| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Train with coupled trucks/carriages |
| Difficulty | Moderate -0.3 This is a straightforward connected particles question requiring application of Newton's second law to the system and individual particles. Part (i) involves simple F=ma for the whole system, while parts (ii) and (iii) require standard techniques of considering the system then individual trucks. The calculations are routine with no conceptual challenges beyond basic M1 content. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P - 800 = 20000 \times 0.2\) | M1 A1 | N2L. Allow \(F = mga\). Allow wrong or zero resistance. No extra forces. Allow sign errors. If done as 1 equation need \(m = 20\,000\). If A and B analysed separately, must have 2 equations with \(T\). |
| \(P = 4800\) | A1 | N2L correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| New accn: \(4800 - 2800 = 20000a\) | M1 | \(F = ma\). Finding new acceleration. No extra forces. Allow 500 N but not 300 N omitted. Allow sign errors. |
| \(a = 0.1\) | A1 | FT their \(P\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T - 2500 = 10000 \times 0.1\) | M1 | N2L with new \(a\). Mass 10000. All forces present for A or B except allow 500 N omitted on A. No extra forces |
| \(T = 3500\) so \(3500\) N | A1 | cao |
## Question 4:
**(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P - 800 = 20000 \times 0.2$ | M1 A1 | N2L. Allow $F = mga$. Allow wrong or zero resistance. No extra forces. Allow sign errors. If done as 1 equation need $m = 20\,000$. If A and B analysed separately, must have 2 equations with $T$. |
| $P = 4800$ | A1 | N2L correct |
**(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| New accn: $4800 - 2800 = 20000a$ | M1 | $F = ma$. Finding new acceleration. No extra forces. Allow 500 N but not 300 N omitted. Allow sign errors. |
| $a = 0.1$ | A1 | FT their $P$ |
**(iii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T - 2500 = 10000 \times 0.1$ | M1 | N2L with new $a$. Mass 10000. All forces present for A or B except allow 500 N omitted on A. No extra forces |
| $T = 3500$ so $3500$ N | A1 | cao |
---4 Two trucks, A and B, each of mass 10000 kg , are pulled along a straight, horizontal track by a constant, horizontal force of $P \mathrm {~N}$. The coupling between the trucks is light and horizontal. This situation and the resistances to motion of the trucks are shown in Fig. 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-2_198_948_1454_592}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
The acceleration of the system is $0.2 \mathrm {~ms} ^ { 2 }$ in the direction of the pulling force of magnitude $P$.\\
(i) Calculate the value of $P$.
Truck A is now subjected to an extra resistive force of 2000 N while $P$ does not change.\\
(ii) Calculate the new acceleration of the trucks.\\
(iii) Calculate the force in the coupling between the trucks.
\hfill \mbox{\textit{OCR MEI M1 Q4 [7]}}