| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.8 This is a straightforward three-part SUVAT question requiring direct application of standard kinematic equations with constant acceleration (g = 9.8 m/s²). Part (a) uses v² = u² + 2as to find maximum height, part (b) applies the same equation for the full descent, and part (c) uses v = u + at. All parts follow standard textbook procedures with no problem-solving insight required, making it easier than average but not trivial due to the multi-step nature and need for careful sign conventions. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \((\uparrow)v^2 = u^2 + 2as\) | M1A1 | |
| \(0 = 14.7^2 - 2x 9.8 x s\) | A1 | |
| \(s = 11.025\) (or 11 or 11.0 or 11.03) m | A1ft | |
| Height is 60 m or 60.0 m ft | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \((\downarrow)w^2 = u^2 + 2as\) | M1 A1 | |
| \(v^2 = (-14.7)^2 + 2x 9.8 x 49\) | A1 | |
| \(v = 34.3\) or 34 m s\(^{-1}\) | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \((\uparrow)v = u + at\) OR \((\downarrow)s = ut + \frac{1}{2}at^2\) | M1 A1 | |
| \(34.3 = -14.7 + 9.8t\) | ||
| \(t = 5\) | ||
| \(49 = -14.7t + 4.9t^2\) | ||
| \(t = 5\) | (3) [10] |
## (a)
$(\uparrow)v^2 = u^2 + 2as$ | M1A1 |
$0 = 14.7^2 - 2x 9.8 x s$ | A1 |
$s = 11.025$ (or 11 or 11.0 or 11.03) m | A1ft |
Height is 60 m or 60.0 m ft | | (4)
## (b)
$(\downarrow)w^2 = u^2 + 2as$ | M1 A1 |
$v^2 = (-14.7)^2 + 2x 9.8 x 49$ | A1 |
$v = 34.3$ or 34 m s$^{-1}$ | | (3)
## (c)
$(\uparrow)v = u + at$ OR $(\downarrow)s = ut + \frac{1}{2}at^2$ | M1 A1 |
$34.3 = -14.7 + 9.8t$ | |
$t = 5$ | |
$49 = -14.7t + 4.9t^2$ | |
$t = 5$ | | (3) [10]A ball is projected vertically upwards with a speed of 14.7 m s$^{-1}$ from a point which is 49 m above horizontal ground. Modelling the ball as a particle moving freely under gravity, find
\begin{enumerate}[label=(\alph*)]
\item the greatest height, above the ground, reached by the ball,
[4]
\item the speed with which the ball first strikes the ground,
[3]
\item the total time from when the ball is projected to when it first strikes the ground.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q6 [10]}}