| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find power and resistance simultaneously |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring calculation of average acceleration from tabulated data, then applying F=ma and P=Fv with constant power and resistance. The question guides students through the method and involves standard M1 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.02l Power and velocity: P = Fv |
| \(t\) | 0 | 0.5 | 16.3 | 24.5 |
| \(v\) | 4 | 6 | 19 | 21 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(4 < v < 6\), \(a_{\text{ave}} = (6-4)/(0.5-0)\); when \(19 < v < 21\), \(a_{\text{ave}} = (21-19)/(24.5-16.3)\) | M1 | For using \(a \approx \frac{\Delta v}{\Delta t}\) |
| Average accelerations are \(4\ \text{ms}^{-2}\) and \(0.244\ \text{ms}^{-2}\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(DF(5) = P/5\) and \(DF(20) = P/20\) | B1 | |
| \([DF - R = ma]\) | M1 | For using Newton's 2nd law |
| \(P/5 - R = 1230 \times 4\) and \(P/20 - R = 1230 \times 0.244\) | A1ft | ft incorrect average values |
| \(P = 30800\) (or \(R = 1240\)) | B1 | |
| \(R = 1240\) (or \(P = 30800\)) | B1ft [5] | ft \(P/5 - 1230a_1\) or \(P/20 - 1230a_2\) or \(5(1230a_1 + R)\) or \(20(1230a_2 + R)\) |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $4 < v < 6$, $a_{\text{ave}} = (6-4)/(0.5-0)$; when $19 < v < 21$, $a_{\text{ave}} = (21-19)/(24.5-16.3)$ | M1 | For using $a \approx \frac{\Delta v}{\Delta t}$ |
| Average accelerations are $4\ \text{ms}^{-2}$ and $0.244\ \text{ms}^{-2}$ | A1 [2] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $DF(5) = P/5$ and $DF(20) = P/20$ | B1 | |
| $[DF - R = ma]$ | M1 | For using Newton's 2nd law |
| $P/5 - R = 1230 \times 4$ and $P/20 - R = 1230 \times 0.244$ | A1ft | ft incorrect average values |
| $P = 30800$ (or $R = 1240$) | B1 | |
| $R = 1240$ (or $P = 30800$) | B1ft [5] | ft $P/5 - 1230a_1$ or $P/20 - 1230a_2$ or $5(1230a_1 + R)$ or $20(1230a_2 + R)$ |
---4 A car of mass 1230 kg increases its speed from $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $21 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in 24.5 s . The table below shows corresponding values of time $t \mathrm {~s}$ and speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$t$ & 0 & 0.5 & 16.3 & 24.5 \\
\hline
$v$ & 4 & 6 & 19 & 21 \\
\hline
\end{tabular}
\end{center}
(i) Using the values in the table, find the average acceleration of the car for $0 < t < 0.5$ and for $16.3 < t < 24.5$.
While the car is increasing its speed the power output of its engine is constant and equal to $P \mathrm {~W}$, and the resistance to the car's motion is constant and equal to $R \mathrm {~N}$.\\
(ii) Assuming that the values obtained in part (i) are approximately equal to the accelerations at $v = 5$ and at $v = 20$, find approximations for $P$ and $R$.
\hfill \mbox{\textit{CAIE M1 2012 Q4 [7]}}