| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding when particle at rest |
| Difficulty | Moderate -0.8 This is a straightforward application of SUVAT equations for vertical motion under gravity. Part (a) uses v²=u²+2as with standard values, and part (b) requires solving a quadratic equation using s=ut+½at². Both parts are routine textbook exercises with no problem-solving insight required, making this easier than average but not trivial due to the algebraic manipulation needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| \(0^2 = u^2 - 2 \times 9.8 \times 40\) | M1 A1 | |
| \(u = 28 \text{ m s}^{-1}\) | A1 | GIVEN ANSWER |
| (3) |
| Answer | Marks |
|---|---|
| \(33.6 = 28t - \frac{1}{2} \times 9.8t^2\) | M1 A1 |
| \(4.9t^2 - 28t + 33.6 = 0\) | |
| \(t = \frac{28 \pm \sqrt{28^2 - 4 \times 4.9 \times 33.6}}{9.8}\) | M1 |
| \(= 4 \text{ s or } (1.7 \text{ s or } 1.71 \text{ s})\) | A1 A1 |
| (5) | |
| 8 |
## (a)
$0^2 = u^2 - 2 \times 9.8 \times 40$ | M1 A1 |
$u = 28 \text{ m s}^{-1}$ | A1 | **GIVEN ANSWER**
| (3) |
## (b)
$33.6 = 28t - \frac{1}{2} \times 9.8t^2$ | M1 A1 |
$4.9t^2 - 28t + 33.6 = 0$ | |
$t = \frac{28 \pm \sqrt{28^2 - 4 \times 4.9 \times 33.6}}{9.8}$ | M1 |
$= 4 \text{ s or } (1.7 \text{ s or } 1.71 \text{ s})$ | A1 A1 |
| (5) |
| **8** |
---At time $t = 0$ a ball is projected vertically upwards from a point $O$ and rises to a maximum height of 40 m above $O$. The ball is modelled as a particle moving freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of projection is 28 m s$^{-1}$. [3]
\item Find the times, in seconds, when the ball is 33.6 m above $O$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2011 Q1 [8]}}