| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Moderate -0.3 This is a standard M1 kinematics question requiring routine integration of acceleration to velocity and velocity to displacement, with straightforward substitution of boundary conditions. While it has multiple parts (13 marks total), each step follows a predictable template with no conceptual surprises—slightly easier than average due to its mechanical nature, though the final part requires recognizing when the particle returns to O by solving s=0. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(v = \int -\frac{1}{10} t \, dt = -\frac{1}{20} t^2 + c\) | M1, A1 | For integrating the acceleration formula; For \(v = -\frac{1}{20} t^2\), with or without \(c\) |
| \(V = 0 + c\) | M1 | For using \(v = V\) when \(t = 0\) to find \(c\) |
| Hence \(v = V - \frac{1}{20} t^2\) | A1 | For correct equation for \(v\) in terms of \(t\) and \(V\) |
| Subtotal: 4 marks | ||
| (ii) \(0 = V - \frac{10^2}{20}\) \(\Rightarrow V = 5\) | M1, A1 | For use of given values to find \(V\); For correct value 5 |
| Subtotal: 2 marks | ||
| (iii) \(s = \int (5 - \frac{1}{20} t^2) \, dt = 5t - \frac{1}{60} t^3 + k\) | M1, A1 | For any attempt to integrate velocity; For correct integration (ignoring \(k\)) |
| Hence displacement is \(50 - \frac{1000}{60} = 33\frac{1}{3}\) m | M1, A1 | For evaluation of \(s\) when \(t = 10\); For correct value \(33\frac{1}{3}\); allow omission of \(k\) |
| Subtotal: 4 marks | ||
| (iv) Returns to O when \(0 = -\frac{1}{20} t^3 + 5t \Rightarrow t^2 = 300\) | M1 | For attempting non-zero root of \(s = 0\) |
| When \(t^2 = 300\), \(v = -\frac{1}{20} \times 300 + 5\) | M1 | For consequent evaluation of \(v\) |
| i.e. speed is 10 m s\(^{-1}\) | A1 | For correct value 3 (allow negative here) |
| Subtotal: 3 marks | ||
| Total: 13 marks |
**(i)** $v = \int -\frac{1}{10} t \, dt = -\frac{1}{20} t^2 + c$ | M1, A1 | For integrating the acceleration formula; For $v = -\frac{1}{20} t^2$, with or without $c$
$V = 0 + c$ | M1 | For using $v = V$ when $t = 0$ to find $c$
Hence $v = V - \frac{1}{20} t^2$ | A1 | For correct equation for $v$ in terms of $t$ and $V$
| | **Subtotal: 4 marks**
**(ii)** $0 = V - \frac{10^2}{20}$ $\Rightarrow V = 5$ | M1, A1 | For use of given values to find $V$; For correct value 5
| | **Subtotal: 2 marks**
**(iii)** $s = \int (5 - \frac{1}{20} t^2) \, dt = 5t - \frac{1}{60} t^3 + k$ | M1, A1 | For any attempt to integrate velocity; For correct integration (ignoring $k$)
Hence displacement is $50 - \frac{1000}{60} = 33\frac{1}{3}$ m | M1, A1 | For evaluation of $s$ when $t = 10$; For correct value $33\frac{1}{3}$; allow omission of $k$
| | **Subtotal: 4 marks**
**(iv)** Returns to O when $0 = -\frac{1}{20} t^3 + 5t \Rightarrow t^2 = 300$ | M1 | For attempting non-zero root of $s = 0$
When $t^2 = 300$, $v = -\frac{1}{20} \times 300 + 5$ | M1 | For consequent evaluation of $v$
i.e. speed is 10 m s$^{-1}$ | A1 | For correct value 3 (allow negative here)
| | **Subtotal: 3 marks**
| | **Total: 13 marks**A particle $P$ moves in a straight line so that, at time $t$ seconds after leaving a fixed point $O$, its acceleration is $-\frac{1}{10}t \text{ m s}^{-2}$. At time $t = 0$, the velocity of $P$ is $V \text{ m s}^{-1}$.
\begin{enumerate}[label=(\roman*)]
\item Find, by integration, an expression in terms of $t$ and $V$ for the velocity of $P$. [4]
\item Find the value of $V$, given that $P$ is instantaneously at rest when $t = 10$. [2]
\item Find the displacement of $P$ from $O$ when $t = 10$. [4]
\item Find the speed with which the particle returns to $O$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 Q5 [13]}}