| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: max height |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question testing standard vertical projection formulas. Part (a) uses v=u+at with v=0, part (b) uses v²=u²+2as (bookwork), and part (c) applies s=ut+½at² with s=-3H. All steps are routine applications of memorized equations with no problem-solving insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Max ht: \(v=0\). \(v = u - gt \Rightarrow T = \frac{u}{g}\) | M1A1 | M1 for use of suvat to obtain equation in \(T\), \(u\) and \(g\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Max ht \(H = ut + \frac{1}{2}at^2 = \frac{u^2}{g} - \frac{u^2}{2g} = \frac{u^2}{2g}\) | M1A1 | M1 for suvat to obtain equation in \(H\), \(u\) and \(g\) only. A1 for \(H = \frac{u^2}{2g}\) correctly obtained (given answer). Or use of \(v^2 = u^2 + 2as\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(-3 \times \frac{u^2}{2g} = ut - \frac{1}{2}gt^2\) | M1 | First M1 for complete method to find total time in terms of \(u\), \(g\), \(H\) or \(T\): e.g. \(3H = -ut + \frac{1}{2}gt^2\), or \(4H = \frac{1}{2}gt^2\) and \(t+T\), or \(v^2 = u^2 + 6gH\) and \(v = -u+gt\) |
| \(-3u^2 = 2ugt - g^2t^2\) | ||
| \(g^2t^2 - 2ugt - 3u^2 = 0\), \(\quad gt = \frac{2u \pm \sqrt{4u^2+12u^2}}{2}\) | DM1 A1 | Second M1 dependent on first M1, for producing expression for total time by solving quadratic |
| \(t = \frac{3u}{g} = 3T\) | A1 | Second A1 for \(3T\) cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(-4H = -\frac{1}{2}gt^2\) | M1 | |
| Total time \(= T + \sqrt{\frac{8H}{g}} = T + \sqrt{\frac{8u^2}{2g^2}}\) | DM1 A1 | |
| \(= T + 2T = 3T\) | A1 |
# Question 4:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| Max ht: $v=0$. $v = u - gt \Rightarrow T = \frac{u}{g}$ | M1A1 | M1 for use of suvat to obtain equation in $T$, $u$ and $g$ only |
## Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| Max ht $H = ut + \frac{1}{2}at^2 = \frac{u^2}{g} - \frac{u^2}{2g} = \frac{u^2}{2g}$ | M1A1 | M1 for suvat to obtain equation in $H$, $u$ and $g$ only. A1 for $H = \frac{u^2}{2g}$ correctly obtained (**given answer**). Or use of $v^2 = u^2 + 2as$ |
## Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| $-3 \times \frac{u^2}{2g} = ut - \frac{1}{2}gt^2$ | M1 | First M1 for complete method to find total time in terms of $u$, $g$, $H$ or $T$: e.g. $3H = -ut + \frac{1}{2}gt^2$, or $4H = \frac{1}{2}gt^2$ and $t+T$, or $v^2 = u^2 + 6gH$ and $v = -u+gt$ |
| $-3u^2 = 2ugt - g^2t^2$ | | |
| $g^2t^2 - 2ugt - 3u^2 = 0$, $\quad gt = \frac{2u \pm \sqrt{4u^2+12u^2}}{2}$ | DM1 A1 | Second M1 dependent on first M1, for producing expression for total time by solving quadratic |
| $t = \frac{3u}{g} = 3T$ | A1 | Second A1 for $3T$ cso |
### Part (c) Alternative
| Working | Marks | Guidance |
|---------|-------|----------|
| $-4H = -\frac{1}{2}gt^2$ | M1 | |
| Total time $= T + \sqrt{\frac{8H}{g}} = T + \sqrt{\frac{8u^2}{2g^2}}$ | DM1 A1 | |
| $= T + 2T = 3T$ | A1 | |
---\begin{enumerate}
\item At time $t = 0$, a particle is projected vertically upwards with speed $u$ from a point $A$. The particle moves freely under gravity. At time $T$ the particle is at its maximum height $H$ above $A$.\\
(a) Find $T$ in terms of $u$ and $g$.\\
(b) Show that $H = \frac { u ^ { 2 } } { 2 g }$
\end{enumerate}
The point $A$ is at a height $3 H$ above the ground.\\
(c) Find, in terms of $T$, the total time from the instant of projection to the instant when the particle hits the ground.\\
\hfill \mbox{\textit{Edexcel M1 2014 Q4 [8]}}