| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find constant speed |
| Difficulty | Moderate -0.3 This is a straightforward application of the power-force-velocity relationship (P=Fv) with standard mechanics principles. Part (i) involves direct substitution and Newton's second law with constant resistance, while part (ii) requires solving a quadratic equation. All steps are routine M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(16000 = F \times 40\) | M1 | Using \(P = Fv\) |
| Resistance is \(400\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(22500 = F \times 45\), \(F = 500\) | B1 | |
| \(500 - 400 = 1200a\) | M1 | Applying Newton's Second Law |
| \(a = \frac{1}{12} = 0.0833 \text{ ms}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(16000 = (590 + 2v)v\) | M1 | Using \(P = Fv\) |
| \([2v^2 + 590v - 16000 = 0] \rightarrow v = \ldots\) | M1 | Solving for \(v\) |
| \(v = 25 \text{ ms}^{-1}\) | A1 |
## Question 6(i)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $16000 = F \times 40$ | M1 | Using $P = Fv$ |
| Resistance is $400$ N | A1 | |
## Question 6(i)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $22500 = F \times 45$, $F = 500$ | B1 | |
| $500 - 400 = 1200a$ | M1 | Applying Newton's Second Law |
| $a = \frac{1}{12} = 0.0833 \text{ ms}^{-2}$ | A1 | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $16000 = (590 + 2v)v$ | M1 | Using $P = Fv$ |
| $[2v^2 + 590v - 16000 = 0] \rightarrow v = \ldots$ | M1 | Solving for $v$ |
| $v = 25 \text{ ms}^{-1}$ | A1 | |
---6 A car of mass 1200 kg is travelling along a horizontal road.\\
(i) It is given that there is a constant resistance to motion.
\begin{enumerate}[label=(\alph*)]
\item The engine of the car is working at 16 kW while the car is travelling at a constant speed of $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the resistance to motion.
\item The power is now increased to 22.5 kW . Find the acceleration of the car at the instant it is travelling at a speed of $45 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) It is given instead that the resistance to motion of the car is $( 590 + 2 v ) \mathrm { N }$ when the speed of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car travels at a constant speed with the engine working at 16 kW . Find this speed.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2017 Q6 [8]}}