| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Car towing trailer, inclined road |
| Difficulty | Standard +0.3 This is a standard M1 connected particles problem requiring application of Newton's second law to a two-body system on an incline. Part (a) involves setting up F=ma for the whole system (straightforward algebra), part (b) requires considering the trailer alone (routine application), and part (c) uses constant acceleration equations. All steps are textbook-standard with no novel insight required, making it slightly easier than average for M1. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Whole system: \(3000 - 1200g\sin\alpha - 600g\sin\alpha - 2R - R = 1800(0.75)\) | M1, A1, A1 | Equation of motion for whole system (or car AND trailer with \(T\) eliminated) giving equation in \(R\) only. Correct number of terms, condone sign errors and sin/cos confusion |
| \(R = 60\) N | A1* cso | Reach given answer with at least one intermediate line of working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Trailer: \(T - 600g\sin\alpha - 60 = 600(0.75)\) OR Car: \(3000 - 1200g\sin\alpha - 2(60) - T = 1200(0.75)\) | M1, A1 | Equation of motion for trailer or car. \(\sin\alpha\) need not be substituted but \(R=60\) does. (\(T\) could be replaced by \(-T\) leading to \(T=-1000\), so tension is 1000) |
| \(T = 1000\) N | A1 | Correct answer for \(T\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-60 - 600g\sin\alpha = 600a\) (or \(-600a\)) | M1, A1 | Form equation of motion for trailer to find new acceleration. \(\sin\alpha\) need not be substituted but \(R=60\) does |
| \(a = -\frac{11}{12} = -0.9166\ldots\) | ||
| \(0 = 12^2 + 2\left(-\frac{11}{12}\right)d\) | M1 | Complete method with calculated acceleration (not \(g\)) to find distance |
| \(d = 78.5, 79\) m | A1 | cao 2 or 3sf. Must be positive. N.B. Allow negative value of \(d\) made positive for distance |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Whole system: $3000 - 1200g\sin\alpha - 600g\sin\alpha - 2R - R = 1800(0.75)$ | M1, A1, A1 | Equation of motion for whole system (or car AND trailer with $T$ eliminated) giving equation in $R$ only. Correct number of terms, condone sign errors and sin/cos confusion |
| $R = 60$ N | A1* cso | Reach given answer with at least one intermediate line of working |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Trailer: $T - 600g\sin\alpha - 60 = 600(0.75)$ **OR** Car: $3000 - 1200g\sin\alpha - 2(60) - T = 1200(0.75)$ | M1, A1 | Equation of motion for trailer or car. $\sin\alpha$ need not be substituted but $R=60$ does. ($T$ could be replaced by $-T$ leading to $T=-1000$, so tension is 1000) |
| $T = 1000$ N | A1 | Correct answer for $T$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-60 - 600g\sin\alpha = 600a$ (or $-600a$) | M1, A1 | Form equation of motion for trailer to find new acceleration. $\sin\alpha$ need not be substituted but $R=60$ does |
| $a = -\frac{11}{12} = -0.9166\ldots$ | | |
| $0 = 12^2 + 2\left(-\frac{11}{12}\right)d$ | M1 | Complete method with calculated acceleration (not $g$) to find distance |
| $d = 78.5, 79$ m | A1 | cao 2 or 3sf. Must be positive. N.B. Allow negative value of $d$ made positive for distance |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2737a11-4a15-41e9-9f87-31a705a8948b-18_326_1107_246_479}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A car of mass 1200 kg is towing a trailer of mass 600 kg up a straight road, as shown in Figure 4.
The road is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 12 }$\\
The driving force produced by the engine of the car is 3000 N .\\
The car moves with acceleration $0.75 \mathrm {~m} \mathrm {~s} ^ { - 2 }$\\
The non-gravitational resistance to motion of
\begin{itemize}
\item the car is modelled as a constant force of magnitude $2 R$ newtons
\item the trailer is modelled as a constant force of magnitude $R$ newtons
\end{itemize}
The car and the trailer are modelled as particles.\\
The tow bar between the car and trailer is modelled as a light rod that is parallel to the direction of motion.
Using the model,
\begin{enumerate}[label=(\alph*)]
\item show that the value of $R$ is 60
\item find the tension in the tow bar.
When the car and trailer are moving at a speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the tow bar breaks.\\
Given that the non-gravitational resistance to motion of the trailer remains unchanged,
\item use the model to find the further distance moved by the trailer before it first comes to rest.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2023 Q7 [11]}}