| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| 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 -0.5 This is a straightforward application of Newton's second law to a lift system. Part (a) requires a single F=ma equation for the whole system, and part (b) requires another F=ma equation for box P alone. Both parts involve simple algebraic manipulation with given values, making this slightly easier than average for M1. |
| 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 |
| \(\frac{42mg}{5} - (m+M)g = (m+M)\frac{2g}{5}\) where \(M = (p+q)\) OR \(\frac{42mg}{5} - Mg = M\frac{2g}{5}\) where \(M = p+q+m\) | M1 | Form equation of motion for whole system with combined mass of \(P\) and \(Q\). Correct terms, condone sign errors. May consider 2 masses (\(M\)) and lift separately and eliminate normal reaction. |
| Correct equation in \(M\) and \(m\) for their \(M\) | A1 | N.B. Award marks for correct equation only if no wrong working seen. |
| Rearrange to find expression for combined mass of \(P\) and \(Q\) | dM1 | Must be a multiple of \(m\) |
| \((p+q) = 5m\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{14mg}{5} - pg = p\left(\frac{2g}{5}\right)\) | M1 | Form equation of motion for box \(P\). Correct terms, condone sign errors. |
| Fully correct equation | A1 | cao |
| \(p = 2m\) | A1 | cao |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{42mg}{5} - (m+M)g = (m+M)\frac{2g}{5}$ where $M = (p+q)$ **OR** $\frac{42mg}{5} - Mg = M\frac{2g}{5}$ where $M = p+q+m$ | M1 | Form equation of motion for whole system with combined mass of $P$ and $Q$. Correct terms, condone sign errors. May consider 2 masses ($M$) and lift separately and eliminate normal reaction. |
| Correct equation in $M$ and $m$ for their $M$ | A1 | N.B. Award marks for correct equation only if no wrong working seen. |
| Rearrange to find expression for combined mass of $P$ and $Q$ | dM1 | Must be a multiple of $m$ |
| $(p+q) = 5m$ | A1 | cao |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{14mg}{5} - pg = p\left(\frac{2g}{5}\right)$ | M1 | Form equation of motion for box $P$. Correct terms, condone sign errors. |
| Fully correct equation | A1 | cao |
| $p = 2m$ | A1 | cao |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{84c0eead-0a87-4d87-b33d-794a94bb466c-22_341_316_283_877}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A simple lift operates by means of a vertical cable which is attached to the top of the lift. The lift has mass $m$\\
A box $Q$ is placed on the floor of the lift.\\
A box $P$ is placed directly on top of box $Q$, as shown in Figure 4.\\
The cable is modelled as being light and inextensible and air resistance is modelled as being negligible.\\
The tension in the cable is $\frac { 42 m g } { 5 }$\\
The lift and its contents move vertically upwards with acceleration $\frac { 2 g } { 5 }$\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item find, in terms of $m$, the combined mass of boxes $P$ and $Q$
During the motion of the lift, the force exerted on box $P$ by box $Q$ is $\frac { 14 m g } { 5 }$ Using the model,
\item find, in terms of $m$, the mass of box $P$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2023 Q7 [7]}}