| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Non-Earth gravity contexts |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT application with non-standard gravity. Part (a) requires only conceptual understanding that gravity differs on the moon. Parts (b) and (c) are routine calculations using s=ut+½at² and v²=u²+2as with given values—no problem-solving insight needed, just direct formula application. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02g Two-dimensional variable acceleration3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Acceleration due to gravity is not constant but depends on location in the universe | B1 [1] | Allow any sensible comment that \(g\) might be different on the moon |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(s=2\), \(u=0\), \(t=1.6\) and \(s = ut + \frac{1}{2}at^2\); downwards as positive direction: \(2 = \frac{1}{2} \times 1.6^2 \cdot a\) giving \([a = 1.5625]\) | M1 | Allow any sign convention |
| \(1.6\ \text{m s}^{-2}\) to 2 sf | A1 [2] | Allow \(-1.6\ \text{m s}^{-2}\) if upwards clearly indicated as positive. Must be 2 significant figures |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(u=15\), \(v=0\), \(a=-1.6\) and \(v^2 = u^2 + 2as\); upwards as positive: \(0 = 15^2 - 2 \times 1.6s\) | M1 | Use of suvat equation(s) leading to a value for \(s\). Allow sign errors |
| \(s = 70.3\) m | A1 [2] | Must follow from correct working. Allow answers in range 70 to 72.2 |
## Question 10:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Acceleration due to gravity is not constant but depends on location in the universe | B1 [1] | Allow any sensible comment that $g$ might be different on the moon |
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $s=2$, $u=0$, $t=1.6$ and $s = ut + \frac{1}{2}at^2$; downwards as positive direction: $2 = \frac{1}{2} \times 1.6^2 \cdot a$ giving $[a = 1.5625]$ | M1 | Allow any sign convention |
| $1.6\ \text{m s}^{-2}$ to 2 sf | A1 [2] | Allow $-1.6\ \text{m s}^{-2}$ if upwards clearly indicated as positive. Must be 2 significant figures |
**Part (c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $u=15$, $v=0$, $a=-1.6$ and $v^2 = u^2 + 2as$; upwards as positive: $0 = 15^2 - 2 \times 1.6s$ | M1 | Use of suvat equation(s) leading to a value for $s$. Allow sign errors |
| $s = 70.3$ m | A1 [2] | Must follow from correct working. Allow answers in range 70 to 72.2 |
---10 An astronaut on the surface of the moon drops a ball from a point 2 m above the surface.
\begin{enumerate}[label=(\alph*)]
\item Without any calculations, explain why a standard model using $g = 9.8 \mathrm {~ms} ^ { - 2 }$ will not be appropriate to model the fall of the ball.
The ball takes 1.6s to hit the surface.
\item Find the acceleration of the ball which best models its motion. Give your answer correct to 2 significant figures.
\item Use this value to predict the maximum height of the ball above the point of projection when thrown vertically upwards with an initial velocity of $15 \mathrm {~ms} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2020 Q10 [5]}}