| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | SUVAT single equation: straightforward find |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring Newton's second law (F=ma) to find constant deceleration, then applying SUVAT equations (v=u+at and s=ut+½at²) to derive the given relationship. Part (b) involves simple substitution using μ=F/(mg). All steps are standard textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Let acceleration be \(a\) in direction of motion; N2L gives \(-F = ma\), so \(a = -\frac{F}{m}\), which is constant | M1, A1 | Decide to use N2L to find acceleration; no need to say 'constant' |
| Answer | Marks | Guidance |
|---|---|---|
| (As \(a\) constant) use suvat: \(S = 0 \times T - \frac{1}{2} \times \left(-\frac{F}{m}\right)T^2\) | B1 | Use appropriate sequence of suvat |
| so \(S = \left(\frac{F}{2m}\right)T^2\) and \(k = \frac{F}{2m}\) | E1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| As sliding, friction is limiting and \(F = \mu R\) | M1 | |
| \(R = mg\) | A1 | |
| \(k = \frac{F}{2m}\) so \(k = \frac{\mu mg}{2m}\) | M1 | In \(F = \mu R\), substitute for \(F\) and \(R\) in terms of \(m\) and \(g\) |
| Hence \(\mu = \frac{2k}{g} = \frac{2 \times 1.4}{9.8} = \frac{2}{7}\) | A1 [4] | Or 0.286 (3s.f.) |
## Question 9:
### Part (a)(i):
Let acceleration be $a$ in direction of motion; N2L gives $-F = ma$, so $a = -\frac{F}{m}$, which is constant | **M1, A1** | Decide to use N2L to find acceleration; no need to say 'constant'
### Part (a)(ii):
(As $a$ constant) use suvat: $S = 0 \times T - \frac{1}{2} \times \left(-\frac{F}{m}\right)T^2$ | **B1** | Use appropriate sequence of suvat
so $S = \left(\frac{F}{2m}\right)T^2$ and $k = \frac{F}{2m}$ | **E1** [4] |
### Part (b):
As sliding, friction is limiting and $F = \mu R$ | **M1** |
$R = mg$ | **A1** |
$k = \frac{F}{2m}$ so $k = \frac{\mu mg}{2m}$ | **M1** | In $F = \mu R$, substitute for $F$ and $R$ in terms of $m$ and $g$
Hence $\mu = \frac{2k}{g} = \frac{2 \times 1.4}{9.8} = \frac{2}{7}$ | **A1** [4] | Or 0.286 (3s.f.)
---9 In an experiment, a small box is hit across a floor. After it has been hit, the box slides without rotation.
The box passes a point A. The distance the box travels after passing A before coming to rest is $S$ metres and the time this takes is $T$ seconds.
The only resistance to the box's motion is friction due to the floor. The mass of the box is $m \mathrm {~kg}$ and the frictional force is a constant $F$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the equation of motion for the box while it is sliding.
\item Show that $S = k T ^ { 2 }$ where $k = \frac { F } { 2 m }$.
\end{enumerate}\item Given that $k = 1.4$, find the value of the coefficient of friction between the box and the floor.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 Q9 [8]}}