| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Engine power on road constant/variable speed |
| Difficulty | Standard +0.3 This is a standard Further Mechanics 1 power-resistance problem requiring application of P=Fv to find driving force, then resolving forces on the trailer to find tension. It involves multiple steps (finding driving force, considering weight components, resistances) but uses routine FM1 techniques without requiring novel insight. Slightly easier than average A-level due to straightforward setup and clear method. |
| Spec | 3.03d Newton's second law: 2D vectors3.03o Advanced connected particles: and pulleys6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion for system or van | M1 | Need all terms, no extras; dimensionally correct; condone sign errors and sin/cos confusion; must have non-zero acceleration and include driving force |
| \(F-(100+200)-(150+600)g\sin\alpha=(150+600)a\) | A1 | Unsimplified equation in \(F\) (and \(T\) if relevant) with at most one error |
| or \(F-200-T-600g\sin\alpha=600a\) | A1 | Correct unsimplified equation in \(F\) (and \(T\) if relevant) |
| Equation of motion for trailer | M1 | Need all terms; dimensionally correct; condone sign errors and sin/cos confusion; or a second equation of motion involving driving force |
| \(T-100-150g\sin\alpha=150a\) | A1 | Correct unsimplified equation in \(T\) and/or \(F\) |
| Use of \(F=\frac{12000}{9}\) | M1 | Use of \(P=Fv\) seen or implied |
| Solve for \(T\) | M1 | Complete method to find \(T\) (FYI: \(a=0.72(4)\)) |
| \(T=307\ (310)\ \text{(N)}\) | A1 | Tension correct to 3 sf or 2 sf; fractional answer \(\frac{920}{3}\) not acceptable as follows from \(g=9.8\) |
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion for system or van | M1 | Need all terms, no extras; dimensionally correct; condone sign errors and sin/cos confusion; must have non-zero acceleration and include driving force |
| $F-(100+200)-(150+600)g\sin\alpha=(150+600)a$ | A1 | Unsimplified equation in $F$ (and $T$ if relevant) with at most one error |
| or $F-200-T-600g\sin\alpha=600a$ | A1 | Correct unsimplified equation in $F$ (and $T$ if relevant) |
| Equation of motion for trailer | M1 | Need all terms; dimensionally correct; condone sign errors and sin/cos confusion; or a second equation of motion involving driving force |
| $T-100-150g\sin\alpha=150a$ | A1 | Correct unsimplified equation in $T$ and/or $F$ |
| Use of $F=\frac{12000}{9}$ | M1 | Use of $P=Fv$ seen or implied |
| Solve for $T$ | M1 | Complete method to find $T$ (FYI: $a=0.72(4)$) |
| $T=307\ (310)\ \text{(N)}$ | A1 | Tension correct to 3 sf or 2 sf; fractional answer $\frac{920}{3}$ not acceptable as follows from $g=9.8$ |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{86a37170-046f-46e5-9c8c-06d5f98ca4fe-06_287_846_246_612}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A van of mass 600 kg is moving up a straight road which is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 15 }$. The van is towing a trailer of mass 150 kg . The van is attached to the trailer by a towbar which is parallel to the direction of motion of the van and the trailer, as shown in Figure 1.
The resistance to the motion of the van from non-gravitational forces is modelled as a constant force of magnitude 200 N .\\
The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 100 N .
The towbar is modelled as a light rod.\\
The engine of the van is working at a constant rate of 12 kW .\\
Find the tension in the towbar at the instant when the speed of the van is $9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
\hfill \mbox{\textit{Edexcel FM1 2022 Q2 [8]}}