| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2021 |
| Session | October |
| Marks | 8 |
| 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 standard M2 power-speed-acceleration problem requiring two applications of F=ma (one for uphill, one for downhill) and the power equation P=Fv. The setup is straightforward with clearly stated conditions, and solving the simultaneous equations is routine. Slightly above average difficulty due to the two-scenario setup and careful sign management, but well within typical M2 expectations. |
| Spec | 3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion down the slope | M1 | First equation (either direction). Condone sign errors and sin/cos confusion |
| \(F_1 + 450g \times \frac{1}{15} - R = 450 \times 0.5\) | A1 | Unsimplified equation with at most one error |
| \(\left(\frac{P}{12} + 30g - R = 225\right)\) \(\left(\frac{P}{12} - R = -69\right)\) | A1 | Correct unsimplified equation in \(P\) or \(F_1\) |
| Equation of motion up the slope | M1 | Second equation. Condone sin/cos confusion. Signs consistent with first equation and change in direction of motion |
| \(F_2 - 450g \times \frac{1}{15} - R = 450 \times -0.5\) | A1 | Correct unsimplified equation in \(P\) or \(F_2\) |
| \(\left(\frac{P}{6} - 30g - R = -225\right)\) \(\left(\frac{P}{6} - R = 69\right)\) | ||
| \(F_1 = \frac{P}{12}\) or \(F_2 = \frac{P}{6}\left(= \frac{2P}{12}\right)\) | M1 | Use of \(P = Fv\) at least once |
| Solve for \(P\) | DM1 | Dependent on all previous M marks |
| \(\left(R = \frac{P}{8}\right)\) \(P = 1660\) or \(P = 1700\) | A1 | 3 sf or 2 sf (follows use of 9.8). Allow 1.66 kW but not 1.66 |
| Total: (8) [8] |
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion down the slope | M1 | First equation (either direction). Condone sign errors and sin/cos confusion |
| $F_1 + 450g \times \frac{1}{15} - R = 450 \times 0.5$ | A1 | Unsimplified equation with at most one error |
| $\left(\frac{P}{12} + 30g - R = 225\right)$ $\left(\frac{P}{12} - R = -69\right)$ | A1 | Correct unsimplified equation in $P$ or $F_1$ |
| Equation of motion up the slope | M1 | Second equation. Condone sin/cos confusion. Signs consistent with first equation and change in direction of motion |
| $F_2 - 450g \times \frac{1}{15} - R = 450 \times -0.5$ | A1 | Correct unsimplified equation in $P$ or $F_2$ |
| $\left(\frac{P}{6} - 30g - R = -225\right)$ $\left(\frac{P}{6} - R = 69\right)$ | | |
| $F_1 = \frac{P}{12}$ or $F_2 = \frac{P}{6}\left(= \frac{2P}{12}\right)$ | M1 | Use of $P = Fv$ at least once |
| Solve for $P$ | DM1 | Dependent on all previous M marks |
| $\left(R = \frac{P}{8}\right)$ $P = 1660$ or $P = 1700$ | A1 | 3 sf or 2 sf (follows use of 9.8). Allow 1.66 kW but not 1.66 |
| **Total: (8) [8]** | | |
---2. A vehicle of mass 450 kg is moving on a straight road that is inclined at angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 15 }$
At the instant when the vehicle is moving down the road at $12 \mathrm {~ms} ^ { - 1 }$
\begin{itemize}
\item the engine of the vehicle is working at a rate of $P$ watts
\item the acceleration of the vehicle is $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
\item the resistance to the motion of the vehicle is modelled as a constant force of magnitude $R$ newtons
\end{itemize}
At the instant when the vehicle is moving up the road at $12 \mathrm {~ms} ^ { - 1 }$
\begin{itemize}
\item the engine of the vehicle is working at a rate of $2 P$ watts
\item the deceleration of the vehicle is $0.5 \mathrm {~ms} ^ { - 2 }$
\item the resistance to the motion of the vehicle from non-gravitational forces is modelled as a constant force of magnitude $R$ newtons
\end{itemize}
Find the value of $P$.\\
\hfill \mbox{\textit{Edexcel M2 2021 Q2 [8]}}