| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: max height |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question requiring basic application of kinematic equations with constant acceleration. Part (a) uses symmetry or v=u+at, part (b) requires finding max height then calculating distance for remaining time, and part (c) is a standard modelling statement. All techniques are routine for AS-level mechanics with no problem-solving insight needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(14.7 = -14.7 + 9.8T\) or \(0 = 14.7T - \frac{1}{2} \times 9.8T^2\) or \(0 = 14.7 - 9.8 \times \left(\frac{1}{2}T\right)\) oe | M1 | Complete method to find \(T\), condone sign errors (M0 if only find time to top) |
| \(T = 3\) | A1 | \(T = 3\) correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s_1 = \frac{(14.7+0)}{2} \times 1.5\) (11.025 or \(\frac{441}{40}\)) | M1 | Complete method to find one key distance |
| \(s_2 = \frac{1}{2} \times 9.8 \times 2.5^2\) (30.625 or \(\frac{245}{8}\)) OR \(s_3 = 14.7\times1 + \frac{1}{2}\times9.8\times1^2\) (19.6 or \(\frac{98}{5}\)) OR \(-s_3 = 14.7\times4 - \frac{1}{2}\times9.8\times4^2\) (-19.6) | M1 | Correct method to find another key distance |
| Total distance \(= s_1 + s_2\) OR \(2s_1 + s_3\) | M1 | Complete method to find total distance |
| \(= 41.7\) m or \(42\) m | A1 | 41.7 or 42 (after use of \(g = 9.8\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. Take account of the dimensions of the stone (e.g. allow for spin), do not model the stone as a particle, use a more accurate value for \(g\) | B1 | B0 if there are incorrect extra refinements but ignore extra incorrect statements |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $14.7 = -14.7 + 9.8T$ or $0 = 14.7T - \frac{1}{2} \times 9.8T^2$ or $0 = 14.7 - 9.8 \times \left(\frac{1}{2}T\right)$ oe | M1 | Complete method to find $T$, condone sign errors (M0 if only find time to top) |
| $T = 3$ | A1 | $T = 3$ correctly obtained |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s_1 = \frac{(14.7+0)}{2} \times 1.5$ (11.025 or $\frac{441}{40}$) | M1 | Complete method to find one key distance |
| $s_2 = \frac{1}{2} \times 9.8 \times 2.5^2$ (30.625 or $\frac{245}{8}$) **OR** $s_3 = 14.7\times1 + \frac{1}{2}\times9.8\times1^2$ (19.6 or $\frac{98}{5}$) **OR** $-s_3 = 14.7\times4 - \frac{1}{2}\times9.8\times4^2$ (-19.6) | M1 | Correct method to find another key distance |
| Total distance $= s_1 + s_2$ **OR** $2s_1 + s_3$ | M1 | Complete method to find total distance |
| $= 41.7$ m or $42$ m | A1 | 41.7 or 42 (after use of $g = 9.8$) |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. Take account of the dimensions of the stone (e.g. allow for spin), do not model the stone as a particle, use a more accurate value for $g$ | B1 | B0 if there are incorrect extra refinements but ignore extra incorrect statements |
> **Note:** If they use $g = 9.81$ or $10$, penalise once for whole question.
---\begin{enumerate}
\item At time $t = 0$, a small stone is thrown vertically upwards with speed $14.7 \mathrm {~ms} ^ { - 1 }$ from a point $A$.
\end{enumerate}
At time $t = T$ seconds, the stone passes through $A$, moving downwards.\\
The stone is modelled as a particle moving freely under gravity throughout its motion.\\
Using the model,\\
(a) find the value of $T$,\\
(b) find the total distance travelled by the stone in the first 4 seconds of its motion.\\
(c) State one refinement that could be made to the model, apart from air resistance, that would make the model more realistic.
\hfill \mbox{\textit{Edexcel AS Paper 2 2021 Q1 [7]}}