| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time at height |
| Difficulty | Moderate -0.3 This is a straightforward SUVAT question requiring standard application of kinematic equations with gravity. Part (a) uses s=ut+½at² directly, part (b) requires finding two times when v=24.5 using v=u+at, then calculating the interval. The refinement in (c) is standard bookwork. Slightly easier than average due to being routine mechanics with no problem-solving insight needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to find displacement after 10 s | M1 | AO 3.1b |
| \(39.2 \times 10 - \frac{1}{2}g \times 10^2\) OR \(-39.2 \times 10 + \frac{1}{2}g \times 10^2\) | A1 | AO 1.1b |
| 98 (m) (must be positive) | A1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete method to find either half the time or the full time | M1 | AO 3.1b |
| \(0 = 24.5 - gt\) OR \(-24.5 = 24.5 - gt\) | A1 | AO 1.1b |
| 5 (s) | A1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. (include) air resistance | B1 | AO 3.5c |
## Question 2:
### Part 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to find displacement after 10 s | M1 | AO 3.1b |
| $39.2 \times 10 - \frac{1}{2}g \times 10^2$ **OR** $-39.2 \times 10 + \frac{1}{2}g \times 10^2$ | A1 | AO 1.1b |
| 98 (m) (must be positive) | A1 | AO 1.1b |
Notes: Complete method using $s = ut + \frac{1}{2}at^2$ or $s = vt - \frac{1}{2}at^2$ with motion reversed, or 'up and down' method with appropriate equations combined to give total distance. For 'up and down': distance up = 78.4, time up = 4, time down = 6, distance down = 176.4, combining correctly i.e. $(176.4 - 78.4)$ or $(78.4 - 176.4)$ for $g = 9.8$
### Part 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to find either half the time or the full time | M1 | AO 3.1b |
| $0 = 24.5 - gt$ **OR** $-24.5 = 24.5 - gt$ | A1 | AO 1.1b |
| 5 (s) | A1 | AO 1.1b |
Notes: Allow inequalities. E.g. for half the time may find $t = 4$ and $t = 1.5$ and subtract; for full time may find $t = 6.5$ and $t = 1.5$ and subtract.
### Part 2(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. (include) air resistance | B1 | AO 3.5c |
Notes: Accept e.g. spin of the stone, shape of the stone, size of the stone, wind effects, rotation. B0 if incorrect extras included e.g. mass or weight. DO NOT ALLOW negatives e.g. "there is no air resistance".
---\begin{enumerate}
\item A small stone is projected vertically upwards with speed $39.2 \mathrm {~ms} ^ { - 1 }$ from a point $O$.
\end{enumerate}
The stone is modelled as a particle moving freely under gravity from when it is projected until it hits the ground 10s later.
Using the model, find\\
(a) the height of $O$ above the ground,\\
(b) the total length of time for which the speed of the stone is less than or equal to $24.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(c) State one refinement that could be made to the model that would make your answer to part (a) more accurate.
\hfill \mbox{\textit{Edexcel AS Paper 2 2023 Q2 [7]}}