| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time to ground |
| Difficulty | Moderate -0.5 This is a straightforward two-part SUVAT question requiring standard application of kinematic equations with gravity. Part (a) uses v²=u²+2as to find speed at a given height, and part (b) uses s=ut+½at² to find time to ground. Both are routine M1 exercises with clear setup and no conceptual challenges, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v^2 = 20^2 - 2g \times (-3)\) | M1 | Complete method to find speed, must use 3 or -3. Allow 9.81 for \(g\) or just \(g\), condone sign errors |
| \(v = 21\) or \(21.4\) (m s\(^{-1}\)) | A1 | Correct answer, must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Complete method to find total time e.g. \(-5 = 20t - \frac{1}{2}gt^2\) | M1 | Complete method, condone sign errors |
| Correct equations used e.g. \(0 = 20 - gt_1 \Rightarrow t_1 = \frac{100}{49} = 2.040816..\) | A1 | -1 for each error, allow 9.81 for \(g\) or just \(g\) |
| \(s_1 = \left(\frac{20+0}{2}\right)t_1 = \frac{1000}{49} = 20.40816...\) then \(25.408.. = \frac{1}{2}gt_2^2 \Rightarrow t_2 = 2.2771..\) | M(A)1 | Second M mark treated as A mark, -1 for each error |
| \(t = t_1 + t_2 = 4.31795..\) so \(t = 4.3\) or \(4.32\) (s) | A1 | Must have used 9.8. No isw — if correct quadratic but roots added, loses M mark |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v^2 = 20^2 - 2g \times (-3)$ | M1 | Complete method to find speed, must use 3 or -3. Allow 9.81 for $g$ or just $g$, condone sign errors |
| $v = 21$ or $21.4$ (m s$^{-1}$) | A1 | Correct answer, must be positive |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Complete method to find total time e.g. $-5 = 20t - \frac{1}{2}gt^2$ | M1 | Complete method, condone sign errors |
| Correct equations used e.g. $0 = 20 - gt_1 \Rightarrow t_1 = \frac{100}{49} = 2.040816..$ | A1 | -1 for each error, allow 9.81 for $g$ or just $g$ |
| $s_1 = \left(\frac{20+0}{2}\right)t_1 = \frac{1000}{49} = 20.40816...$ then $25.408.. = \frac{1}{2}gt_2^2 \Rightarrow t_2 = 2.2771..$ | M(A)1 | Second M mark treated as A mark, -1 for each error |
| $t = t_1 + t_2 = 4.31795..$ so $t = 4.3$ or $4.32$ (s) | A1 | Must have used 9.8. No isw — if correct quadratic but roots added, loses M mark |
---\begin{enumerate}
\item A small stone is projected vertically upwards with speed $20 \mathrm {~ms} ^ { - 1 }$ from a point $O$ which is 5 m above horizontal ground. The stone is modelled as a particle moving freely under gravity.
\end{enumerate}
Find\\
(a) the speed of the stone at the instant when it is 2 m above the ground,\\
(b) the total time between the instant when the stone is projected from $O$ and the instant when it first strikes the ground.\\
\hfill \mbox{\textit{Edexcel M1 2021 Q1 [6]}}