| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| 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-power question requiring straightforward application of P=Fv, F=ma, and constant acceleration equations. Part (i) uses power formula to find driving force then Newton's second law; part (ii) is a simple 'show that' using equilibrium at constant speed; part (iii) requires combining two motion phases with basic kinematics. All techniques are routine for M1 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| DF = 17280/12 (= 1440 N) | B1 | |
| \([DF - R = ma \Rightarrow 1440 - 960 = 1200a]\) | M1 | For using Newton's 2nd law |
| Acceleration is 0.4 ms\(^{-2}\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \([17280/V - 960 = 0]\) | M1 | For using P/v - R = 0 |
| \(V = 18\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| For BC, -960 = 1200a (a = -0.8) | B1 | |
| M1 | For using \(0 = 18 + at\) and \(0 = 18^2 + 2as\) for BC | |
| \(t_{BC} = (0 - 18)(-0.8)\) and \(s_{BC} = (0 - 18^2)(-1.6)\) (= 22.5 s and 202.5 m) | A1 | |
| Distance AB = 18(52.5 - 22.5) | B1 | |
| Distance is AC is 742.5 m | A1 | 5 |
**(i)**
DF = 17280/12 (= 1440 N) | B1 |
$[DF - R = ma \Rightarrow 1440 - 960 = 1200a]$ | M1 | For using Newton's 2nd law
Acceleration is 0.4 ms$^{-2}$ | A1 | 3
**(ii)**
$[17280/V - 960 = 0]$ | M1 | For using P/v - R = 0
$V = 18$ | A1 | 2 | AG
**(iii)**
For BC, -960 = 1200a (a = -0.8) | B1 |
| M1 | For using $0 = 18 + at$ and $0 = 18^2 + 2as$ for BC
$t_{BC} = (0 - 18)(-0.8)$ and $s_{BC} = (0 - 18^2)(-1.6)$ (= 22.5 s and 202.5 m) | A1 |
Distance AB = 18(52.5 - 22.5) | B1 |
Distance is AC is 742.5 m | A1 | 5 | Accept 742 or 7437 A car of mass 1200 kg moves in a straight line along horizontal ground. The resistance to motion of the car is constant and has magnitude 960 N . The car's engine works at a rate of 17280 W .\\
(i) Calculate the acceleration of the car at an instant when its speed is $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
The car passes through the points $A$ and $B$. While the car is moving between $A$ and $B$ it has constant speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Show that $V = 18$.
At the instant that the car reaches $B$ the engine is switched off and subsequently provides no energy. The car continues along the straight line until it comes to rest at the point $C$. The time taken for the car to travel from $A$ to $C$ is 52.5 s .\\
(iii) Find the distance $A C$.
\hfill \mbox{\textit{CAIE M1 2012 Q7 [10]}}