| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Free fall: time or distance |
| Difficulty | Easy -1.2 This is a straightforward SUVAT question with three standard parts requiring direct application of kinematic equations. Part (a) uses v=u+at with given values, part (b) uses s=ut+½at², and part (c) uses v²=u²+2as. All parts involve single-step calculations with no problem-solving or conceptual challenges—purely routine mechanics practice below average A-level difficulty. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 0 + 1.5 \times 9.8 = 14.7 \text{ ms}^{-1}\) | M1 | Use of constant acceleration equation to find \(v\) |
| A1 | AG Correct \(v\) from correct working. \(1.5 \times 9.8 = 14.7\) is not enough on its own | |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = \frac{1}{2} \times 9.8 \times 1.5^2 = 11.0 \text{ m (to 3 sf)}\) | M1 | Use of constant acceleration equation with \(a = 9.8\) to find \(h\) |
| A1 | Correct \(h\). Allow 11 m; ignore negative signs | |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(s^2 = 0^2 + 2 \times 9.8s\) | M1 | Use of constant acceleration equation with \(u = 0\) to find \(s\) |
| A1 | Correct equation | |
| \(s = \frac{25}{19.6} = 1.28 \text{ m (to 3 sf)}\) | A1 | Correct \(s\). Accept 1.27 |
| OR | ||
| \(t = \frac{5}{9.8} = 0.510\) | ||
| \(s = \frac{1}{2}(0 + 5) \times \frac{5}{9.8} = 1.28 \text{ m}\) | ||
| OR | ||
| \(s = 0 + \frac{1}{2} \times 9.8 \times \left(\frac{5}{9.8}\right)^2 = 1.28 \text{ m}\) | ||
| Total: 3 marks |
**1(a)**
| $v = 0 + 1.5 \times 9.8 = 14.7 \text{ ms}^{-1}$ | M1 | Use of constant acceleration equation to find $v$ |
| | A1 | AG Correct $v$ from correct working. $1.5 \times 9.8 = 14.7$ is not enough on its own |
| **Total: 2 marks** | | |
**1(b)**
| $h = \frac{1}{2} \times 9.8 \times 1.5^2 = 11.0 \text{ m (to 3 sf)}$ | M1 | Use of constant acceleration equation with $a = 9.8$ to find $h$ |
| | A1 | Correct $h$. Allow 11 m; ignore negative signs |
| **Total: 2 marks** | | |
**1(c)**
| $s^2 = 0^2 + 2 \times 9.8s$ | M1 | Use of constant acceleration equation with $u = 0$ to find $s$ |
| | A1 | Correct equation |
| $s = \frac{25}{19.6} = 1.28 \text{ m (to 3 sf)}$ | A1 | Correct $s$. Accept 1.27 |
| **OR** | | |
| $t = \frac{5}{9.8} = 0.510$ | | |
| $s = \frac{1}{2}(0 + 5) \times \frac{5}{9.8} = 1.28 \text{ m}$ | | |
| **OR** | | |
| $s = 0 + \frac{1}{2} \times 9.8 \times \left(\frac{5}{9.8}\right)^2 = 1.28 \text{ m}$ | | |
| **Total: 3 marks** | | |1 A ball is released from rest at a height $h$ metres above ground level. The ball hits the ground 1.5 seconds after it is released. Assume that the ball is a particle that does not experience any air resistance.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of the ball is $14.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it hits the ground.
\item Find $h$.
\item Find the distance that the ball has fallen when its speed is $5 \mathrm {~ms} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q1 [7]}}