| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question requiring direct application of kinematic equations with given values. Parts (a) and (b) involve standard two-step calculations using v² = u² + 2as and v = u + at. Parts (c) and (d) test basic understanding of modeling assumptions. The simplified value g = 10 m/s² further reduces computational difficulty. This is easier than average for AS mechanics. |
| 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 |
| Complete method to produce an equation in \(U\) only | M1 | AO 3.4 |
| e.g. \(10^2 = U^2 + 2 \times g \times 1.8\) | A1 | AO 1.1b; correct equation in \(U\) only, \(g\) does not need to be substituted; allow \(g = 9.8\) or \(9.81\) |
| OR complete method finding \(T\) first, then equation in \(U\) only | M1 | |
| Correct equation in \(U\) only | A1 | |
| \(U = 8\) (only this answer) | A1 | AO 1.1b; cao (A0 if \(g = 10\) has not been used) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete method to find equation in \(T\) only: \(10 = -8 + gT\) or \(1.8 = 10T - \frac{1}{2}gT^2\) or \(1.8 = \frac{(-8+10)}{2}T\) or \(1.8 = -8T + \frac{1}{2}gT^2\) | M1 | AO 3.4; M1 earned on final line when adding two times; \(g\) does not need substituted; condone sign errors; M0 if incorrect formula; follow through on their \(U\) |
| ALT 1: \(0 = 8 - gt_{\text{UP}} \Rightarrow t_{\text{UP}} = 0.8\); \(h_{\text{UP}} = \frac{(8+0)}{2} \times 0.8 = 3.2\); \((h_{\text{UP}}+1.8) = \frac{(0+10)}{2} \times t_{\text{DOWN}} \Rightarrow t_{\text{DOWN}} = 1\); \(T = t_{\text{UP}} + t_{\text{DOWN}}\) | M1 | |
| ALT 2: \(8 = -8 + gt_A \Rightarrow t_A = 1.6\); \(1.8 = \frac{(8+10)}{2} \times t_{AG} \Rightarrow t_{AG} = 0.2\); \(T = t_A + t_{AG}\) | M1 | |
| \(T = 1.8\) oe e.g. \(\frac{9}{5}\) | A1 | AO 1.1b; cao (A0 if \(g = 10\) not used); A0 if two answers given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use more accurate value for \(g\) (e.g. \(g = 9.8\) or \(g = 9.81\)); allow for wind effects; allow for spin of stone; include dimensions of stone (not a particle); allow for variable acceleration. If air resistance mentioned as an extra, ignore it. | B1 | AO 3.5c; any appropriate refinement; B0 if incorrect extra given e.g. mass or weight mentioned |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(U\) would be greater. Allow without \(U\), e.g. "it would be greater" or just "greater" | B1 | AO 3.5a; ISW; cao |
# Question 1:
## Part 1(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to produce an equation in $U$ only | M1 | AO 3.4 |
| e.g. $10^2 = U^2 + 2 \times g \times 1.8$ | A1 | AO 1.1b; correct equation in $U$ only, $g$ does not need to be substituted; allow $g = 9.8$ or $9.81$ |
| **OR** complete method finding $T$ first, then equation in $U$ only | M1 | |
| Correct equation in $U$ only | A1 | |
| $U = 8$ (only this answer) | A1 | AO 1.1b; cao (A0 if $g = 10$ has not been used) |
## Part 1(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to find equation in $T$ only: $10 = -8 + gT$ **or** $1.8 = 10T - \frac{1}{2}gT^2$ **or** $1.8 = \frac{(-8+10)}{2}T$ **or** $1.8 = -8T + \frac{1}{2}gT^2$ | M1 | AO 3.4; M1 earned on final line when adding two times; $g$ does not need substituted; condone sign errors; M0 if incorrect formula; follow through on their $U$ |
| **ALT 1:** $0 = 8 - gt_{\text{UP}} \Rightarrow t_{\text{UP}} = 0.8$; $h_{\text{UP}} = \frac{(8+0)}{2} \times 0.8 = 3.2$; $(h_{\text{UP}}+1.8) = \frac{(0+10)}{2} \times t_{\text{DOWN}} \Rightarrow t_{\text{DOWN}} = 1$; $T = t_{\text{UP}} + t_{\text{DOWN}}$ | M1 | |
| **ALT 2:** $8 = -8 + gt_A \Rightarrow t_A = 1.6$; $1.8 = \frac{(8+10)}{2} \times t_{AG} \Rightarrow t_{AG} = 0.2$; $T = t_A + t_{AG}$ | M1 | |
| $T = 1.8$ oe e.g. $\frac{9}{5}$ | A1 | AO 1.1b; cao (A0 if $g = 10$ not used); A0 if two answers given |
## Part 1(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use more accurate value for $g$ (e.g. $g = 9.8$ or $g = 9.81$); allow for wind effects; allow for spin of stone; include dimensions of stone (not a particle); allow for variable acceleration. If air resistance mentioned as an extra, ignore it. | B1 | AO 3.5c; any appropriate refinement; B0 if incorrect extra given e.g. mass or weight mentioned |
## Part 1(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $U$ would be greater. Allow without $U$, e.g. "it would be greater" or just "greater" | B1 | AO 3.5a; ISW; cao |
---\begin{enumerate}
\item The point $A$ is 1.8 m vertically above horizontal ground.
\end{enumerate}
At time $t = 0$, a small stone is projected vertically upwards with speed $U \mathrm {~ms} ^ { - 1 }$ from the point $A$.
At time $t = T$ seconds, the stone hits the ground.\\
The speed of the stone as it hits the ground is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
In an initial model of the motion of the stone as it moves from $A$ to where it hits the ground
\begin{itemize}
\item the stone is modelled as a particle moving freely under gravity
\item the acceleration due to gravity is modelled as having magnitude $\mathbf { 1 0 } \mathbf { m ~ s } ^ { \mathbf { - 2 } }$
\end{itemize}
Using the model,\\
(a) find the value of $U$,\\
(b) find the value of $T$.\\
(c) Suggest one refinement, apart from including air resistance, that would make the model more realistic.
In reality the stone will not move freely under gravity and will be subject to air resistance.\\
(d) Explain how this would affect your answer to part (a).
\hfill \mbox{\textit{Edexcel AS Paper 2 2022 Q1 [7]}}