| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2023 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Acceleration from power and speed |
| Difficulty | Standard +0.3 This is a standard M2 work-energy-power question requiring straightforward application of P=Fv, work-energy principle, and resolving forces on an incline. Part (a) uses P=Fv to find driving force then F=ma; part (b) applies work-energy theorem with gravitational PE; part (c) finds power at constant speed on incline. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.02c Work by variable force: using integration6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(P = Fv\ \left(F = \dfrac{500}{6}\right)\) | M1 | |
| Equation of motion | M1 | Dimensionally correct. Required terms and no extras |
| \(F - 60 = 80a\) | A1 | Correct unsimplified equation in \(F\) |
| \(a = \dfrac{7}{24}\ (\text{ms}^{-2})\) | A1 | 0.29 or better (0.291666666..) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gain in \(\text{KE} = \frac{1}{2} \times 80 \times 8^2\ \text{(J)}\ (= 2560\ \text{J})\) | B1 | Any one correct (seen or implied) |
| Gain in \(\text{GPE} = 80 \times 9.8 \times 300\ \text{(J)}\ (= 235200\ \text{J})\) | B1 | A second term correct (seen or implied) |
| Work done against resistance \(= 20000 \times 60\) | \((\text{KE gain} + \text{GPE gain} = 237760\ \text{J})\) | |
| Use of *suvat* and \(F = ma\) is M0A0A0 | ||
| Expression for combined work and energy | M1 | All terms required and no double counting. Mass replaced with 80. Condone sign errors. Dimensionally correct. Condone error in zeros in 20000 |
| Total work done \(= 40\times 64 + 80\times 9.8\times 300 + 20000\times 60\) | A1 | Correct unsimplified expression for work done |
| \(1440\ \text{(kJ)}\) or \(1400\ \text{(kJ)}\) | A1 | Accept answers in joules. 3 sf or 2 sf (1437760) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion | M1 | Dimensionally correct. Required terms and no extras |
| \(F - 60 - 80g \times \sin\alpha = 0\) | A1 | Unsimplified equation in \(P\) or \(F\) with at most one error |
| \(\dfrac{P}{7} - 60 - 80g \times \dfrac{1}{20} = 0\) | A1 | Correct unsimplified equation in \(P\) |
| \(P = 694\) or \(P = 690\) | A1 | 3 sf or 2 sf only |
| Total: [4] | Overall: (13) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(A\): M0 if there is no resolving | M1 | Need all terms and no extras. Dimensionally consistent. Condone sign errors and sine/cosine confusion. |
| \(4a\cos 30°\times W + 8a\cos 30°\times\dfrac{W}{4} = 5a\cos 30°\times T\) | A1 | Correct unsimplified equation |
| \(6W = 5T \Rightarrow T = \dfrac{6}{5}W\) * | A1* | Obtain given answer from correct working, e.g. show cancelling of the common factors or some simplification of the moments equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| First equation dimensionally correct. Condone sine/cosine confusion and sign errors | M1 | e.g. Resolve horizontally |
| \(H = T\cos 30°\) | A1 | \(H = \dfrac{3\sqrt{3}}{5}W\) |
| Second equation dimensionally correct. Condone sine/cosine confusion and sign errors | M1 | e.g. Resolve vertically |
| \(V + T\cos 60° = W + \dfrac{W}{4}\) | A1 | \(V = \dfrac{13}{20}W\) |
| \(\ | R\ | = \sqrt{V^2 + H^2}\) or \(\ |
| \(\ | R\ | = \dfrac{W}{20}\sqrt{3\times144 + 169} = \dfrac{\sqrt{601}}{20}W\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation for CLM | M1 | Dimensionally correct. All terms required. Condone sign errors. |
| \(8mu - 6mu = 2my - 4mx\) giving \((u = y - 2x)\) | A1 | Correct unsimplified equation |
| Equation for kinetic energy \(\left(\frac{1}{2}\text{ or }2\text{ must be used}\right)\) | M1 | Dimensionally correct. Correct masses paired with correct velocities. All terms required. No sign errors. Condone 2 on the wrong side. |
| \(2mx^2 + my^2 = \frac{1}{2}\left(2m\times 4u^2 + m\times 9u^2\right)\) giving \((17u^2 = 4x^2 + 2y^2)\) | A1 | Correct unsimplified equation |
