration at A is \(0.336\ \text{ms}^{-2}\) | A1 | |
| The speed at B is \(24\ \text{ms}^{-1}\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| WD by \(DF = 5800 \times 500\) | ||
| WD against res'ce \(= 4800 \times 500\) | B1 | |
| Loss in \(KE = \frac{1}{2}12500(24^2 - 16^2)\) | B1 | |
| M1 | For using WD by DF = PE gain – KE loss + WD against res'ce | |
| \(5800 \times 500 = 12500gh - \frac{1}{2}12500(24^2 - 16^2) + 4800 \times 500\) | A1 | |
| Height of C is 20 m | A1 | 5 marks |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[144000/v - 4800 = 12500a]$ | M1 | For using $DF = P/v$ and Newton's 2nd law at A or at B |
| Acceleration at A is $0.336\ \text{ms}^{-2}$ | A1 | |
| The speed at B is $24\ \text{ms}^{-1}$ | A1 | 3 marks | AG |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| WD by $DF = 5800 \times 500$ | | |
| WD against res'ce $= 4800 \times 500$ | B1 | |
| Loss in $KE = \frac{1}{2}12500(24^2 - 16^2)$ | B1 | |
| | M1 | For using WD by DF = PE gain – KE loss + WD against res'ce |
| $5800 \times 500 = 12500gh - \frac{1}{2}12500(24^2 - 16^2) + 4800 \times 500$ | A1 | |
| Height of C is 20 m | A1 | 5 marks |6 A lorry of mass 12500 kg travels along a road from $A$ to $C$ passing through a point $B$. The resistance to motion of the lorry is 4800 N for the whole journey from $A$ to $C$.\\
(i) The section $A B$ of the road is straight and horizontal. On this section of the road the power of the lorry's engine is constant and equal to 144 kW . The speed of the lorry at $A$ is $16 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its acceleration at $B$ is $0.096 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find the acceleration of the lorry at $A$ and show that its speed at $B$ is $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) The section $B C$ of the road has length 500 m , is straight and inclined upwards towards $C$. On this section of the road the lorry's driving force is constant and equal to 5800 N . The speed of the lorry at $C$ is $16 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the height of $C$ above the level of $A B$.
\hfill \mbox{\textit{CAIE M1 2013 Q6 [8]}}