| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Lift with passenger or load |
| Difficulty | Moderate -0.3 This is a straightforward two-part connected particles problem requiring standard application of F=ma to find acceleration from the parcel, then using that result to find the reaction force on the man. The setup is clear, the method is routine (resolve forces, apply Newton's second law twice), and no novel insight is required—slightly easier than average due to its direct approach. |
| Spec | 3.03d Newton's second law: 2D vectors3.03f Weight: W=mg3.03i Normal reaction force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For the parcel | M1 | Applying N2L to parcel. Correct mass. Allow \(F=mga\). Condone missing force but do not allow spurious forces |
| \(\uparrow\) N2L \(55 - 5g = 5a\) | A1 | Allow only sign error(s) |
| \(a = 1.2\) so \(1.2\ \text{m s}^{-2}\) | A1 | Allow \(-1.2\) only if sign convention is clear |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R - 80g = 80\times1.2\) or \(R - 75g - 55 = 75\times1.2\) | M1 | N2L. Must have correct mass. Allow only sign errors. FT their \(a\) |
| \(R = 880\) so \(880\) N | A1 | cao. [NB beware spurious methods giving 880 N] |
| 2 |
# Question 5:
**(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| For the parcel | M1 | Applying N2L to parcel. Correct mass. Allow $F=mga$. Condone missing force but do not allow spurious forces |
| $\uparrow$ N2L $55 - 5g = 5a$ | A1 | Allow only sign error(s) |
| $a = 1.2$ so $1.2\ \text{m s}^{-2}$ | A1 | Allow $-1.2$ only if sign convention is clear |
| | **3** | |
**(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R - 80g = 80\times1.2$ or $R - 75g - 55 = 75\times1.2$ | M1 | N2L. Must have correct mass. Allow only sign errors. FT their $a$ |
| $R = 880$ so $880$ N | A1 | cao. [NB beware spurious methods giving 880 N] |
| | **2** | |
---5 A man of mass 75 kg is standing in a lift. He is holding a parcel of mass 5 kg by means of a light inextensible string, as shown in Fig. 5. The tension in the string is 55 N .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{93a5d409-ade4-418b-9c09-620d97df97de-3_456_476_833_833}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
(i) Find the upward acceleration.\\
(ii) Find the reaction on the man of the lift floor.
\hfill \mbox{\textit{OCR MEI M1 2009 Q5 [5]}}