| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Vertically connected particles, air resistance |
| Difficulty | Moderate -0.8 This is a standard two-body connected particles problem with clearly specified forces and straightforward application of Newton's second law. Students need to draw force diagrams, write F=ma equations for each mass, and solve simultaneously—all routine M1 techniques with no conceptual surprises or problem-solving insight required. |
| Spec | 3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Diagram for P or Q | B1 | Must be properly labelled with arrows |
| Other diagram | B1 | Must be properly labelled with arrows consistent with 1st diagram. Accept single diagram if clear. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let tension in rope be \(T\) N and accn \(\uparrow\) \(a \text{ m s}^{-2}\) | M1 | N2L applied correctly to either part. Allow \(F = mga\) and sign errors. Do not condone missing or extra forces. |
| For box P: N2L \(\uparrow\): \(1030 - 75g - T = 75a\) | A1 | |
| For box Q: N2L \(\uparrow\): \(T - 25g = 25a\) | A1 | 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | Solving for \(T\) their simultaneous equations with 2 variables. | |
| tension is \(257.5 \text{ N}\) | A1 | cao CWO |
# Question 4:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram for P or Q | B1 | Must be properly labelled with arrows |
| Other diagram | B1 | Must be properly labelled with arrows consistent with 1st diagram. Accept single diagram if clear. |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let tension in rope be $T$ N and accn $\uparrow$ $a \text{ m s}^{-2}$ | M1 | N2L applied correctly to either part. Allow $F = mga$ and sign errors. Do not condone missing or extra forces. |
| For box P: N2L $\uparrow$: $1030 - 75g - T = 75a$ | A1 | |
| For box Q: N2L $\uparrow$: $T - 25g = 25a$ | A1 | 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)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | Solving for $T$ their simultaneous equations with 2 variables. |
| tension is $257.5 \text{ N}$ | A1 | 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]{6cca1e5e-82b0-487d-8048-b9db7745dea6-3_341_364_210_1489}
\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 2010 Q4 [7]}}