| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Vector form projectile motion |
| Difficulty | Standard +0.3 This is a standard projectile motion problem requiring finding initial velocity components from given landing position and time, then using energy/kinematic equations to find speed at a given height. Part (a) involves routine application of SUVAT equations and Pythagoras, while parts (b) and (c) test understanding that horizontal velocity remains constant at maximum height and that air resistance was neglected. The multi-step nature and 6 marks for part (a) place it slightly above average, but all techniques are standard A-level mechanics. |
| Spec | 1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| 17(a) | Obtains correct horizontal | |
| component of the initial velocity | AO1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| component of initial velocity | AO3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| of initial velocity | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| and V | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| velocity | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| have been awarded | AO3.2a | A1F |
| (b) | States ‘their’ value of horizontal |
| Answer | Marks | Guidance |
|---|---|---|
| from part (a) | AO3.4 | A1F |
| (c) | Explains that horizontal velocity |
| Answer | Marks | Guidance |
|---|---|---|
| reasoning | AO3.5b | E1 |
| Answer | Marks |
|---|---|
| Total | 8 |
| TOTAL | 100 |
Question 17:
--- 17(a) ---
17(a) | Obtains correct horizontal
component of the initial velocity | AO1.1b | B1 | 2.5U 40
U 16
102.5V 0.59.812.52
V 8.2625
v2 8.26252 2(9.81)3
y
v 3.067...
y
v 162 3.0672 16.3 m s1
Forms equation to find vertical
component of initial velocity | AO3.3 | M1
Obtains correct vertical component
of initial velocity | AO1.1b | A1
Forms equation for vertical
component of velocity at height 3
using ‘their’ derived values for U
and V | AO3.4 | M1
Obtains correct component of
velocity | AO1.1b | A1
Correct final speed with units,
correct for ‘their’ U and v
y
FT applies only if both M1 marks
have been awarded | AO3.2a | A1F
(b) | States ‘their’ value of horizontal
component of the initial velocity
from part (a) | AO3.4 | A1F | 16 m s–1
(c) | Explains that horizontal velocity
has been assumed to be constant
in their model and that this is not
likely to be true, with valid
reasoning | AO3.5b | E1 | It was assumed that there were
no resistance forces acting on
the ball which is unlikely to be
true in reality. The horizontal
speed of the ball is likely to
vary… air resistance would slow
the ball down, wind might speed
the ball up
Total | 8
TOTAL | 100In this question use $g = 9.81$ m s$^{-2}$.
A ball is projected from the origin. After 2.5 seconds, the ball lands at the point with position vector $(40\mathbf{i} - 10\mathbf{j})$ metres.
The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertical respectively.
Assume that there are no resistance forces acting on the ball.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of the ball when it is at a height of 3 metres above its initial position.
[6 marks]
\item State the speed of the ball when it is at its maximum height.
[1 mark]
\item Explain why the answer you found in part (b) may not be the actual speed of the ball when it is at its maximum height.
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 Q17 [8]}}