| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| 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 a lift problem with two parts requiring F=ma calculations. The setup is standard (lift with occupant), the acceleration is constant, and students simply need to apply F=ma to the system (part a) and to the block alone (part b). No problem-solving insight required, just routine application of a well-practiced technique. |
| Spec | 3.03d Newton's second law: 2D vectors3.03f Weight: W=mg |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Translate situation into model and set up equation of motion for cage and block | M1 | Correct number of terms, condone sign errors |
| \(T - 40g - 10g = 50 \times 0.2\) | A1 | Correct equation in \(T\) only |
| \(T = 500 \text{ (N)}\), must be positive | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Set up equation of motion for the block | M1 | Correct number of terms, condone sign errors |
| \(R - 10g = 10 \times 0.2\) (allow \(-R\) instead of \(R\)) | A1 | Correct equation in \(R\) only |
| \(R = 100 \text{ (N)}\), must be positive | A1 | cao |
| OR equation of motion for the cage: \(T - 40g - R = 40 \times 0.2\) with \(T\) substituted | M1 | — |
| \(T - 40g - R = 40 \times 0.2\) | A1 | Correct equation in \(R\) only |
| \(R = 100 \text{ (N)}\), must be positive | A1 | cao |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Translate situation into model and set up equation of motion for cage and block | M1 | Correct number of terms, condone sign errors |
| $T - 40g - 10g = 50 \times 0.2$ | A1 | Correct equation in $T$ only |
| $T = 500 \text{ (N)}$, must be positive | A1 | cao |
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## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Set up equation of motion for the block | M1 | Correct number of terms, condone sign errors |
| $R - 10g = 10 \times 0.2$ (allow $-R$ instead of $R$) | A1 | Correct equation in $R$ only |
| $R = 100 \text{ (N)}$, must be positive | A1 | cao |
| **OR** equation of motion for the cage: $T - 40g - R = 40 \times 0.2$ with $T$ substituted | M1 | — |
| $T - 40g - R = 40 \times 0.2$ | A1 | Correct equation in $R$ only |
| $R = 100 \text{ (N)}$, must be positive | A1 | cao |
> **Note:** Only penalise incorrect value of $g$ once for whole question; max (a) M1A1A0 (b) M1A1A14.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{278a61e9-e27f-4fd5-895a-db01886aadfe-14_545_314_248_877}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A vertical rope $P Q$ has its end $Q$ attached to the top of a small lift cage.\\
The lift cage has mass 40 kg and carries a block of mass 10 kg , as shown in Figure 1.\\
The lift cage is raised vertically by moving the end $P$ of the rope vertically upwards with constant acceleration $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
The rope is modelled as being light and inextensible and air resistance is ignored.\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item find the tension in the rope $P Q$
\item find the magnitude of the force exerted on the block by the lift cage.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 2 2022 Q4 [6]}}