| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Lift with passenger or load |
| Difficulty | Moderate -0.3 Question 6(i-ii) involves standard kinematics: setting v=0 to find stationary points (solving a quadratic) and integrating velocity to find displacement with an initial condition. These are routine M1 techniques requiring straightforward application of formulas. Question 7 is a standard statics problem with forces in equilibrium, requiring resolution of forces and basic trigonometry. Both are typical textbook exercises with no novel problem-solving required, making them slightly easier than average A-level mechanics questions. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 0\): \(24 - 18t + 3t^2 = 0 \Rightarrow t^2 - 6t + 8 = 0\) | M1 | |
| \((t-2)(t-4)=0 \Rightarrow T_1 = 2, T_2 = 4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \int v \, dt = 24t - 9t^2 + t^3 + c\) | M1 | |
| \(x=0\) when \(t=0 \Rightarrow c=0\), so \(x = 24t - 9t^2 + t^3\) | A1 | |
| When \(t = T_1 = 2\): \(x = 48 - 36 + 8 = 20\) ✓ | A1 | Shown |
| When \(t = T_2 = 4\): \(x = 96 - 144 + 64 = 16\) | A1 |
# Question 6:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 0$: $24 - 18t + 3t^2 = 0 \Rightarrow t^2 - 6t + 8 = 0$ | M1 | |
| $(t-2)(t-4)=0 \Rightarrow T_1 = 2, T_2 = 4$ | A1 | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \int v \, dt = 24t - 9t^2 + t^3 + c$ | M1 | |
| $x=0$ when $t=0 \Rightarrow c=0$, so $x = 24t - 9t^2 + t^3$ | A1 | |
| When $t = T_1 = 2$: $x = 48 - 36 + 8 = 20$ ✓ | A1 | Shown |
| When $t = T_2 = 4$: $x = 96 - 144 + 64 = 16$ | A1 | |
---6 A particle moves along a straight line through an origin O . Initially the particle is at O .\\
At time $t \mathrm {~s}$, its displacement from O is $x \mathrm {~m}$ and its velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, is given by
$$v = 24 - 18 t + 3 t ^ { 2 } .$$
(i) Find the times, $T _ { 1 } \mathrm {~s}$ and $T _ { 2 } \mathrm {~s}$ (where $T _ { 2 } > T _ { 1 }$ ), at which the particle is stationary.\\
(ii) Find an expression for $x$ at time $t \mathrm {~s}$.
Show that when $t = T _ { 1 } , x = 20$ and find the value of $x$ when $t = T _ { 2 }$.
Section B (36 marks)\\
$7 \quad$ Abi and Bob are standing on the ground and are trying to raise a small object of weight 250 N to the top of a building. They are using long light ropes. Fig. 7.1 shows the initial situation.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{83e69140-4abf-4713-85da-922ce7530e47-4_773_1071_429_497}
\captionsetup{labelformat=empty}
\caption{Fig. 7.1}
\end{center}
\end{figure}
Abi pulls vertically downwards on the rope A with a force $F$ N. This rope passes over a small smooth pulley and is then connected to the object. Bob pulls on another rope, B, in order to keep the object away from the side of the building.
In this situation, the object is stationary and in equilibrium. The tension in rope B, which is horizontal, is 25 N . The pulley is 30 m above the object. The part of rope A between the pulley and the object makes an angle $\theta$ with the vertical.\\
\hfill \mbox{\textit{OCR MEI M1 2013 Q6 [6]}}