| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against air resistance - vertical motion |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem requiring students to calculate initial KE, final PE, find the energy lost, then use work done = force × distance to find R. Part (c) is basic modelling discussion. While it's Further Maths content, the mechanics is standard and requires only direct application of formulas with no novel problem-solving insight. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Energy Loss = KE Loss – PE Gain | M1 | For a difference in KE and PE |
| \(= \frac{1}{2} \times 0.5 \times 25^2 - 0.5g \times 20\) | A1 | For a correct expression |
| \(= 58.25 = 58 \text{ (J) or } 58.3 \text{ (J)}\) | A1 | For either 58 (2sf) or 58.3 (3sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using work-energy principle, \(20R = 58.25\) | M1 | For use of work-energy principle |
| \(R = 2.9125 = 2.9 \text{ or } 2.91\) | A1ft | For either 2.9 (2sf) or 2.91 (3sf), follow through on answer to (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Make resistance variable (dependent on speed) | B1 | For variable resistance oe |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Energy Loss = KE Loss – PE Gain | M1 | For a difference in KE and PE |
| $= \frac{1}{2} \times 0.5 \times 25^2 - 0.5g \times 20$ | A1 | For a correct expression |
| $= 58.25 = 58 \text{ (J) or } 58.3 \text{ (J)}$ | A1 | For either 58 (2sf) or 58.3 (3sf) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using work-energy principle, $20R = 58.25$ | M1 | For use of work-energy principle |
| $R = 2.9125 = 2.9 \text{ or } 2.91$ | A1ft | For either 2.9 (2sf) or 2.91 (3sf), follow through on answer to (a) |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Make resistance variable (dependent on speed) | B1 | For variable resistance oe |
---\begin{enumerate}
\item A small stone of mass 0.5 kg is thrown vertically upwards from a point A with an initial speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The stone first comes to instantaneous rest at the point B which is 20 m vertically above the point A . As the stone moves it is subject to air resistance. The stone is modelled as a particle.\\
(a) Find the energy lost due to air resistance by the stone, as it moves from A to B
\end{enumerate}
The air resistance is modelled as a constant force of magnitude $R$ newtons.\\
(b) Find the value of R .\\
(c) State how the model for air resistance could be refined to make it more realistic.
\hfill \mbox{\textit{Edexcel FM1 AS Q2 [6]}}