| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find constant speed |
| Difficulty | Standard +0.3 This is a standard M2 power-speed-force question requiring P=Fv and F=ma in part (a), then equilibrium on an incline in part (b). The steps are routine and follow textbook methods with straightforward arithmetic, making it slightly easier than average for M2 content. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks |
|---|---|
| \(F = \frac{10000}{15}\) | B1 |
| Use of "\(F = ma\)": \(F - R = 1800 \times 0.25(= 450)\) with their \(F\) | M1A1 ft |
| \(R = \frac{10000}{15} - 450 = 220\) or better, \(650/3\) oe (216.6...) | A1 (4) |
| Notes: B1 for \((F) = \frac{10000}{15}\) seen or implied; B0 for \(\frac{10}{15}\); M1 for resolving horizontally; First A1 ft for a correct equation with their \(F\); Second A1 for 650/3 oe, 220 or better i.e. 217, 216.7,.. |
| Answer | Marks |
|---|---|
| \(F = \frac{12000}{V}\) | B1 |
| Moving at constant speed: \(D = mg \sin\theta + 30V\) | M1 |
| \(\frac{12000}{V} = 1800g \times \frac{1}{14} + 30V\) | A2 |
| Quadratic in \(V\): \(30V^2 + 1260V - 12000 = 0\) | |
| Solve for \(V\): \(V^2 + 42V - 400 = 0 = (V + 50)(V - 8)\) | M1 |
| \(V = 8\) | A1 (6) [10] |
| Notes: B1 for \((D) = 12000/V\) seen or implied; B0 for \(12/V\); First M1 for resolving parallel to the plane with \(a = 0\) but neither \(D\) nor \(\sin\theta\) need to be substituted; First A1 and Second A1 for an equation in \(V\) only, A1A0 if one error; Second M1 for solving a 3-term quadratic in \(V\) by attempting to factorise or use the formula; this M1 mark can be implied by a correct answer but an incorrect answer scores M0A0; Third A1 for \(V = 8\); N.B. Penalise use of kW ONCE per question |
## 2a
| $F = \frac{10000}{15}$ | B1 |
| Use of "$F = ma$": $F - R = 1800 \times 0.25(= 450)$ with their $F$ | M1A1 ft |
| $R = \frac{10000}{15} - 450 = 220$ or better, $650/3$ oe (216.6...) | A1 (4) |
| **Notes:** B1 for $(F) = \frac{10000}{15}$ seen or implied; B0 for $\frac{10}{15}$; M1 for resolving horizontally; First A1 ft for a correct equation with their $F$; Second A1 for 650/3 oe, 220 or better i.e. 217, 216.7,.. |
## 2b
| $F = \frac{12000}{V}$ | B1 |
| Moving at constant speed: $D = mg \sin\theta + 30V$ | M1 |
| $\frac{12000}{V} = 1800g \times \frac{1}{14} + 30V$ | A2 |
| Quadratic in $V$: $30V^2 + 1260V - 12000 = 0$ | |
| Solve for $V$: $V^2 + 42V - 400 = 0 = (V + 50)(V - 8)$ | M1 |
| $V = 8$ | A1 (6) [10] |
| **Notes:** B1 for $(D) = 12000/V$ seen or implied; B0 for $12/V$; First M1 for resolving parallel to the plane with $a = 0$ but neither $D$ nor $\sin\theta$ need to be substituted; First A1 and Second A1 for an equation in $V$ only, A1A0 if one error; Second M1 for solving a 3-term quadratic in $V$ by attempting to factorise or use the formula; this M1 mark can be implied by a correct answer but an incorrect answer scores M0A0; Third A1 for $V = 8$; N.B. Penalise use of kW ONCE per question |
---2. A truck of mass 1800 kg is moving along a straight horizontal road. The engine of the truck is working at a constant rate of 10 kW . The non-gravitational resistance to motion is modelled as a constant force of magnitude $R$ newtons. At the instant when the truck is moving with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the acceleration of the truck is $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $R$.\\
(4)
The truck now moves up a straight road at a constant speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The road is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 14 }$. The non-gravitational resistance to motion is now modelled as a constant force of magnitude 30 V newtons. The engine of the truck is now working at a constant rate of 12 kW .
\item Find the value of $V$.\\
DO NOT WIRITE IN THIS AREA\\
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{790546bf-38a4-4eb7-876e-941fe58f4a48-05_529_1040_118_450}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2016 Q2 [10]}}