## Question 7:

### Part 7(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| EPE at $A = \frac{\lambda a^2}{2a}$ or EPE at $B = \frac{\lambda(2a)^2}{2a}$ | M1 | Correct method for EPE seen or implied. Need $\frac{1}{2}kx^2$ where $k = \frac{\lambda}{a}$. Must use EPE formula correctly at least once |
| Form work-energy equation | M1 | Require all terms. Dimensionally correct. Condone their EPE. Condone sign errors |
| $\frac{\lambda a^2}{2a} + mg \times 3a = \frac{\lambda(2a)^2}{2a}$ $\left(\frac{\lambda a}{2} + 3mga = 2\lambda a\right)$ | A1 | Unsimplified equation with at most one error. A repeated error in EPE formula is one error |
| | A1 | Correct unsimplified equation |
| $3mg = \frac{3\lambda}{2} \Rightarrow \lambda = 2mg$ * | A1* | Obtain **given answer** from correct working |

**(5 marks)**

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### Part 7(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Extension at equilibrium: $\frac{2mgx}{a} = mg \Rightarrow x = \frac{a}{2}$ * | M1 | Use correct method for tension to find extension at equilibrium. Need to see formula for tension **used**. Allow verification with appropriate conclusion. If SHM used must use $F = ma$ to prove $P$ is moving with SHM, otherwise 0/2 |
| | A1* | Correct answer from correct working. Allow verification **with appropriate conclusion** |
| **Alternative:** Use work-energy equation to obtain $\frac{dV^2}{dx}$ and set derivative equal to zero | M1 | Or equivalent method for finding turning point of a quadratic |
| $\frac{1}{a} \times 2x - 1 = 0 \Rightarrow x = \frac{a}{2}$ | A1 | Correct answer from correct working |
| Use work-energy equation to find max speed | M1 | Use given $x$ to form work-energy equation. Need all terms, dimensionally correct. Condone sign errors. Accept with $\lambda$ and $x$ not substituted |
| $\frac{2mgx^2}{2a} + mg(2a - x) + \frac{1}{2}mV^2 = \frac{2mg(2a)^2}{2a}$ | A1 | Unsimplified equation with at most one error. Need given $\lambda$ and given $x$ substituted at some point. Repeated error in EPE formula is one error |
| $\left(\frac{ag}{4} + \frac{3ag}{2} + \frac{1}{2}V^2 = 4ag\right)$ | A1 | Correct unsimplified equation with given $\lambda$ and $x$ substituted |
| $V = 3\sqrt{\frac{ag}{2}}$ | A1 | Use correct method for tension to find equilibrium extension. Any equivalent form. $2.1\sqrt{ag}$ or better |

**(6 marks)**

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### Part 7(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Any valid response, e.g.: the GPE of the system changes; need to include KE of the spring in energy equation; need to include GPE of the spring in energy equation; the extension would be different at equilibrium; it has weight; there would be tension in the spring as well; the velocity would decrease as energy is converted; the mass of the spring would drag down and the EPE would change | B1 | Any valid response. B0 if answer includes an additional incorrect factor. Must be specific (e.g. not just "the GPE changes" but "the GPE of the system changes"). Must relate to an effect on the energy equation |

**(1 mark)**

**(Total 12 marks)**

## Question 8a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = 6\mathbf{i} + \ldots$ | B1 | AO3.4 |
| $\ldots + 8e\mathbf{j}$ | B1 | AO3.4 |
| Impact with $ST \Rightarrow \frac{8e}{6} < \frac{1}{2}$, $0 < e < \frac{3}{8}$ | B1 | AO3.1b — Use direction to determine range for $e$ (could come via $e\tan\alpha = \tan\beta < \frac{1}{2}$) |

**Total: 3 marks**

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## Question 8b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Perpendicular to $ST$: direction $\pm\mu(-\mathbf{i}+2\mathbf{j})$ | B1 | AO1.2 — Correct vector perpendicular to $ST$ seen or implied; $\mu$ can have any scalar value |
| Component parallel to $ST$: $(6\mathbf{i}+2\mathbf{j})\cdot\lambda(2\mathbf{i}+\mathbf{j})$ | M1 | AO3.1b — Use scalar product to find component of $\mathbf{v}$ parallel to $ST$; $\lambda$ can have any scalar value |
| $= \left((6\mathbf{i}+2\mathbf{j})\cdot\frac{1}{\sqrt{5}}(2\mathbf{i}+\mathbf{j})\right)\frac{1}{\sqrt{5}}(12+2)$ | A1 | AO1.1b — Correct unsimplified expression for the magnitude |
| Component perpendicular to $ST$: $\pm\left(\frac{1}{2}(6\mathbf{i}+2\mathbf{j})\cdot\gamma(-\mathbf{i}+2\mathbf{j})\right)$ | M1 | AO3.4 — Use scalar product and impact law perpendicular to $ST$; must clearly be using $e=\frac{1}{2}$; $\gamma$ can have any scalar value |
| $= \frac{1}{2\sqrt{5}}(-6+4)$ | A1 | AO1.1b — Correct unsimplified expression for perpendicular component; allow $\pm$ |
| $\mathbf{w} = \frac{14}{\sqrt{5}}\frac{1}{\sqrt{5}}(2\mathbf{i}+\mathbf{j}) + \frac{1}{\sqrt{5}}\frac{1}{\sqrt{5}}(-\mathbf{i}+2\mathbf{j})$ | M1 | AO3.1b — Combine magnitudes and directions to obtain velocity; perpendicular should now be in correct direction |
| $\mathbf{w} = \left(\frac{28}{5}-\frac{1}{5}\right)\mathbf{i}+\left(\frac{14}{5}+\frac{2}{5}\right)\mathbf{j} = \left(\frac{27}{5}\mathbf{i}+\frac{16}{5}\mathbf{j}\right)$ or $(5.4\mathbf{i}+3.2\mathbf{j})$ (m s$^{-1}$) | A1 | AO2.2a — Correct simplified velocity |

