| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| 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 mechanics question involving integration of acceleration to find velocity and position, plus a routine collision problem with constant acceleration. Both parts require only straightforward application of kinematic equations with no novel problem-solving insight, making it slightly easier than average for A-level. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(8 = \frac{1}{2}(u + 2.25) \times 32\) | M1 | Using \(s = \frac{1}{2}(u+v)t\) |
| \(u = -1.75\), so \(1.75 \text{ m s}^{-1}\) | A1 | |
| \(2.25 = -1.75 + 32a\) | M1 | Use of any appropriate suvat with their values and correct signs |
| \(a = 0.125\), so \(0.125 \text{ m s}^{-2}\) | F1 | Sign must be consistent with their \(u\), FT from their value of \(u\) |
| Directions of \(u\) and \(a\) are defined | F1 | Establish directions of both \(u\) and \(a\) in terms of A and B. May be shown by diagram. Without a diagram, wording must be absolutely clear. Dependent on both M marks |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(8 = 2.25 \times 32 - \frac{1}{2} \times a \times 32^2\) | M1 | Using \(s = vt - \frac{1}{2}at^2\) |
| \(a = 0.125\), so \(0.125 \text{ m s}^{-2}\) | A1 | |
| \(2.25 = u + 32 \times 0.125\) | M1 | Use of any appropriate suvat with their values and correct signs |
| \(u = -1.75\), so \(1.75 \text{ m s}^{-1}\) | F1 | Sign must be consistent with their \(a\), FT from their value of \(a\) |
| Directions of \(u\) and \(a\) are defined | F1 | Establish directions of both \(u\) and \(a\) in terms of A and B. Dependent on both M marks |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Set up one relevant equation with \(a\) and \(u\) | M1 | Using one of \(v = u + at\), \(s = ut + \frac{1}{2}at^2\) and \(v^2 = u^2 + 2as\) |
| Set up second relevant equation with \(a\) and \(u\) | M1 | Using another of \(v = u + at\), \(s = ut + \frac{1}{2}at^2\) and \(v^2 = u^2 + 2as\) |
| Solving to find \(u = -1.75\), so \(1.75 \text{ m s}^{-1}\) | A1 | |
| Solving to find \(a = 0.125\), so \(0.125 \text{ m s}^{-2}\) | F1 | FT from their value of \(u\) or \(a\), whichever found first |
| Directions of \(u\) and \(a\) are defined | F1 | Establish directions of both \(u\) and \(a\) in terms of A and B. Dependent on both M marks |
| [5] |
## Question 5:
**Either** (finding $u$ first):
$8 = \frac{1}{2}(u + 2.25) \times 32$ | M1 | Using $s = \frac{1}{2}(u+v)t$ |
$u = -1.75$, so $1.75 \text{ m s}^{-1}$ | A1 | |
$2.25 = -1.75 + 32a$ | M1 | Use of any appropriate suvat with **their** values and correct signs |
$a = 0.125$, so $0.125 \text{ m s}^{-2}$ | F1 | Sign must be consistent with their $u$, FT from their value of $u$ |
Directions of $u$ and $a$ are defined | F1 | Establish directions of both $u$ and $a$ in terms of A and B. May be shown by diagram. Without a diagram, wording must be absolutely clear. Dependent on both M marks |
| [5] | |
**Or** (finding $a$ first):
$8 = 2.25 \times 32 - \frac{1}{2} \times a \times 32^2$ | M1 | Using $s = vt - \frac{1}{2}at^2$ |
$a = 0.125$, so $0.125 \text{ m s}^{-2}$ | A1 | |
$2.25 = u + 32 \times 0.125$ | M1 | Use of any appropriate suvat with **their** values and correct signs |
$u = -1.75$, so $1.75 \text{ m s}^{-1}$ | F1 | Sign must be consistent with their $a$, FT from their value of $a$ |
Directions of $u$ and $a$ are defined | F1 | Establish directions of both $u$ and $a$ in terms of A and B. Dependent on both M marks |
| [5] | |
**Or** (simultaneous equations):
Set up one relevant equation with $a$ and $u$ | M1 | Using **one** of $v = u + at$, $s = ut + \frac{1}{2}at^2$ and $v^2 = u^2 + 2as$ |
Set up second relevant equation with $a$ and $u$ | M1 | Using **another** of $v = u + at$, $s = ut + \frac{1}{2}at^2$ and $v^2 = u^2 + 2as$ |
Solving to find $u = -1.75$, so $1.75 \text{ m s}^{-1}$ | A1 | |
Solving to find $a = 0.125$, so $0.125 \text{ m s}^{-2}$ | F1 | FT from their value of $u$ or $a$, whichever found first |
Directions of $u$ and $a$ are defined | F1 | Establish directions of both $u$ and $a$ in terms of A and B. Dependent on both M marks |
| [5] | |5 A particle is moving along a straight line and its position is relative to an origin on the line. At time $t \mathrm {~s}$, the particle's acceleration, $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, is given by
$$a = 6 t - 12 .$$
At $t = 0$ the velocity of the particle is $+ 9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its position is - 2 m .\\
(i) Find an expression for the velocity of the particle at time $t \mathrm {~s}$ and verify that it is stationary when $t = 3$.\\
(ii) Find the position of the particle when $t = 2$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b9e41fac-9f4b-4165-af03-67ebdcb326de-3_349_987_375_623}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
Particles P and Q move in the same straight line. Particle P starts from rest and has a constant acceleration towards $Q$ of $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Particle $Q$ starts 125 m from $P$ at the same time and has a constant speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ away from $P$. The initial values are shown in Fig. 4.\\
(i) Write down expressions for the distances travelled by P and by Q at time $t$ seconds after the start of the motion.\\
(ii) How much time does it take for P to catch up with Q and how far does P travel in this time?
\hfill \mbox{\textit{OCR MEI M1 Q5 [7]}}