| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Projectile energy - finding speed or height |
| Difficulty | Moderate -0.3 This is a standard M2 projectile motion question with energy calculations. Parts (a)-(b) use routine SUVAT equations for projectile motion, (c) is direct PE calculation, (d) applies energy conservation, and (e) asks for standard modelling assumptions. All techniques are textbook exercises requiring no novel insight, though the multi-part structure and energy method in part (d) elevate it slightly above pure recall. |
| Spec | 3.02i Projectile motion: constant acceleration model6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| \(7 = \frac{1}{2}gt^2\) | M1 A1 |
| \(l^2 = 14 + 9.8\) | |
| \(t = 1.195\) | |
| In \(1.195\) s, stone travels \(4 \times 1.195 = 4.78\) m | M1 A1 |
| Answer | Marks |
|---|---|
| When \(x = 2.39\), \(t = 0.598\) | M1 A1 M1 A1 |
| \(y = 7 - \frac{1}{2}gt^2 = 5.25\) m |
| Answer | Marks |
|---|---|
| \(mgh = 1.5 \times 9.8 \times 7 = 102.9\) J | M1 A1 |
| Answer | Marks |
|---|---|
| \(\frac{1}{2}mv^2 = mgh\) | M1 A1 M1 A1 |
| \(v = \sqrt{(2gh)} = \sqrt{(14g)} = 11.7 \, \text{ms}^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Modelled stone as particle, ignored air resistance, etc. | B1 B1 | Total: 15 marks |
### (a)
$7 = \frac{1}{2}gt^2$ | M1 A1 |
$l^2 = 14 + 9.8$ | |
$t = 1.195$ | |
In $1.195$ s, stone travels $4 \times 1.195 = 4.78$ m | M1 A1 |
### (b)
When $x = 2.39$, $t = 0.598$ | M1 A1 M1 A1 |
$y = 7 - \frac{1}{2}gt^2 = 5.25$ m | |
### (c)
$mgh = 1.5 \times 9.8 \times 7 = 102.9$ J | M1 A1 |
### (d)
$\frac{1}{2}mv^2 = mgh$ | M1 A1 M1 A1 |
$v = \sqrt{(2gh)} = \sqrt{(14g)} = 11.7 \, \text{ms}^{-1}$ | |
### (e)
Modelled stone as particle, ignored air resistance, etc. | B1 B1 | **Total: 15 marks**A stone, of mass 1.5 kg, is projected horizontally with speed 4 ms$^{-1}$ from a height of 7 m above horizontal ground.
\begin{enumerate}[label=(\alph*)]
\item Show that the stone travels about 4.78 m horizontally before it hits the ground. [4 marks]
\item Find the height of the stone above the ground when it has travelled half of this horizontal distance. [4 marks]
\item Calculate the potential energy lost by the stone as it moves from its point of projection to the ground. [2 marks]
\item Showing your method clearly, use your answer to part (c) to find the speed with which the stone hits the ground. [3 marks]
\item State two modelling assumptions that you have made in answering this question. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q8 [15]}}