| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Average speed or total distance calculation |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question combining equilibrium of forces (resolving forces/triangle of forces) and projectile motion with SUVAT. Part (i) is routine diagram drawing, (ii) uses standard triangle of forces with given angles, (iii) requires recognizing geometric impossibility, and the projectile parts involve straightforward application of kinematic equations with given initial conditions. All techniques are textbook exercises requiring minimal problem-solving insight, though the multi-part nature and calculation steps place it slightly below average difficulty overall. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Taking whole system: net force \(= \pm(30-F)\), total mass \(=10\) kg | M1 | Newton's second law for system |
| \(30 - F = 10 \times 2\) or \(F - 30 = 10 \times 2\) | A1 | Either equation |
| \(F = 10\) N or \(F = 50\) N | A1 A1 | Both values required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For \(F=10\): consider 6 kg block: \(30-T=6\times2\), so \(T=18\) N | M1 A1 | |
| For \(F=50\): consider 6 kg block: \(30-T=6\times(-2)\), so \(T=42\) N | A1 |
# Question 5:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Taking whole system: net force $= \pm(30-F)$, total mass $=10$ kg | M1 | Newton's second law for system |
| $30 - F = 10 \times 2$ or $F - 30 = 10 \times 2$ | A1 | Either equation |
| $F = 10$ N or $F = 50$ N | A1 A1 | Both values required |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| For $F=10$: consider 6 kg block: $30-T=6\times2$, so $T=18$ N | M1 A1 | |
| For $F=50$: consider 6 kg block: $30-T=6\times(-2)$, so $T=42$ N | A1 | |
---5 A small box B of weight 400 N is held in equilibrium by two light strings AB and BC . The string BC is fixed at C . The end A of string AB is fixed so that AB is at an angle $\alpha$ to the vertical where $\alpha < 60 ^ { \circ }$. String BC is at $60 ^ { \circ }$ to the vertical. This information is shown in Fig. 5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9a79f274-1a3f-4d11-9775-313d82075035-003_424_472_1599_774}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
(i) Draw a labelled diagram showing all the forces acting on the box.\\
(ii) In one situation string AB is fixed so that $\alpha = 30 ^ { \circ }$.
By drawing a triangle of forces, or otherwise, calculate the tension in the string BC and the tension in the string AB .\\
(iii) Show carefully, but briefly, that the box cannot be in equilibrium if $\alpha = 60 ^ { \circ }$ and BC remains at $60 ^ { \circ }$ to the vertical.
7 The trajectory ABCD of a small stone moving with negligible air resistance is shown in Fig. 7. AD is horizontal and BC is parallel to AD .
The stone is projected from A with speed $40 \mathrm {~ms} ^ { - 1 }$ at $50 ^ { \circ }$ to the horizontal.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9a79f274-1a3f-4d11-9775-313d82075035-004_341_1107_484_498}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Write down an expression for the horizontal displacement from A of the stone $t$ seconds after projection. Write down also an expression for the vertical displacement at time $t$.\\
(ii) Show that the stone takes 6.253 seconds (to three decimal places) to travel from A to D . Calculate the range of the stone.
You are given that $X = 30$.\\
(iii) Calculate the time it takes the stone to reach B . Hence determine the time for it to travel from A to C.\\
(iv) Calculate the direction of the motion of the stone at $\mathbf { C }$.
Section B (36 marks)
\hfill \mbox{\textit{OCR MEI M1 Q5}}