| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Modelling assumptions and refinements |
| Difficulty | Standard +0.3 This is a standard two-part mechanics question on inclined plane motion using energy methods. Part (a) requires straightforward application of conservation of energy on a smooth slope, while part (b) adds friction requiring the work-energy principle. Both parts follow textbook procedures with given values that simplify nicely (tan α = 3/4 gives sin α = 3/5, cos α = 4/5, and μ = 3/5), making calculations cleaner than typical. The question is slightly easier than average due to its routine structure and helpful numerical choices. |
| Spec | 3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(mg \times \frac{25}{6}\sin\alpha\) | B1 | Seen anywhere |
| Use of the principle of conservation of mechanical energy | M1 | Correct no. of terms, dimensionally correct, condone sign errors and sin/cos confusion. M0 for non-energy methods. Allow max M1A0A0 if \(\frac{25}{6}\) not resolved correctly in PE term |
| \(\frac{1}{2}m \times 25^2 - \frac{1}{2}mv^2 = mg \times \frac{25}{6}\sin\alpha\) | A1 | Correct equation in \(m\), \(g\), \(v\) and \(\alpha\) |
| \(v = 24\ (\text{ms}^{-1})\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve perpendicular to the plane | M1 | Correct no. of terms, dimensionally correct, condone sign errors and sin/cos confusion |
| \(R = mg\cos\alpha\) | A1 | Correct equation |
| \(F = \frac{3}{5}R\) | B1 | Seen anywhere |
| WD against friction \(= F \times \frac{25}{6}\) | B1 | Seen anywhere |
| Use of work-energy principle | M1 | M0 for non work-energy methods. Allow max M1A1A0A0 if \(\frac{25}{6}\) not resolved correctly in PE term |
| \(\frac{1}{2}m\times 25^2 - \frac{1}{2}mv^2 = mg\times\frac{25}{6}\sin\alpha + \frac{3}{5}\times mg\cos\alpha\times\frac{25}{6}\) | A1 | Equation in \(m\), \(g\), \(v\) and \(\alpha\) with at most one error. N.B. If KE terms reversed, only penalise ONCE |
| A1 | Correct equation in \(m\), \(g\), \(v\) and \(\alpha\) | |
| \(v = 23.2\) or \(23\ (\text{ms}^{-1})\) | A1 | cao |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $mg \times \frac{25}{6}\sin\alpha$ | B1 | Seen anywhere |
| Use of the principle of conservation of mechanical energy | M1 | Correct no. of terms, dimensionally correct, condone sign errors and sin/cos confusion. M0 for non-energy methods. Allow max M1A0A0 if $\frac{25}{6}$ not resolved correctly in PE term |
| $\frac{1}{2}m \times 25^2 - \frac{1}{2}mv^2 = mg \times \frac{25}{6}\sin\alpha$ | A1 | Correct equation in $m$, $g$, $v$ and $\alpha$ |
| $v = 24\ (\text{ms}^{-1})$ | A1 | cao |
---
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve perpendicular to the plane | M1 | Correct no. of terms, dimensionally correct, condone sign errors and sin/cos confusion |
| $R = mg\cos\alpha$ | A1 | Correct equation |
| $F = \frac{3}{5}R$ | B1 | Seen anywhere |
| WD against friction $= F \times \frac{25}{6}$ | B1 | Seen anywhere |
| Use of work-energy principle | M1 | M0 for non work-energy methods. Allow max M1A1A0A0 if $\frac{25}{6}$ not resolved correctly in PE term |
| $\frac{1}{2}m\times 25^2 - \frac{1}{2}mv^2 = mg\times\frac{25}{6}\sin\alpha + \frac{3}{5}\times mg\cos\alpha\times\frac{25}{6}$ | A1 | Equation in $m$, $g$, $v$ and $\alpha$ with at most one error. N.B. If KE terms reversed, only penalise ONCE |
| | A1 | Correct equation in $m$, $g$, $v$ and $\alpha$ |
| $v = 23.2$ or $23\ (\text{ms}^{-1})$ | A1 | cao |\begin{enumerate}
\item A plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$
\end{enumerate}
A particle $P$ is held at rest at a point $A$ on the plane.\\
The particle $P$ is then projected with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from $A$, up a line of greatest slope of the plane.
In an initial model, the plane is modelled as being smooth and air resistance is modelled as being negligible.
Using this model and the principle of conservation of mechanical energy,\\
(a) find the speed of $P$ at the instant when it has travelled a distance $\frac { 25 } { 6 } \mathrm {~m}$ up the plane from $A$.
In a refined model, the plane is now modelled as being rough, with the coefficient of friction between $P$ and the plane being $\frac { 3 } { 5 }$
Air resistance is still modelled as being negligible.\\
Using this refined model and the work-energy principle,\\
(b) find the speed of $P$ at the instant when it has travelled a distance $\frac { 25 } { 6 } \mathrm {~m}$ up the plane from $A$.
\hfill \mbox{\textit{Edexcel FM1 AS 2022 Q3 [12]}}