| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Projectile energy - basic KE/PE calculation |
| Difficulty | Moderate -0.3 This is a straightforward energy conservation problem with standard mechanics calculations. Part (a) is direct KE formula application, part (b) uses basic conservation of energy (KE + PE), and part (c) calculates energy difference. All parts follow routine A-level mechanics procedures with no novel problem-solving required, making it slightly easier than average despite being Further Maths content. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| KE = 1825·2 (J) or \(\frac{9126}{5}\) | M1, A1[2] | Used; cao |
| Answer | Marks |
|---|---|
| Using expression for PE or KE | M1 |
| Answer | Marks |
|---|---|
| PE = \(60g(10)\) (= 600\(g\) = 5880 J) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| KE = \(\frac{1}{2}(60)v^2\) (= 30\(v^2\)) | A1 | |
| Conservation of energy | M1 | Used, all terms, allow sign errors |
| Answer | Marks | Guidance |
|---|---|---|
| (7705·2 = 30\(v^2\)) | A1 | All correct, oe; FT KE from (a) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 16 \cdot 0262 \ldots \approx 16\) (ms\(^{-1}\)) | A1 | Convincing, cso |
| Answer | Marks | Guidance |
|---|---|---|
| Work-energy principle | M1 | Used, all terms, allow sign errors |
| Answer | Marks | Guidance |
|---|---|---|
| (7705·2 = 5070 + \(E\)) | A1 | All correct, oe; FT KE from (a); FT PE from (b) |
| \(E_{lost} = 2635 \cdot 2\) (J) or \(\frac{13176}{5}\) | A1 | FT their KE and PE |
| Answer | Marks | Guidance |
|---|---|---|
| Taking a difference in KE | (M1) | At least one \(v^2\) correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(E_{lost} = 2635 \cdot 2\) (J) or \(\frac{13176}{5}\) | (A1) | All correct, oe; Accept \(\frac{1}{2}(60)(16)^2 = 7680\) |
| (A1) | \(E_{lost} = 2610\) (J) for \(v = 16\) |
## Part (a)
Using KE = $\frac{1}{2}mv^2$ with $m = 60$, $v = 7 \cdot 8$
KE = $\frac{1}{2}(60)(7 \cdot 8)^2$
KE = 1825·2 (J) or $\frac{9126}{5}$ | M1, A1[2] | Used; cao
## Part (b)
Using expression for PE or KE | M1 |
At start (platform),
PE = $60g(10)$ (= 600$g$ = 5880 J) | A1 |
At end (water),
KE = $\frac{1}{2}(60)v^2$ (= 30$v^2$) | A1 |
Conservation of energy | M1 | Used, all terms, allow sign errors
1825·2 + 5880 = 30$v^2$
(7705·2 = 30$v^2$) | A1 | All correct, oe; FT KE from (a)
$v^2 = 256 \cdot 84$ or $\frac{6421}{25}$
$v = 16 \cdot 0262 \ldots \approx 16$ (ms$^{-1}$) | A1 | Convincing, cso
**[6]**
## Part (c)
Work-energy principle | M1 | Used, all terms, allow sign errors
1825·2 + 5880 = $\frac{1}{2}(60)(13)^2 + E_{lost}$
(7705·2 = 5070 + $E$) | A1 | All correct, oe; FT KE from (a); FT PE from (b)
$E_{lost} = 2635 \cdot 2$ (J) or $\frac{13176}{5}$ | A1 | FT their KE and PE
**[3]**
### Alternative Solution
Taking a difference in KE | (M1) | At least one $v^2$ correct
$E_{lost} = \frac{1}{60}\left(\frac{6421}{25}\right) - \frac{1}{2}(60)(13)^2$
$E_{lost} = 2635 \cdot 2$ (J) or $\frac{13176}{5}$ | (A1) | All correct, oe; Accept $\frac{1}{2}(60)(16)^2 = 7680$
| (A1) | $E_{lost} = 2610$ (J) for $v = 16$
**[(3)]**
**Total for Question 2: 11**
---The diagram below shows a woman standing at the end of a diving platform. She is about to dive into the water below.
The woman has mass 60 kg and she may be modelled as a particle positioned at the end of the platform which is 10 m above the water.
\includegraphics{figure_2}
When the woman dives, she projects herself from the platform with a speed of $7.8\text{ ms}^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find the kinetic energy of the woman when she leaves the platform. [2]
\item Initially, the situation is modelled ignoring air resistance. By using conservation of energy, show that the model predicts that the woman enters the water with an approximate speed of $16\text{ ms}^{-1}$. [6]
\item Suppose that this model is refined to include air resistance so that the speed with which the woman enters the water is now predicted to be $13\text{ ms}^{-1}$.
Determine the amount of energy lost to air resistance according to the refined model. [3]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2022 Q2 [11]}}