Edexcel M2 2023 October — Question 5 13 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2023
SessionOctober
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeAcceleration from power and speed
DifficultyStandard +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.
Spec3.03d Newton's second law: 2D vectors6.02c Work by variable force: using integration6.02l Power and velocity: P = Fv

  1. A cyclist is travelling on a straight horizontal road and working at a constant rate of 500 W .
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 .
  1. 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 .
  2. 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 .
  3. Using this model, find the value of \(P\)

Question 5:
Part 5a:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Use of \(P = Fv\ \left(F = \dfrac{500}{6}\right)\)M1
Equation of motionM1 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:
AnswerMarks Guidance
Answer/WorkingMark 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 energyM1 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:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Equation of motionM1 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:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Moments about \(A\): M0 if there is no resolvingM1 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:
AnswerMarks Guidance
Answer/WorkingMark Guidance
First equation dimensionally correct. Condone sine/cosine confusion and sign errorsM1 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 errorsM1 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 \(\
\(\R\ = \dfrac{W}{20}\sqrt{3\times144 + 169} = \dfrac{\sqrt{601}}{20}W\)
Question 7a:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Equation for CLMM1 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:
AnswerMarks Guidance
Answer/WorkingMark 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:
AnswerMarks Guidance
Answer/WorkingMark 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:
AnswerMarks Guidance
Answer/WorkingMark 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]}}