| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.8 This is a standard AS-level mechanics question testing routine SUVAT equations with vertical motion under gravity. Parts (a)-(d) involve straightforward application of kinematic equations with given values, while parts (e)-(f) test basic conceptual understanding of modeling assumptions. The multi-part structure and 'show that' in part (a) are typical, but no step requires novel insight or complex problem-solving beyond textbook exercises. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(19^2 = (-U)^2 + 2 \times 10 \times 16.8\) | M1 | Complete method to find \(U\), condone sign errors and use of \(g = 9.8\) |
| \(U = 5\) | A1* | \(U = 5\) cao; allow \(U^2\) instead of \((-U)^2\); allow verification |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(19 = -5 + 10T\) OR \(16.8 = \frac{(-5+19)}{2}T\) OR \(16.8 = -5T + \frac{1}{2} \times 10T^2\) OR \(16.8 = 19T - \frac{1}{2} \times 10T^2\) | M1 | Complete method to find \(T\), condone sign errors |
| \(T = 2.4\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(1.2 = -5t + \frac{1}{2} \times 10 \times t^2\) | M1 | Complete method to find \(t\), condone sign errors |
| \(5t^2 - 5t - 1.2 = 0\) | A1 | Correct equation with at most one error |
| \(5t^2 - 5t - 1.2 = 0\) | M(A)1 | Correct equation |
| \(t = 1.2\) (s) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Graph: straight line from \((0,5)\) sloping down through \(t\)-axis to \((2.4, -19)\) | B1 shape | Graph could be reflected in the \(t\)-axis |
| Points \((0, 5)\) and \((2.4, -19)\) marked | B1ft | Follow through on their \(T\) value; if reflected: \((0,-5)\) and \((2.4,19)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Greater since air resistance would slow the ball down | B1 | Any similar appropriate comment |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Take into account: spin, wind effects, use a more accurate value of \(g\), not model the ball as a particle | B1 | B0 if any incorrect extras e.g. weight/mass included |
# Question 1:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $19^2 = (-U)^2 + 2 \times 10 \times 16.8$ | M1 | Complete method to find $U$, condone sign errors and use of $g = 9.8$ |
| $U = 5$ | A1* | $U = 5$ cao; allow $U^2$ instead of $(-U)^2$; allow verification |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $19 = -5 + 10T$ **OR** $16.8 = \frac{(-5+19)}{2}T$ **OR** $16.8 = -5T + \frac{1}{2} \times 10T^2$ **OR** $16.8 = 19T - \frac{1}{2} \times 10T^2$ | M1 | Complete method to find $T$, condone sign errors |
| $T = 2.4$ | A1 | cao |
## Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| $1.2 = -5t + \frac{1}{2} \times 10 \times t^2$ | M1 | Complete method to find $t$, condone sign errors |
| $5t^2 - 5t - 1.2 = 0$ | A1 | Correct equation with at most one error |
| $5t^2 - 5t - 1.2 = 0$ | M(A)1 | Correct equation |
| $t = 1.2$ (s) | A1 | cao |
## Part (d)
| Working | Mark | Guidance |
|---------|------|----------|
| Graph: straight line from $(0,5)$ sloping down through $t$-axis to $(2.4, -19)$ | B1 shape | Graph could be reflected in the $t$-axis |
| Points $(0, 5)$ and $(2.4, -19)$ marked | B1ft | Follow through on their $T$ value; if reflected: $(0,-5)$ and $(2.4,19)$ |
## Part (e)
| Working | Mark | Guidance |
|---------|------|----------|
| Greater since air resistance would slow the ball down | B1 | Any similar appropriate comment |
## Part (f)
| Working | Mark | Guidance |
|---------|------|----------|
| Take into account: spin, wind effects, use a more accurate value of $g$, not model the ball as a particle | B1 | B0 if any incorrect extras e.g. weight/mass included |
---\begin{enumerate}
\item At time $t = 0$, a small ball is projected vertically upwards with speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $A$ that is 16.8 m above horizontal ground.
\end{enumerate}
The speed of the ball at the instant immediately before it hits the ground for the first time is $19 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
The ball hits the ground for the first time at time $t = T$ seconds.\\
The motion of the ball, from the instant it is projected until the instant just before it hits the ground for the first time, is modelled as that of a particle moving freely under gravity.
The acceleration due to gravity is modelled as having magnitude $10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$\\
Using the model,\\
(a) show that $U = 5$\\
(b) find the value of $T$,\\
(c) find the time from the instant the ball is projected until the instant when the ball is 1.2 m below $A$.\\
(d) Sketch a velocity-time graph for the motion of the ball for $0 \leqslant t \leqslant T$, stating the coordinates of the start point and the end point of your graph.
In a refinement of the model of the motion of the ball, the effect of air resistance on the ball is included and this refined model is now used to find the value of $U$.\\
(e) State, with a reason, how this new value of $U$ would compare with the value found in part (a), using the initial unrefined model.\\
(f) Suggest one further refinement that could be made to the model, apart from including air resistance, that would make the model more realistic.
\hfill \mbox{\textit{Edexcel AS Paper 2 2020 Q1 [12]}}