| 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 | Vertically connected particles, air resistance |
| Difficulty | Moderate -0.3 This is a standard two-body connected particles problem requiring force diagrams and Newton's second law applied to each mass separately. While it involves multiple steps (drawing diagrams, writing two equations, solving simultaneously), the method is routine and commonly practiced in M1. The setup is straightforward with clearly given values and no conceptual subtleties, making it slightly easier than average. |
| 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 |
|---|---|---|
| Diagram for P or Q; Other diagram | B1, B1 | Must be properly labelled with arrows. Must be properly labelled with arrows consistent with 1st diagram. Accept single diagram if clear. |
| Answer | Marks | Guidance |
|---|---|---|
| For box Q: N2L \(\uparrow\): \(T - 25g = 25a\) | M1, A1, A1 | N2L applied correctly to either part. Allow \(F=mga\) and sign errors. Do not condone missing or extra forces. Direction of \(a\) consistent with equation for P. [Condone taking \(+\)ve downwards in either equation. \(+\)ve direction must be consistent in both equations to receive both A1s] |
| Answer | Marks | Guidance |
|---|---|---|
| Tension is \(257.5\text{ N}\) | M1, A1 | Solving for \(T\) their simultaneous equations with 2 variables. cao CWO. |
## Question 4:
### Part (i)
Diagram for P or Q; Other diagram | B1, B1 | Must be properly labelled with arrows. Must be properly labelled with arrows consistent with 1st diagram. Accept single diagram if clear.
### Part (ii)
Let tension in rope be $T\text{ N}$ and acceleration $\uparrow\; a\text{ m s}^{-2}$
For box P: N2L $\uparrow$: $1030 - 75g - T = 75a$
For box Q: N2L $\uparrow$: $T - 25g = 25a$ | M1, A1, A1 | N2L applied correctly to either part. Allow $F=mga$ and sign errors. Do not condone missing or extra forces. Direction of $a$ consistent with equation for P. [Condone taking $+$ve downwards in either equation. $+$ve direction must be consistent in both equations to receive both A1s]
### Part (iii)
Solving their simultaneous equations with 2 variables.
Tension is $257.5\text{ N}$ | M1, A1 | Solving for $T$ their simultaneous equations with 2 variables. cao CWO.
---4 As shown in Fig. 4, boxes P and Q are descending vertically supported by a parachute. Box P has mass 75 kg . Box Q has mass 25 kg and hangs from box P by means of a light vertical wire. Air resistance on the boxes should be neglected.
At one stage the boxes are slowing in their descent with the parachute exerting an upward vertical force of 1030 N on box P . The acceleration of the boxes is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ upwards and the tension in the wire is $T \mathrm {~N}$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5a1895e1-abe3-4739-876a-f19458f0f6ed-3_332_358_1504_1526}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) Draw a labelled diagram showing all the forces acting on box P and another diagram showing all the forces acting on box Q .\\
(ii) Write down separate equations of motion for box P and for box Q .\\
(iii) Calculate the tension in the wire.
\hfill \mbox{\textit{OCR MEI M1 Q4 [7]}}