| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time at height |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question requiring standard application of kinematic equations with constant acceleration. Part (a) uses v²=u²+2as at maximum height where v=0, and part (b) requires solving a quadratic equation to find two times when height equals 14.7m. Both are routine textbook exercises with no problem-solving insight needed, making this easier than average but not trivial due to the two-part structure. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0^2 = 19.6^2 - 2 \times g \times H\) | M1 | M1 for complete method for finding \(H\), equation in \(H\) only. Condone sign errors |
| \(H = 19.6\text{m}\) (20) | A1 | A1 for 19.6 or 20 correctly obtained (\(2g\) is A0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(14.7 = 19.6t - \frac{1}{2}gt^2\) | M1 A1 | First M1 for quadratic in \(t\) only (where \(t\) is time at 14.7 above \(O\)). First A1 for correct equation |
| \(t^2 - 4t + 3 = 0\) | ||
| \((t-1)(t-3) = 0\) | DM1 | Second DM1 (dependent on first M1) for solving for \(t\) |
| \(t = 1\) or \(3\); Answer \(2\) s | A1; A1 | Second A1 for both values 1 and 3. Third A1 for 2 s. N.B. Obtaining \(t=1\) at \(s=14.7\) only scores max M1 A1 |
| ALT 1: \((h - 14.7) = \frac{1}{2}gt^2\), \(t=1\), Total \(= 2 \times\) their \(1 = 2\) s | M1 A1, DM1, A1 |
| Answer | Marks |
|---|---|
| - \(v^2 = 19.6^2 - 2g \times 14.7 \Rightarrow v = \pm 9.8\) | M1 A1 |
| - EITHER: \(-9.8 = 9.8 - gT\), \(T = 2\) | DM1 A1 |
| - OR: \(0 = 9.8t - \frac{1}{2}gt^2\), \(t = 0\) or \(2\) | DM1 A1 |
## Question 2:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0^2 = 19.6^2 - 2 \times g \times H$ | M1 | M1 for complete method for finding $H$, equation in $H$ only. Condone sign errors |
| $H = 19.6\text{m}$ (20) | A1 | A1 for 19.6 or 20 correctly obtained ($2g$ is A0) |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $14.7 = 19.6t - \frac{1}{2}gt^2$ | M1 A1 | First M1 for quadratic in $t$ only (where $t$ is time at 14.7 above $O$). First A1 for correct equation |
| $t^2 - 4t + 3 = 0$ | | |
| $(t-1)(t-3) = 0$ | DM1 | Second DM1 (dependent on first M1) for solving for $t$ |
| $t = 1$ or $3$; Answer $2$ s | A1; A1 | Second A1 for both values 1 and 3. Third A1 for 2 s. N.B. Obtaining $t=1$ at $s=14.7$ only scores max M1 A1 |
**ALT 1:** $(h - 14.7) = \frac{1}{2}gt^2$, $t=1$, Total $= 2 \times$ their $1 = 2$ s | M1 A1, DM1, A1 |
**ALT 2/3:**
- $v^2 = 19.6^2 - 2g \times 14.7 \Rightarrow v = \pm 9.8$ | M1 A1 |
- EITHER: $-9.8 = 9.8 - gT$, $T = 2$ | DM1 A1 |
- OR: $0 = 9.8t - \frac{1}{2}gt^2$, $t = 0$ or $2$ | DM1 A1 |
---2. A small stone is projected vertically upwards from a point $O$ with a speed of $19.6 \mathrm {~ms} ^ { - 1 }$. Modelling the stone as a particle moving freely under gravity,
\begin{enumerate}[label=(\alph*)]
\item find the greatest height above $O$ reached by the stone,
\item find the length of time for which the stone is more than 14.7 m above $O$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2015 Q2 [7]}}