| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Lift with passenger or load |
| Difficulty | Moderate -0.8 This is a straightforward application of Newton's second law to two connected bodies with given acceleration. Students simply write F=ma twice (once for person, once for lift), then solve two simple linear equations. No problem-solving insight required—purely routine mechanics with clear setup and standard method. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(560 - mg = 1.4m\) | M1 A1 | Equation of motion for person only; at least one value (560 or 1.4) substituted. Credit only if appears in (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2800 - Mg - 560 = 1.4M\) | M1 A1 | Equation of motion for lift only; at least one value (2800, 560 or 1.4) substituted. Credit only if appears in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(560 = 11.2m\) | DM1 | Dependent on appropriate previous M mark; solving one of their equations |
| \(m = 50\) | A1 | |
| \(2240 = 11.2M\) | ||
| \(M = 200\) | A1 |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $560 - mg = 1.4m$ | M1 A1 | Equation of motion for person only; at least one value (560 or 1.4) substituted. Credit only if appears in (a) |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2800 - Mg - 560 = 1.4M$ | M1 A1 | Equation of motion for lift only; at least one value (2800, 560 or 1.4) substituted. Credit only if appears in (b) |
## Part (c)(i) and (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $560 = 11.2m$ | DM1 | Dependent on appropriate previous M mark; solving one of their equations |
| $m = 50$ | A1 | |
| $2240 = 11.2M$ | | |
| $M = 200$ | A1 | |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0d5a56ba-6a33-4dc8-b612-d2957211124f-08_387_204_251_872}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A lift of mass $M \mathrm {~kg}$ is being raised by a vertical cable attached to the top of the lift. A person of mass $m \mathrm {~kg}$ stands on the floor inside the lift, as shown in Figure 1. The lift ascends vertically with constant acceleration $1.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The tension in the cable is 2800 N and the person experiences a constant normal reaction of magnitude 560 N from the floor of the lift. The cable is modelled as being light and inextensible, the person is modelled as a particle and air resistance is negligible.
\begin{enumerate}[label=(\alph*)]
\item Write down an equation of motion for the person only.
\item Write down an equation of motion for the lift only.
\item Hence, or otherwise, find
\begin{enumerate}[label=(\roman*)]
\item the value of $m$,
\item the value of $M$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2019 Q3 [7]}}