| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.3 This is a standard SUVAT projectile question with straightforward application of kinematic equations. Part (a) requires finding two positions where speed equals 14.7 m/s, part (b) uses symmetry of motion, and part (c) is a routine sketch. While multi-part with 9 marks total, each step follows textbook methods without requiring problem-solving insight, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a) | 0=14.72 −2gs | M1A1 |
| 22 or 22.1 (m) | A1 |
| Answer | Marks |
|---|---|
| 5(a) | M1 Complete method to find distance UP |
| Answer | Marks |
|---|---|
| 5(b) | 1 |
| Answer | Marks |
|---|---|
| from the positive root | M1A1 |
| t = 0.61 or 0.606 (s) | A1 |
| Answer | Marks |
|---|---|
| 5(b) | M1 Complete method to find required time |
| Answer | Marks |
|---|---|
| 5(c) | v |
| Answer | Marks |
|---|---|
| 3 | B1 shape |
| Answer | Marks |
|---|---|
| 5(c) | B1 V shape with v coord of end pt > 29.4 and each half roughly equally |
Question 5:
--- 5(a) ---
5(a) | 0=14.72 −2gs | M1A1
22 or 22.1 (m) | A1
(3)
5(a) | M1 Complete method to find distance UP
N.B. They may find time UP (1.5s ) AND use it to find distance UP
OR: (Distance from A to top – Distance from ‘14.7’ to top)
= (44.1 – 33.075)
A1 Correct equation(s) used
A1 cao
--- 5(b) ---
5(b) | 1
1 9 .6 = 2 9 .4 t + g t 2
2
1
N.B. 1 9 .6 = 2 9 .4 t − g t 2 is M0A0
2
1
− 1 9 .6 = 2 9 .4 t + g t 2 is M0A0
2
1
− 1 9 .6 = 2 9 .4 t − g t 2 is M0A0 unless they go on to subtract 6
2
from the positive root | M1A1
t = 0.61 or 0.606 (s) | A1
(3)
5(b) | M1 Complete method to find required time
N.B. They may find the speed as it hits the ground ( g 1 3 = 3 5 .3 3 4 ... )
AND use it to find the time.
A1 Correct equation(s) used
A1 cao
N.B. If they add to or subtract from 0.606, it’s M0 for an incorrect
method.
--- 5(c) ---
5(c) | v
29.4
0 t
3 | B1 shape
B1 29.4
B1 3
(3)
(9)
Notes for question 5
5(c) | B1 V shape with v coord of end pt > 29.4 and each half roughly equally
inclined to the t-axis. B0 if a vertical line is included at the end.
B1 29.4 independent
B1 3 independentA small ball is projected vertically upwards with speed $29.4\text{ ms}^{-1}$ from a point $A$ which is $19.6\text{ m}$ above horizontal ground.
The ball is modelled as a particle moving freely under gravity until it hits the ground. It is assumed that the ball does not rebound.
\begin{enumerate}[label=(\alph*)]
\item Find the distance travelled by the ball while its speed is less than $14.7\text{ ms}^{-1}$ [3]
\item Find the time for which the ball is moving with a speed of more than $29.4\text{ ms}^{-1}$ [3]
\item Sketch a speed-time graph for the motion of the ball from the instant when it is projected from $A$ to the instant when it hits the ground. Show clearly where your graph meets the axes. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2022 Q5 [9]}}