| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Sketch velocity-time graph |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question with standard vertical motion under gravity. Part (a) is direct application of v=u+at, part (b) is sketching a basic speed-time graph, part (c) uses area under graph equals distance. Parts (d) and (e) test conceptual understanding but require minimal calculation. Easier than average A-level due to simple numbers, clear structure, and routine application of standard techniques. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(V = 30\) (m s\(^{-1}\)) | B1 | cao |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Overall shape of graph, starting at origin | B1 | Dotted vertical line at end is OK but solid vertical line is B0 |
| 3, 5 and \(T\) marked on \(t\)-axis; 30 marked on speed axis | B1ft | 3 must be where graph reaches a peak. Allow delineators: 3, 2 and \(T-5\) or a mixture |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using total area \(= 550\) to set up equation in one unknown | M1 | Need all sections included with correct structure. e.g. triangle + trapezium + rectangle \(= 550\) giving equation in one unknown (may not be \(T\)) |
| \(\frac{1}{2}\times3\times30+\frac{(30+6)}{2}\times2+6(T-5)=550\) OR \(\frac{1}{2}\times3\times30+\frac{1}{2}\times2\times24+6(T-3)=550\) OR \(\frac{1}{2}\times3\times30+\frac{1}{2}\times2\times24+(2\times6)+6(T-5)=550\) | A2ft | ft on their answer to (a). \(-1\) each error. N.B. If '6' is incorrect, treat as one error unless correct ft from their 30 |
| Solve for \(T\) | M1 | Attempt to solve for \(T\) provided they have tried to find area using at least 3 sections. M0 if they only solve for unknown and never try to find \(T\) |
| \(T = 83\) (nearest whole number) | A1 | 83 is the only answer |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| New value of \(T\) would be bigger | B1 | Clear statement about the value of \(T\). Allow '*it* would increase, get larger etc'. B0 for 'Takes longer' or 'the value of \(T\) would be longer' |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| e.g. effect of wind; allow for dimensions of parachutist; use more accurate value for \(g\); parachutist does not fall vertically after chute opens; smooth changes in \(v\); time for parachute to open; deceleration not constant; smooth changes in \(a\) | B1 | Any appropriate refinement of the model. B0 if incorrect or irrelevant extras. B0 for: moves horizontally; mass/weight of parachutist; upthrust; air pressure; air resistance; terminal velocity |
| (1) | ||
| (10 marks) |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $V = 30$ (m s$^{-1}$) | B1 | cao |
| | **(1)** | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Overall shape of graph, starting at origin | B1 | Dotted vertical line at end is OK but solid vertical line is B0 |
| 3, 5 and $T$ marked on $t$-axis; 30 marked on speed axis | B1ft | 3 must be where graph reaches a peak. Allow delineators: 3, 2 and $T-5$ or a mixture |
| | **(2)** | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Using total area $= 550$ to set up equation in **one unknown** | M1 | Need all sections included with correct structure. e.g. triangle + trapezium + rectangle $= 550$ giving equation in **one unknown (may not be $T$)** |
| $\frac{1}{2}\times3\times30+\frac{(30+6)}{2}\times2+6(T-5)=550$ **OR** $\frac{1}{2}\times3\times30+\frac{1}{2}\times2\times24+6(T-3)=550$ **OR** $\frac{1}{2}\times3\times30+\frac{1}{2}\times2\times24+(2\times6)+6(T-5)=550$ | A2ft | ft on their answer to (a). $-1$ each error. N.B. If '6' is incorrect, treat as one error unless correct ft from their 30 |
| Solve for $T$ | M1 | Attempt to solve for $T$ provided they have tried to find area using at least 3 sections. M0 if they only solve for unknown and never try to find $T$ |
| $T = 83$ (nearest whole number) | A1 | 83 is the only answer |
| | **(5)** | |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| New value of $T$ would be bigger | B1 | Clear statement about the value of $T$. Allow '*it* would increase, get larger etc'. B0 for 'Takes longer' or 'the value of $T$ would be longer' |
| | **(1)** | |
## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. effect of wind; allow for dimensions of parachutist; use more accurate value for $g$; parachutist does not fall vertically after chute opens; smooth changes in $v$; time for parachute to open; deceleration not constant; smooth changes in $a$ | B1 | Any appropriate refinement of the model. B0 if incorrect or irrelevant extras. B0 for: moves horizontally; mass/weight of parachutist; upthrust; air pressure; air resistance; terminal velocity |
| | **(1)** | |
| | **(10 marks)** | |
---\begin{enumerate}
\item At time $t = 0$, a parachutist falls vertically from rest from a helicopter which is hovering at a height of 550 m above horizontal ground.
\end{enumerate}
The parachutist, who is modelled as a particle, falls for 3 seconds before her parachute opens.\\
While she is falling, and before her parachute opens, she is modelled as falling freely under gravity.
The acceleration due to gravity is modelled as being $10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(a) Using this model, find the speed of the parachutist at the instant her parachute opens.
When her parachute is open, the parachutist continues to fall vertically.\\
Immediately after her parachute opens, she decelerates at $12 \mathrm {~ms} ^ { - 2 }$ for 2 seconds before reaching a constant speed and she reaches the ground with this speed.
The total time taken by the parachutist to fall the 550 m from the helicopter to the ground is $T$ seconds.\\
(b) Sketch a speed-time graph for the motion of the parachutist for $0 \leqslant t \leqslant T$.\\
(c) Find, to the nearest whole number, the value of $T$.
In a refinement of the model of the motion of the parachutist, the effect of air resistance is included before her parachute opens and this refined model is now used to find a new value of $T$.\\
(d) How would this new value of $T$ compare with the value found, using the initial model, in part (c)?\\
(e) Suggest one further refinement to the model, apart from air resistance, to make the model more realistic.
\hfill \mbox{\textit{Edexcel AS Paper 2 2019 Q1 [10]}}