| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | November |
| Marks | 10 |
| 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 systematic application of the work-energy principle. Part (i) uses W = ΔKE + ΔPE + work against resistance; part (ii) uses PE = mgh to find distance; part (iii) applies work-energy again with piecewise resistance. All steps are standard M1 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| KE loss \(= \frac{1}{2} 16000(15^2 - 12^2)\) | B1 | |
| PE gain \(= 16000g(AB/20)\) | B1 | |
| M1 | For using WD by DF = PE gain + WD against resistance – KE loss | |
| \(1200 = 0.8g(AB) + 1.24(AB) - 648\) | A1 | |
| Distance AB is \(200m\) | A1 | 5 |
| (ii) Distance BD is \(300m\) | B1 | 1 |
| (iii) WD against resistance \(=\) | ||
| \(1240(BC) + 1860(300 - BC)\) | B1 ft | ft distance BD |
| M1 | For using KE loss = PE gain + WD against res'ce – WD by DF |
| Answer | Marks | Guidance |
|---|---|---|
| \(2400000 + (558000 - 620BC) - 7200 \times 300\) | A1 | |
| Distance BC is \(61.3 \text{ m}\) | A1 | 4 |
| Answer | Marks |
|---|---|
| For BC 16000a = 7200 – 1240 – 1860(300 – BC) and for CD 16000a = 7200 – 1860 – 8000 | B1 |
| For using \(v^2 = u^2 + 2as\) for both BC and CD | M1 |
| \(v_C^2 = 144 - 2 \times 0.1275(BC)\) and \(49 = v_C^2 - 2 \times 0.16625(300 - BC)\) | A1 |
| For eliminating \(v_C^2\) and obtaining BC = 61.3 m | A1 |
| SR for candidates who assume that the acceleration is constant in part (i), although there is no justification for the assumption (max. 3/5) |
| Answer | Marks |
|---|---|
| \([1200000 = AB - 1240 - 16000020 = 16000a\) and \(a = (12^2 - 15^2)/(2(AB))]\) | M1 |
| For eliminating \(a\) and attempting to solve for AB | M1 |
| Distance AB is \(200m\) | A1 |
**(i)**
KE loss $= \frac{1}{2} 16000(15^2 - 12^2)$ | B1 |
PE gain $= 16000g(AB/20)$ | B1 |
| M1 | For using WD by DF = PE gain + WD against resistance – KE loss
$1200 = 0.8g(AB) + 1.24(AB) - 648$ | A1 |
Distance AB is $200m$ | A1 | 5
**(ii)** Distance BD is $300m$ | B1 | 1
**(iii)** WD against resistance $=$ | |
$1240(BC) + 1860(300 - BC)$ | B1 ft | ft distance BD
| M1 | For using KE loss = PE gain + WD against res'ce – WD by DF
$\frac{1}{2} 16000(12^2 - 7^2) =$
$2400000 + (558000 - 620BC) - 7200 \times 300$ | A1 |
Distance BC is $61.3 \text{ m}$ | A1 | 4
**Alternative for Q6 part (iii).**
For BC 16000a = 7200 – 1240 – 1860(300 – BC) and for CD 16000a = 7200 – 1860 – 8000 | B1 |
For using $v^2 = u^2 + 2as$ for both BC and CD | M1 |
$v_C^2 = 144 - 2 \times 0.1275(BC)$ and $49 = v_C^2 - 2 \times 0.16625(300 - BC)$ | A1 |
For eliminating $v_C^2$ and obtaining BC = 61.3 m | A1 |
SR for candidates who assume that the acceleration is constant in part (i), although there is no justification for the assumption (max. 3/5) | |
For appropriate use of Newton's second law and $v^2 = u^2 + 2as$
$[1200000 = AB - 1240 - 16000020 = 16000a$ and $a = (12^2 - 15^2)/(2(AB))]$ | M1 |
For eliminating $a$ and attempting to solve for AB | M1 |
Distance AB is $200m$ | A1 |
---6 A lorry of mass 16000 kg climbs a straight hill $A B C D$ which makes an angle $\theta$ with the horizontal, where $\sin \theta = \frac { 1 } { 20 }$. For the motion from $A$ to $B$, the work done by the driving force of the lorry is 1200 kJ and the resistance to motion is constant and equal to 1240 N . The speed of the lorry is $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $A$ and $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $B$.\\
(i) Find the distance $A B$.
For the motion from $B$ to $D$ the gain in potential energy of the lorry is 2400 kJ .\\
(ii) Find the distance $B D$.
For the motion from $B$ to $D$ the driving force of the lorry is constant and equal to 7200 N . From $B$ to $C$ the resistance to motion is constant and equal to 1240 N and from $C$ to $D$ the resistance to motion is constant and equal to 1860 N .\\
(iii) Given that the speed of the lorry at $D$ is $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find the distance $B C$.
\hfill \mbox{\textit{CAIE M1 2011 Q6 [10]}}