| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Average speed or total distance calculation |
| Difficulty | Moderate -0.3 This is a multi-part question combining standard calculus (finding maximum velocity by differentiation, integrating to find distance) with routine mechanics (pulley system with tension and forces). Part (b) requires showing that dv/dt = 0 at maximum, part (c) is straightforward integration of a cubic, and the mechanics parts involve standard applications of F=ma and energy conservation. While multi-step, each component uses well-practiced techniques without requiring novel insight, making it slightly easier than average. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration3.03b Newton's first law: equilibrium3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0\ (\text{m s}^{-2})\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v=t(-t^2+11t-24)\) \(\Rightarrow v=-t^3+11t^2-24t\) | B1 | Expand and simplify \(v\) correctly |
| \(\dfrac{dv}{dt}=-3t^2+22t-24\) | M1 | Differentiate their cubic expression for \(v\) correctly |
| \(3t^2-22t+24=0\) | M1 | Sets their three-term quadratic in \(t\) equal to zero |
| \((3t-4)(t-6)=0\) | M1 | Factorises (oe) their three-term quadratic in \(t\); Condone this factorisation for \(-3t^2+22t-24=0\) |
| From sketch \(T>3\) therefore \(T=6\) | A1 | Correct value of \(T\) with reason for why \(T\neq\dfrac{4}{3}\); Any working used to determine the required value of \(T\) must be accurate |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^3 v\,dt\) and \(\int_3^T v\,dt\) | M1 | Need to see attempt at integrals but may be BC; Where \(v\) is a cubic expression |
| \((-){\dfrac{117}{4}}\) and \(\dfrac{261}{4}\) | A1 | |
| Total distance \(= \dfrac{117}{4}+\dfrac{261}{4}=94.5\) (m) | A1 | cao |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0\ (\text{m s}^{-2})$ | B1 | |
---
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v=t(-t^2+11t-24)$ $\Rightarrow v=-t^3+11t^2-24t$ | B1 | Expand and simplify $v$ correctly |
| $\dfrac{dv}{dt}=-3t^2+22t-24$ | M1 | Differentiate their cubic expression for $v$ correctly |
| $3t^2-22t+24=0$ | M1 | Sets their three-term quadratic in $t$ equal to zero |
| $(3t-4)(t-6)=0$ | M1 | Factorises (oe) their three-term quadratic in $t$; Condone this factorisation for $-3t^2+22t-24=0$ |
| From sketch $T>3$ therefore $T=6$ | A1 | Correct value of $T$ with reason for why $T\neq\dfrac{4}{3}$; Any working used to determine the required value of $T$ must be accurate |
---
## Question 11(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^3 v\,dt$ and $\int_3^T v\,dt$ | M1 | Need to see attempt at integrals but may be **BC**; Where $v$ is a cubic expression |
| $(-){\dfrac{117}{4}}$ and $\dfrac{261}{4}$ | A1 | |
| Total distance $= \dfrac{117}{4}+\dfrac{261}{4}=94.5$ (m) | A1 | cao |11\\
\includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-08_586_672_1231_242}
A particle $P$ moves along the $x$-axis. At time $t$ seconds, where $t \geqslant 0$, the velocity of $P$ in the positive $x$-direction is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It is given that $v = t ( t - 3 ) ( 8 - t )$.\\
$P$ attains its maximum velocity at time $T$ seconds. The diagram shows part of the velocity-time graph for the motion of $P$.
\begin{enumerate}[label=(\alph*)]
\item State the acceleration of $P$ at time $T$.
\item In this question you must show detailed reasoning.
Determine the value of $T$.
\item Find the total distance that $P$ travels between times $t = 0$ and $t = T$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-09_524_410_251_242}
Particles $P$ and $Q$, of masses 4 kg and 6 kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley. The system is in equilibrium with $P$ hanging 1.75 m above a horizontal plane and $Q$ resting on the plane. Both parts of the string below the pulley are vertical (see diagram).\\
(a) Find the magnitude of the normal reaction force acting on $Q$.
The mass of $P$ is doubled, and the system is released from rest. You may assume that in the subsequent motion $Q$ does not reach the pulley.\\
(b) Determine the magnitude of the force exerted on the pulley by the string before $P$ strikes the plane.\\
(c) Determine the total distance travelled by $Q$ between the instant when the system is released and the instant when $Q$ first comes momentarily to rest.
When this motion is observed in practice, it is found that the total distance travelled by $Q$ between the instant when the system is released and the instant when $Q$ first comes momentarily to rest is less than the answer calculated in part (c).
\item State one factor that could account for this difference.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q11 [9]}}