| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Energy methods on slope |
| Difficulty | Standard +0.3 This is a standard M2 energy methods question with routine application of work-energy principle on a slope. Part (a) requires basic resolution of forces (given tan α, find sin/cos, then friction = μR). Parts (b) and (c) apply work-energy in straightforward ways with no novel insight required. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 3.03v 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 concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \frac{1}{7} \times mg\cos\alpha \left(= \frac{1}{7} \times mg \times \frac{4}{5}\right)\) | M1 | Condone sin/cos confusion |
| \(= \frac{4mg}{35}\) | A1* | Obtain given answer from correct working. Correct trig value must be seen as it is a given answer – could be against the Q |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Energy equation: PE gain + WD against \(Fr\) = KE lost or equivalent | M1 | NB: The question tells them to use work-energy. Need all terms, dimensionally correct but condone sign errors. Condone sine/cosine confusion |
| \(\frac{4mgd}{35} + mgd\sin\alpha = \frac{1}{2}m \times 10ag\) or \(\frac{4mgd}{35} + mgd \times \frac{3}{5} = \frac{1}{2}m \times 10ag\) | A1 | Unsimplified equation with at most one error |
| A1 | Correct unsimplified equation | |
| \(d (= AB) = 7a\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Energy equation | M1 | NB: The question tells them to use work-energy. Need all terms, dimensionally correct but condone sign errors |
| \(\frac{4mg}{35} \times 14a = \frac{1}{2}m \times 10ag - \frac{1}{2}mV^2\) or \(\frac{4mg}{35} \times 7a = mg \times 7a \times \frac{3}{5} - \frac{1}{2}mV^2\) or \(\frac{4mg}{35} \times d = mg \times d \times \frac{3}{5} - \frac{1}{2}mV^2\) | A1ft | Unsimplified equation with at most one error, ft on their \(AB\) |
| A1ft | Correct unsimplified equation. Allow A1A1 if they have substituted for \(g\) | |
| \(V = \sqrt{\frac{34ag}{5}}\) | A1 | Accept \(2.6\sqrt{ag}\), \(\sqrt{6.8ag}\) or better. Accept \(\sqrt{\frac{170ag}{25}}\) |
# Question 4:
## Part 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \frac{1}{7} \times mg\cos\alpha \left(= \frac{1}{7} \times mg \times \frac{4}{5}\right)$ | M1 | Condone sin/cos confusion |
| $= \frac{4mg}{35}$ | A1* | Obtain **given answer** from correct working. Correct trig value must be seen as it is a given answer – could be against the Q |
## Part 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Energy equation: PE gain + WD against $Fr$ = KE lost or equivalent | M1 | NB: The question tells them to use work-energy. Need all terms, dimensionally correct but condone sign errors. Condone sine/cosine confusion |
| $\frac{4mgd}{35} + mgd\sin\alpha = \frac{1}{2}m \times 10ag$ or $\frac{4mgd}{35} + mgd \times \frac{3}{5} = \frac{1}{2}m \times 10ag$ | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation |
| $d (= AB) = 7a$ | A1 | cao |
## Part 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Energy equation | M1 | NB: The question tells them to use work-energy. Need all terms, dimensionally correct but condone sign errors |
| $\frac{4mg}{35} \times 14a = \frac{1}{2}m \times 10ag - \frac{1}{2}mV^2$ or $\frac{4mg}{35} \times 7a = mg \times 7a \times \frac{3}{5} - \frac{1}{2}mV^2$ or $\frac{4mg}{35} \times d = mg \times d \times \frac{3}{5} - \frac{1}{2}mV^2$ | A1ft | Unsimplified equation with at most one error, ft on their $AB$ |
| | A1ft | Correct unsimplified equation. Allow A1A1 if they have substituted for $g$ |
| $V = \sqrt{\frac{34ag}{5}}$ | A1 | Accept $2.6\sqrt{ag}$, $\sqrt{6.8ag}$ or better. Accept $\sqrt{\frac{170ag}{25}}$ |\begin{enumerate}
\item A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$
\end{enumerate}
A particle $P$ of mass $m$ is held at rest at a point $A$ on the plane.\\
The particle is then projected with speed $u$ up a line of greatest slope of the plane and comes to instantaneous rest at the point $B$.
The coefficient of friction between the particle and the plane is $\frac { 1 } { 7 }$\\
(a) Show that the magnitude of the frictional force acting on the particle, as it moves from $A$ to $B$, is $\frac { 4 m g } { 35 }$
Given that $u = \sqrt { 10 a g }$, use the work-energy principle\\
(b) to find $A B$ in terms of $a$,\\
(c) to find, in terms of $a$ and $g$, the speed of $P$ when it returns to $A$.
\hfill \mbox{\textit{Edexcel M2 2024 Q4 [10]}}