| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down smooth slope |
| Difficulty | Moderate -0.3 This is a straightforward kinematics problem on an inclined plane requiring application of standard SUVAT equations with two unknowns (u and a) from two equations, followed by resolving forces to find the angle. The problem-solving is routine and methodical with no conceptual surprises, making it slightly easier than average but still requiring competent execution of multiple steps. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| M1 | Uses \(s=(u+v)/2\) or a combination of two other formulae; \(5^2=u^2 + 2x6.4a\) M1; \(5 = u +1.6a\) M1 |
| A1 | Accurate equation in one variable |
| A1 | \(u = 3\) m s\(^{-1}\) A1; \(a = 1.25\) m s\(^{-2}\) A1 |
| [5] | Candidates may find \(a\) first (see below); \(s = vt +/- a t^2/2\); Must be from \(s = vt - a t^2/2\) |
| M1 | |
| A1 | |
| A1 | |
| SC | Do not accept \(a = 1.25\) from \(6.4=5x1.6+a1.6^2/2\) but allow subsequent use of \(a = 1.25\) in \(5 = u +1.25x1.6\) |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| M1 | Resolves \(g\) or weight, \(a \neq g\) |
| A1√ | ft cv(1.25) from (i) |
| A1 | Must be angle with vertical |
| [3] |
## Part i
**Answer/Working:**
$6.4 = (u+5)/2 \times 1.6$
$u = 3$ m s$^{-1}$
$5 = 3+1.6a$
$a = 1.25$ m s$^{-2}$
OR
$6.4=5x1.6-a1.6^2/2$
$a = 1.25$ m s$^{-2}$
OR
$5 = u+1.25x1.6$
$u = 3$ m s$^{-1}$
| **Marks** | **Guidance** |
|-----------|-------------|
| M1 | Uses $s=(u+v)/2$ or a combination of two other formulae; $5^2=u^2 + 2x6.4a$ M1; $5 = u +1.6a$ M1 |
| A1 | Accurate equation in one variable |
| A1 | $u = 3$ m s$^{-1}$ A1; $a = 1.25$ m s$^{-2}$ A1 |
| [5] | Candidates may find $a$ first (see below); $s = vt +/- a t^2/2$; Must be from $s = vt - a t^2/2$ |
| M1 | |
| A1 | |
| A1 | |
| SC | Do not accept $a = 1.25$ from $6.4=5x1.6+a1.6^2/2$ but allow subsequent use of $a = 1.25$ in $5 = u +1.25x1.6$ |
## Part ii
**Answer/Working:**
$1.25(m) = (m)g\cos\theta$
$1.25(m) = (m)g\cos\theta$ OR $1.25(m) = (m)g\sin\theta$
Angle with vertical = 82.7°
| **Marks** | **Guidance** |
|-----------|-------------|
| M1 | Resolves $g$ or weight, $a \neq g$ |
| A1√ | ft cv(1.25) from (i) |
| A1 | Must be angle with vertical |
| [3] | |
---A particle $P$ is projected down a line of greatest slope on a smooth inclined plane. $P$ has velocity $5\text{ m s}^{-1}$ at the instant when it has been in motion for $1.6\text{ s}$ and travelled a distance of $6.4\text{ m}$. Calculate
\begin{enumerate}[label=(\roman*)]
\item the initial speed and the acceleration of $P$, [5]
\item the inclination of the plane to the vertical. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2016 Q2 [8]}}