| 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 | Maximum speed on horizontal road |
| Difficulty | Standard +0.3 This is a standard M1 work-energy question requiring application of P=Fv for constant speed, work-energy principle for deceleration, and integration of power equation for variable speed. All three parts use routine mechanics techniques with straightforward calculations, making it slightly easier than average but requiring multiple connected steps. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(DF = 30000/v\) or \(WD\) by \(DF = 30000 \times 100\) | B1 | |
| \(DF = R = 750\) \((v = 40)\) or \(WD\) by \(DF = WD\) by \(R = 750 \times AB\) | B1 | |
| Distance AB is 4000 m | B1 | 3 |
| (ii) \(-750 = 600a\) \((a = -1.25)\) | B1 | |
| \(20^2 = 40^2 + 2(-1.25)BC\) | M1 | For using \(v^2 = u^2 + 2as\) |
| Distance BC is 480 m | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For using 'Loss of energy = WD against resistance' | |
| \(\frac{1}{2} \times 600(40^2 - 20^2) = 750(BC)\) | A1 | |
| Distance BC is 480 m | A1 | |
| (iii) \(WD\) by engine \(= 30000 \times 14\) | B1 | |
| Gain in KE \(= \frac{1}{2} \times 600(30^2 - 20^2)\) | B1 | |
| \([750 \times CD = 420\,000 - 150\,000]\) | M1 | For using \(750 \times CD = WD\) by engine \(-\) gain in KE |
| Distance CD is 360 m | A1 | 4 |
(i) $DF = 30000/v$ or $WD$ by $DF = 30000 \times 100$ | B1 |
$DF = R = 750$ $(v = 40)$ or $WD$ by $DF = WD$ by $R = 750 \times AB$ | B1 |
Distance AB is 4000 m | B1 | 3
(ii) $-750 = 600a$ $(a = -1.25)$ | B1 |
$20^2 = 40^2 + 2(-1.25)BC$ | M1 | For using $v^2 = u^2 + 2as$
Distance BC is 480 m | A1 | 3
### Alternative for (ii):
| M1 | For using 'Loss of energy = WD against resistance'
$\frac{1}{2} \times 600(40^2 - 20^2) = 750(BC)$ | A1 |
Distance BC is 480 m | A1 |
(iii) $WD$ by engine $= 30000 \times 14$ | B1 |
Gain in KE $= \frac{1}{2} \times 600(30^2 - 20^2)$ | B1 |
$[750 \times CD = 420\,000 - 150\,000]$ | M1 | For using $750 \times CD = WD$ by engine $-$ gain in KE
Distance CD is 360 m | A1 | 47 A car of mass 600 kg travels along a straight horizontal road starting from a point $A$. The resistance to motion of the car is 750 N .\\
(i) The car travels from $A$ to $B$ at constant speed in 100 s . The power supplied by the car's engine is constant and equal to 30 kW . Find the distance $A B$.\\
(ii) The car's engine is switched off at $B$ and the car's speed decreases until the car reaches $C$ with a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the distance $B C$.\\
(iii) The car's engine is switched on at $C$ and the power it supplies is constant and equal to 30 kW . The car takes 14 s to travel from $C$ to $D$ and reaches $D$ with a speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the distance $C D$.
\hfill \mbox{\textit{CAIE M1 2011 Q7 [10]}}