| 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 |
| Type | Constant speed up/down incline |
| Difficulty | Standard +0.3 This is a straightforward energy method problem requiring standard application of work-energy principles. Part (i) uses work done against resistance plus gravitational PE gain. Part (ii) applies work-energy theorem with given values. Part (iii) uses P=Fv with given force ratio. All steps are routine M1 techniques with no novel problem-solving 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 formulae6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (i) PE gain \(= 1200g \times 45\) | B1 | |
| \(WD = 1200g \times 45 + 360\,000\) | M1 | For \(WD\) by car's engine \(=\) PE gain \(+\) WD against resistance |
| Work done is 900 000 J or 900 kJ | A1 | 3 |
| (ii) WD against resistance \(= 360 \times \sin5/\sin1\) (kJ) or \([360000 \div (45/\sin5°)] \times (45/\sin1°)\) (J) or \(697.24... \times 2578.44...\) (J) or 1798 (kJ) | B1 | |
| KE gain \(= 1660 + 540 - 1798\) | B1ft | Accept \(1660 + 540 - 1800\) |
| \([402000 = \frac{1}{2}1200(v^2 - 225)]\) | M1 | For using KE gain \(= \frac{1}{2}m(v^2 - 15^2)\) |
| Speed is 29.9 ms\(^{-1}\) | A1 | 4 AG |
| (iii) \(\frac{P_B}{P_C} = \left(\frac{DF_B}{DF_C}\right) \times \frac{v_B}{v_C} = 1.5 \times 15/29.9\) | M1 | For using \(P = Fv\) |
| A1 | ||
| Ratio is 0.75 | A1 | 3 |
| **(i)** PE gain $= 1200g \times 45$ | B1 | |
| $WD = 1200g \times 45 + 360\,000$ | M1 | For $WD$ by car's engine $=$ PE gain $+$ WD against resistance |
| Work done is 900 000 J or 900 kJ | A1 | 3 |
| **(ii)** WD against resistance $= 360 \times \sin5/\sin1$ (kJ) or $[360000 \div (45/\sin5°)] \times (45/\sin1°)$ (J) or $697.24... \times 2578.44...$ (J) or 1798 (kJ) | B1 | |
| KE gain $= 1660 + 540 - 1798$ | B1ft | Accept $1660 + 540 - 1800$ |
| $[402000 = \frac{1}{2}1200(v^2 - 225)]$ | M1 | For using KE gain $= \frac{1}{2}m(v^2 - 15^2)$ |
| Speed is 29.9 ms$^{-1}$ | A1 | 4 AG |
| **(iii)** $\frac{P_B}{P_C} = \left(\frac{DF_B}{DF_C}\right) \times \frac{v_B}{v_C} = 1.5 \times 15/29.9$ | M1 | For using $P = Fv$ |
| | A1 | |
| Ratio is 0.75 | A1 | 3 |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{28562a1b-ec9a-40d2-bbb3-729770688971-3_218_1280_1146_431}\\
$A B$ and $B C$ are straight roads inclined at $5 ^ { \circ }$ to the horizontal and $1 ^ { \circ }$ to the horizontal respectively. $A$ and $C$ are at the same horizontal level and $B$ is 45 m above the level of $A$ and $C$ (see diagram, which is not to scale). A car of mass 1200 kg travels from $A$ to $C$ passing through $B$.\\
(i) For the motion from $A$ to $B$, the speed of the car is constant and the work done against the resistance to motion is 360 kJ . Find the work done by the car's engine from $A$ to $B$.
The resistance to motion is constant throughout the whole journey.\\
(ii) For the motion from $B$ to $C$ the work done by the driving force is 1660 kJ . Given that the speed of the car at $B$ is $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, show that its speed at $C$ is $29.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, correct to 3 significant figures.\\
(iii) The car's driving force immediately after leaving $B$ is 1.5 times the driving force immediately before reaching $C$. Find, correct to 2 significant figures, the ratio of the power developed by the car's engine immediately after leaving $B$ to the power developed immediately before reaching $C$.
\hfill \mbox{\textit{CAIE M1 2011 Q6 [10]}}