| Solve for \(y\): \(17u^2 = 2y^2 + (y-u)^2 \Rightarrow 3y^2 - 2yu - 16u^2 = 0\) | DM1 | Some working must be shown to obtain the quadratic in \(y\) (and \(u\)). Dependent on preceding M marks. \(\left((3y-8u)(y+2u)=0\right)\) |
| \(\Rightarrow y = \dfrac{8}{3}u\) * | A1* | Obtain given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of Impact Law: \(x + y = e\times 5u\) | M1 | Condone sign errors but must be used the right way round. |
| \(e = \dfrac{\frac{1}{2}\left(\frac{8}{3}u - u\right) + \frac{8}{3}u}{5u}\) | A1 | Correct unsimplified equation. \(\left(x = \dfrac{5u}{6}\right)\) |
| \(= \dfrac{7}{10}\) | A1 | Correct only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Velocity of \(Q\) after impact \(= f\times\dfrac{8}{3}u\) | B1 | Allow \(\pm\) |
| No collision if \(f\times\dfrac{8}{3}u \leq \dfrac{5}{6}u\), i.e. speed of \(P \geq\) speed of \(Q\) | M1 | Correct inequality with their values. Accept strict inequality. Dimensionally correct. |
| \(\Rightarrow 0 < f \leq \dfrac{5}{16}\) | A1 | Both ends required. \((0 < f \leq 0.3125)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(I = \pm 2m\left(y - \left(-\dfrac{1}{4}y\right)\right)\) | M1 | Subtraction seen or implied with their \(\frac{1}{4}y\). Requires correct mass. Requires correct impact law. |
| \(\ | I\ | = \dfrac{20}{3}mu\) |
# Question 5:
## Part 5a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $P = Fv\ \left(F = \dfrac{500}{6}\right)$ | M1 | |
| Equation of motion | M1 | Dimensionally correct. Required terms and no extras |
| $F - 60 = 80a$ | A1 | Correct unsimplified equation in $F$ |
| $a = \dfrac{7}{24}\ (\text{ms}^{-2})$ | A1 | 0.29 or better (0.291666666..) |
**Total: [4]**
## Part 5b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gain in $\text{KE} = \frac{1}{2} \times 80 \times 8^2\ \text{(J)}\ (= 2560\ \text{J})$ | B1 | Any one correct (seen or implied) |
| Gain in $\text{GPE} = 80 \times 9.8 \times 300\ \text{(J)}\ (= 235200\ \text{J})$ | B1 | A second term correct (seen or implied) |
| Work done against resistance $= 20000 \times 60$ | | $(\text{KE gain} + \text{GPE gain} = 237760\ \text{J})$ |
| **Use of *suvat* and $F = ma$ is M0A0A0** | | |
| Expression for combined work and energy | M1 | All terms required and no double counting. Mass replaced with 80. Condone sign errors. Dimensionally correct. Condone error in zeros in 20000 |
| Total work done $= 40\times 64 + 80\times 9.8\times 300 + 20000\times 60$ | A1 | Correct unsimplified expression for work done |
| $1440\ \text{(kJ)}$ or $1400\ \text{(kJ)}$ | A1 | Accept answers in joules. 3 sf or 2 sf (1437760) |
**Total: [5]**
## Part 5c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion | M1 | Dimensionally correct. Required terms and no extras |
| $F - 60 - 80g \times \sin\alpha = 0$ | A1 | Unsimplified equation in $P$ or $F$ with at most one error |
| $\dfrac{P}{7} - 60 - 80g \times \dfrac{1}{20} = 0$ | A1 | Correct unsimplified equation in $P$ |
| $P = 694$ or $P = 690$ | A1 | 3 sf or 2 sf only |
**Total: [4] | Overall: (13)**
## Question 6a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $A$: **M0 if there is no resolving** | M1 | Need all terms and no extras. Dimensionally consistent. Condone sign errors and sine/cosine confusion. |
| $4a\cos 30°\times W + 8a\cos 30°\times\dfrac{W}{4} = 5a\cos 30°\times T$ | A1 | Correct unsimplified equation |
| $6W = 5T \Rightarrow T = \dfrac{6}{5}W$ * | A1* | Obtain given answer from correct working, e.g. show cancelling of the common factors or some simplification of the moments equation |
---
## Question 6b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| First equation dimensionally correct. Condone sine/cosine confusion and sign errors | M1 | e.g. Resolve horizontally |
| $H = T\cos 30°$ | A1 | $H = \dfrac{3\sqrt{3}}{5}W$ |
| Second equation dimensionally correct. Condone sine/cosine confusion and sign errors | M1 | e.g. Resolve vertically |
