| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Lift with passenger or load |
| Difficulty | Standard +0.3 This is a standard two-body connected particles problem requiring application of Newton's second law to find acceleration from given forces, then finding tension. The setup is clearly defined with all necessary information provided, requiring straightforward force resolution and F=ma application across two parts, making it 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 |
|---|---|---|
| Working | Marks | Guidance |
| For crate: \(55g - 473 = 55a\) | M1 A1 | M1 for equation in \(a\) only with usual rules; A1 for correct equation |
| \(a = 1.2 \text{ m s}^{-2}\) | A1 (3) | Allow \(-1.2 \text{ m s}^{-2}\) if appropriate |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| For system: \(55g + 200g \pm T - 150 = 255a\) | M1 A2 | M1 for equation in \(T\) and \(a\); A2 (-1 each error); Allow \(\pm T\) |
| Magnitude \(= 2040\) N or \(2000\) N | A1 (4) | |
| OR For lift: \(200g + 473 - 150 \pm T = 200a\) | M1 A2 | \(a\) does not need to be a numerical value |
| Magnitude \(= 2040\) N or \(2000\) N | A1 (4) | N.B. Use mass to identify which part of system is considered |
## Question 4:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| For crate: $55g - 473 = 55a$ | M1 A1 | M1 for equation in $a$ only with usual rules; A1 for correct equation |
| $a = 1.2 \text{ m s}^{-2}$ | A1 (3) | Allow $-1.2 \text{ m s}^{-2}$ if appropriate |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| For system: $55g + 200g \pm T - 150 = 255a$ | M1 A2 | M1 for equation in $T$ and $a$; A2 (-1 each error); Allow $\pm T$ |
| Magnitude $= 2040$ N or $2000$ N | A1 (4) | |
| OR For lift: $200g + 473 - 150 \pm T = 200a$ | M1 A2 | $a$ does not need to be a numerical value |
| Magnitude $= 2040$ N or $2000$ N | A1 (4) | N.B. Use mass to identify which part of system is considered |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{de3245a7-cf6e-423e-8689-9a074bdbc23b-06_428_373_246_788}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A lift of mass 200 kg is being lowered into a mineshaft by a vertical cable attached to the top of the lift. A crate of mass 55 kg is on the floor inside the lift, as shown in Figure 2. The lift descends vertically with constant acceleration. There is a constant upwards resistance of magnitude 150 N on the lift. The crate experiences a constant normal reaction of magnitude 473 N from the floor of the lift.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the lift.
\item Find the magnitude of the force exerted on the lift by the cable.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2015 Q4 [7]}}