| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Maximum speed on horizontal road |
| Difficulty | Standard +0.3 This is a multi-part mechanics question requiring standard application of F=ma, P=Fv relationships, and work-energy principles. While it has 6 parts, each involves routine techniques (finding resistance from power/speed/acceleration, recognizing terminal velocity, calculating time from work/power). The conceptual demand is modest—understanding that acceleration→0 as speed approaches terminal velocity is a standard M1 concept. Slightly above average due to length and the need to coordinate multiple standard techniques across parts. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(DF = 24000/20\) \([DF - R = 1250x0.32]\) \(R = 800\) | B1, M1, A1 | For using Newton's second law (3 terms) |
| (ii) \(24000/29.9 = 800 + 1250a\) Acceleration is \(0.002 \text{ ms}^{-2}\) | B1, B1 | — |
| (iii) \([a = (24000/30 - 800)/1250\) \(24000\div 800 > 0\) → \(v > −30\) or Speed cannot reach \(30\text{ms}^{-1}\) | M1, A1 | For finding a when \(v = 30\) or for using \(a > 0\) to obtain an inequality for \(v\) |
| (iv) \(29.9 ≤ v < 30\) → speed approximately constant | B1 | — |
| (v) \(30\text{ms}^{-1}\) (max error 0.1) or \(29.95\text{ms}^{-1}\) (max error 0.05) or \(29.9\text{ms}^{-1}\) (max error 0.1) | B1 | — |
| (vi)(a) \([24 = 1200/T]\) Time taken is \(50\text{s}\) | M1, A1 | For using \(P = \Delta WD/\Delta t\) |
| (vi)(b) \([s = 30x50\) or \(29.95x50\) or \(29.9x50]\) Distance BC is \(1500 \text{ m}\) or \(1500 \text{ m}\) or \(1495 \text{ m}\) | M1, A1 | For using \(s = vt\) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \([1200 000 = 800d]\) Distance BC is \(1500 \text{ m}\) | M1, A1 | For using 'no change in KE' → WD by car's engine = WD against resistance (may be implied) |
| (a) \([t = 1500/30\) or \(1500/29.95\) or \(1500/29.9]\) Time taken is \(50\text{s}\) or \(50.1\text{s}\) or \(50.2\text{s}\) | M1, A1 | For using \(t = s/v\) |
(i) $DF = 24000/20$ $[DF - R = 1250x0.32]$ $R = 800$ | B1, M1, A1 | For using Newton's second law (3 terms) | [3]
(ii) $24000/29.9 = 800 + 1250a$ Acceleration is $0.002 \text{ ms}^{-2}$ | B1, B1 | — | [2]
(iii) $[a = (24000/30 - 800)/1250$ $24000\div 800 > 0$ → $v > −30$ or Speed cannot reach $30\text{ms}^{-1}$ | M1, A1 | For finding a when $v = 30$ or for using $a > 0$ to obtain an inequality for $v$ | AG | [2]
(iv) $29.9 ≤ v < 30$ → speed approximately constant | B1 | — | [1]
(v) $30\text{ms}^{-1}$ (max error 0.1) or $29.95\text{ms}^{-1}$ (max error 0.05) or $29.9\text{ms}^{-1}$ (max error 0.1) | B1 | — | [1]
(vi)(a) $[24 = 1200/T]$ Time taken is $50\text{s}$ | M1, A1 | For using $P = \Delta WD/\Delta t$ | —
(vi)(b) $[s = 30x50$ or $29.95x50$ or $29.9x50]$ Distance BC is $1500 \text{ m}$ or $1500 \text{ m}$ or $1495 \text{ m}$ | M1, A1 | For using $s = vt$ | [4]
---
## Question 7 — Alternative for Part (vi)
(b) $[1200 000 = 800d]$ Distance BC is $1500 \text{ m}$ | M1, A1 | For using 'no change in KE' → WD by car's engine = WD against resistance (may be implied)
(a) $[t = 1500/30$ or $1500/29.95$ or $1500/29.9]$ Time taken is $50\text{s}$ or $50.1\text{s}$ or $50.2\text{s}$ | M1, A1 | For using $t = s/v$7 A car of mass 1250 kg travels along a horizontal straight road. The power of the car's engine is constant and equal to 24 kW and the resistance to the car's motion is constant and equal to $R \mathrm {~N}$. The car passes through the point $A$ on the road with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and acceleration $0.32 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Find the value of $R$.
The car continues with increasing speed, passing through the point $B$ on the road with speed $29.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car subsequently passes through the point $C$.\\
(ii) Find the acceleration of the car at $B$, giving the answer in $\mathrm { m } \mathrm { s } ^ { - 2 }$ correct to 3 decimal places.\\
(iii) Show that, while the car's speed is increasing, it cannot reach $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iv) Explain why the speed of the car is approximately constant between $B$ and $C$.\\
(v) State a value of the approximately constant speed, and the maximum possible error in this value at any point between $B$ and $C$.
The work done by the car's engine during the motion from $B$ to $C$ is 1200 kJ .\\
(vi) By assuming the speed of the car is constant from $B$ to $C$, find, in either order,
\begin{enumerate}[label=(\alph*)]
\item the approximate time taken for the car to travel from $B$ to $C$,
\item an approximation for the distance $B C$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2010 Q7 [13]}}