| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Moderate -0.3 This is a straightforward energy conservation problem requiring a single application of PE = KE. Students must find the vertical height change (1.2cos20° - 1.2 = 1.2(cos20° - 1)) and equate mgh to ½mv². While it involves a pendulum setup, the calculation is routine for Further Maths mechanics with no problem-solving insight required, making it slightly easier than average. |
| Spec | 6.02i Conservation of energy: mechanical energy principle |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2} \times 0.5v^2 = 0.5 \times 10 \times 1.2(1-\cos 20°)\) | M1 | Forms equation equating GPE and KE. Do not allow constant acceleration equations. |
| Obtains a correct equation | A1 | |
| \(v = \sqrt{1.447} = 1.203 = 1 \text{ ms}^{-1}\) (1 sf) | A1 | Answer to 1 sf. Condone AWRT 1.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Recognises maximum tension at lowest point | M1 | Forms equation for tension at lowest point using speed from (a) |
| \(T - 0.5 \times 10 = 0.5 \times \frac{1.447}{1.2}\) | A1 | Obtains correct equation |
| \(T = 5.60 = 6\text{ N}\) (1 sf) | A1 | Answer to 1 sf. Condone AWRT 5.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| States sphere will not touch chin again | B1 | |
| No because air resistance will be present and the sphere will not return to its original position | E1F | Follow through if they say ball does touch chin again |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2} \times 0.5v^2 = 0.5 \times 10 \times 1.2(1-\cos 20°)$ | M1 | Forms equation equating GPE and KE. Do not allow constant acceleration equations. |
| Obtains a correct equation | A1 | |
| $v = \sqrt{1.447} = 1.203 = 1 \text{ ms}^{-1}$ (1 sf) | A1 | Answer to 1 sf. Condone AWRT 1.2 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Recognises maximum tension at lowest point | M1 | Forms equation for tension at lowest point using speed from (a) |
| $T - 0.5 \times 10 = 0.5 \times \frac{1.447}{1.2}$ | A1 | Obtains correct equation |
| $T = 5.60 = 6\text{ N}$ (1 sf) | A1 | Answer to 1 sf. Condone AWRT 5.6 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| States sphere will not touch chin again | B1 | |
| No because air resistance will be present and the sphere will not return to its original position | E1F | Follow through if they say ball does touch chin again |
---4 In this question use $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
An inelastic string has length 1.2 metres.\\
One end of the string is attached to a fixed point $O$.\\
A sphere, of mass 500 grams, is attached to the other end of the string.\\
The sphere is held, with the string taut and at an angle of $20 ^ { \circ }$ to the vertical, touching the chin of a student, as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{f2470caa-0f73-4ec1-b08f-525c02ed2e67-04_739_511_676_762}
The sphere is released from rest.\\
Assume that the student stays perfectly still once the sphere has been released.\\
4 (a) Calculate the maximum speed of the sphere.\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
\multicolumn{3}{|l|}{4 (c)} \\
\hline
\end{tabular}
\end{center}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2019 Q4 [8]}}