**Total: 7 marks**

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## Question 8b alt 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Perpendicular to $ST$: direction $\pm\mu(-\mathbf{i}+2\mathbf{j})$ | B1 | AO1.2 — $\mu$ can have any scalar value |
| $\mathbf{w} = a\mathbf{i}+b\mathbf{j} \Rightarrow (6\mathbf{i}+2\mathbf{j})\cdot(2\mathbf{i}+\mathbf{j}) = (a\mathbf{i}+b\mathbf{j})\cdot(2\mathbf{i}+\mathbf{j})$ | M1 | — Correct method for component parallel to $ST$ |
| $14 = 2a+b$ | A1 | — Correct equation in $a$ and $b$ |
| $\pm\frac{1}{2}(6\mathbf{i}+2\mathbf{j})\cdot(-\mathbf{i}+2\mathbf{j}) = (a\mathbf{i}+b\mathbf{j})\cdot(-\mathbf{i}+2\mathbf{j})$ | M1 | — Correct method for component perpendicular to $ST$; allow $\pm$ for their perpendicular vector |
| $2b - a = \pm 1$ | A1 | — Correct equation in $a$ and $b$ |
| Solve simultaneous equations for $a$ and $b$ | M1 | — Using correct direction for perpendicular component |
| $\mathbf{w} = \left(\frac{27}{5}\mathbf{i}+\frac{16}{5}\mathbf{j}\right)$ or $(5.4\mathbf{i}+3.2\mathbf{j})$ (m s$^{-1}$) | A1 | — Correct simplified answer |

**Total: 7 marks**

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## Question 8b alt 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Perpendicular to $ST$: direction $\pm\mu(-\mathbf{i}+2\mathbf{j})$ | B1 | AO1.2 — $\mu$ can have any scalar value |
| $\mathbf{v} = 6\mathbf{i}+2\mathbf{j} = p(2\mathbf{i}+\mathbf{j})+q(-\mathbf{i}+2\mathbf{j})$ | M1 | AO3.1b — Split $\mathbf{v}$ into components parallel and perpendicular to $ST$ |
| $6 = 2p-q$, $2 = p+2q$ giving $p=\frac{14}{5}$, $q=\frac{-2}{5}$ | A1 | AO1.1b — Two equations in $p$ and $q$ |
| Component perpendicular to $ST$: $\pm\frac{1}{2}\times q(-\mathbf{i}+2\mathbf{j})$ | M1 | AO3.4 — Use impact law perpendicular to $ST$ for their perpendicular vector |
| $\pm\frac{1}{2}\times q(-\mathbf{i}+2\mathbf{j})$ | A1 | AO1.1b — Correct unsimplified perpendicular component; with $q$ or their $q$ |
| Solve for $p$ and $q$ to obtain velocity $\mathbf{w} = \frac{14}{5}(2\mathbf{i}+\mathbf{j})+\frac{1}{2}\times\frac{2}{5}(-\mathbf{i}+2\mathbf{j})$ | M1 | AO3.1b — Using correct direction for perpendicular component |
| $\mathbf{w} = \left(\frac{27}{5}\mathbf{i}+\frac{16}{5}\mathbf{j}\right)$ or $(5.4\mathbf{i}+3.2\mathbf{j})$ (m s$^{-1}$) | A1 | AO2.2a — Correct simplified total |

**Total: 7 marks**

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## Question 8b alt 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha - \beta = 8.1\ldots°$ | B1 | — $\sin(\alpha-\beta)=\frac{1}{\sqrt{50}}$, $\cos(\alpha-\beta)=\frac{7}{\sqrt{50}}$, $\tan(\alpha-\beta)=\frac{1}{7}$ seen or implied |
| Component of $\mathbf{w}$ parallel to $ST$ is $|\mathbf{v}|\cos(\alpha-\beta)$ | M1 | — Correct use of their $|\mathbf{v}|$ and their $\alpha-\beta$ |
| $= \sqrt{40}\cos(\alpha-\beta) = \sqrt{40}\times\frac{7}{\sqrt{50}} = 6.26\ldots$ | A1 | — Correct unsimplified |
| Component of $\mathbf{w}$ perpendicular to $ST$ is $\frac{1}{2}|\mathbf{v}|\sin(\alpha-\beta)$ | M1 | — Correct use of $\frac{1}{2}$, their $|\mathbf{v}|$ and their $\alpha-\beta$ |
| $\pm\frac{1}{2}\times\sqrt{40}\sin(\alpha-\beta) = \frac{\sqrt{40}}{2}\times\frac{1}{\sqrt{50}} = 0.447\ldots$ | A1 | — Correct unsimplified |
| $\mathbf{w} = |\mathbf{w}|\cos(\alpha+\theta)\mathbf{i} + |\mathbf{w}|\sin(\alpha+\theta)\mathbf{j}$ | M1 | — Use of Pythagoras and correct method for $\theta+\alpha$; $\cos(\alpha+\theta)=\frac{27}{\sqrt{5}\sqrt{197}}$, $\sin(\alpha+\theta)=\frac{16}{\sqrt{5}\sqrt{197}}$, $\alpha+\theta=30.65°$ |
| $\mathbf{w} = \left(\frac{27}{5}\mathbf{i}+\frac{16}{5}\mathbf{j}\right)$ or $(5.4\mathbf{i}+3.2\mathbf{j})$ (m s$^{-1}$) | A1 | — Correct simplified total |

**Total: 7 marks**