| $V + T\cos 60° = W + \dfrac{W}{4}$ | A1 | $V = \dfrac{13}{20}W$ |
| $\|R\| = \sqrt{V^2 + H^2}$ or $\|R\|^2 = V^2 + H^2$ | DM1 | Correct use of Pythagoras. Dependent on two preceding M marks. |
| $\|R\| = \dfrac{W}{20}\sqrt{3\times144 + 169} = \dfrac{\sqrt{601}}{20}W$ | A1 | $1.2W$ or better $(1.22576\ldots)$ |
---
## Question 7a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation for CLM | M1 | Dimensionally correct. All terms required. Condone sign errors. |
| $8mu - 6mu = 2my - 4mx$ giving $(u = y - 2x)$ | A1 | Correct unsimplified equation |
| Equation for kinetic energy $\left(\frac{1}{2}\text{ or }2\text{ must be used}\right)$ | M1 | Dimensionally correct. Correct masses paired with correct velocities. All terms required. No sign errors. Condone 2 on the wrong side. |
| $2mx^2 + my^2 = \frac{1}{2}\left(2m\times 4u^2 + m\times 9u^2\right)$ giving $(17u^2 = 4x^2 + 2y^2)$ | A1 | Correct unsimplified equation |
| Solve for $y$: $17u^2 = 2y^2 + (y-u)^2 \Rightarrow 3y^2 - 2yu - 16u^2 = 0$ | DM1 | Some working must be shown to obtain the quadratic in $y$ (and $u$). Dependent on preceding M marks. $\left((3y-8u)(y+2u)=0\right)$ |
| $\Rightarrow y = \dfrac{8}{3}u$ * | A1* | Obtain given answer from correct working |
---
## Question 7b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of Impact Law: $x + y = e\times 5u$ | M1 | Condone sign errors but must be used the right way round. |
| $e = \dfrac{\frac{1}{2}\left(\frac{8}{3}u - u\right) + \frac{8}{3}u}{5u}$ | A1 | Correct unsimplified equation. $\left(x = \dfrac{5u}{6}\right)$ |
| $= \dfrac{7}{10}$ | A1 | Correct only |
---
## Question 7c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Velocity of $Q$ after impact $= f\times\dfrac{8}{3}u$ | B1 | Allow $\pm$ |
| No collision if $f\times\dfrac{8}{3}u \leq \dfrac{5}{6}u$, i.e. speed of $P \geq$ speed of $Q$ | M1 | Correct inequality with their values. Accept strict inequality. Dimensionally correct. |
| $\Rightarrow 0 < f \leq \dfrac{5}{16}$ | A1 | Both ends required. $(0 < f \leq 0.3125)$ |
---
## Question 7d:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $I = \pm 2m\left(y - \left(-\dfrac{1}{4}y\right)\right)$ | M1 | Subtraction seen or implied with their $\frac{1}{4}y$. Requires correct mass. Requires correct impact law. |
| $\|I\| = \dfrac{20}{3}mu$ | A1 | Or equivalent. Must be positive. $6.7mu$ or better. Condone $-\dfrac{20}{3}mu \rightarrow \dfrac{20}{3}mu$ with no explanation. |
\begin{enumerate}
\item A cyclist is travelling on a straight horizontal road and working at a constant rate of 500 W .
\end{enumerate}
The total mass of the cyclist and her cycle is 80 kg .\\
The total resistance to the motion of the cyclist is modelled as a constant force of magnitude 60 N .\\
(a) Using this model, find the acceleration of the cyclist at the instant when her speed is $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
On the following day, the cyclist travels up a straight road from a point $A$ to a point $B$.\\
The distance from $A$ to $B$ is 20 km .\\
Point $A$ is 500 m above sea level and point $B$ is 800 m above sea level.\\
The cyclist starts from rest at $A$.\\
At the instant she reaches $B$ her speed is $8 \mathrm {~ms} ^ { - 1 }$\\
The total resistance to the motion of the cyclist from non-gravitational forces is modelled as a constant force of magnitude 60 N .\\
(b) Using this model, find the total work done by the cyclist in the journey from $A$ to $B$.
Later on, the cyclist is travelling up a straight road which is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 20 }$
The cyclist is now working at a constant rate of $P$ watts and has a constant speed of $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
The total resistance to the motion of the cyclist from non-gravitational forces is again modelled as a constant force of magnitude 60 N .\\
(c) Using this model, find the value of $P$
\hfill \mbox{\textit{Edexcel M2 2023 Q5 [13]}}