| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| 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 on an inclined plane requiring two force equations (one for each body) and Newton's second law. The setup is straightforward with given values, requiring routine application of F=ma with resolution parallel to the plane. Slightly easier than average due to clear diagram, given thrust value, and standard two-part structure with no geometric complications. |
| Spec | 3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2000 - 500 - 500g\sin\alpha = 500a\) (truck) | M1 A2 | Using equation(s) of motion to give an equation in \(a\) only; correct number of terms and \(500g\) resolved; condone sign errors. A1 equation with at most one error; A1 correct equation |
| \(a = 0.256\) or \(0.26\) (m s\(^{-2}\)) (\(\frac{32}{125}\) is A0) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(D - 1200 - 500 - 1500g\sin\alpha - 500g\sin\alpha = 2000a\) (system) | M1 A2 | Using equation of motion to give equation in \(D\) and \(a\) only; correct number of terms and \(500g\) (or \(1500g\)) resolved; condone sign errors. A1 at most one error (treat omission of \(g\) as one error); A1 correct equation |
| OR: \(D - 1200 - 1500g\sin\alpha - 2000 = 1500a\) (engine) | ||
| \(D = 7700\) | A1 | Correct answer |
# Question 6:
## Part 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2000 - 500 - 500g\sin\alpha = 500a$ (truck) | M1 A2 | Using equation(s) of motion to give an equation in $a$ only; correct number of terms and $500g$ resolved; condone sign errors. A1 equation with at most one error; A1 correct equation |
| $a = 0.256$ or $0.26$ (m s$^{-2}$) ($\frac{32}{125}$ is A0) | A1 | Correct answer |
## Part 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $D - 1200 - 500 - 1500g\sin\alpha - 500g\sin\alpha = 2000a$ (system) | M1 A2 | Using equation of motion to give equation in $D$ and $a$ only; correct number of terms and $500g$ (or $1500g$) resolved; condone sign errors. A1 at most one error (treat omission of $g$ as one error); A1 correct equation |
| **OR:** $D - 1200 - 1500g\sin\alpha - 2000 = 1500a$ (engine) | | |
| $D = 7700$ | A1 | Correct answer |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{05cf68a3-1ba4-487f-9edd-48a246f4194f-20_328_1082_127_438}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A railway engine of mass 1500 kg is attached to a railway truck of mass 500 kg by a straight rigid coupling. The engine pushes the truck up a straight track, which is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 7 } { 25 }$. The coupling is parallel to the track and parallel to the direction of motion, as shown in Figure 3.
The engine produces a constant driving force of magnitude $D$ newtons. The engine and the truck experience constant resistances to motion, from non-gravitational forces, of magnitude 1200 N and 500 N respectively.
The thrust in the coupling is 2000 N .
The coupling is modelled as a light rod.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the engine and the truck.
\item Find the value of $D$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2020 Q6 [8]}}