# Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve horizontally or vertically | M1 | Allow without friction $= \mu R$ |
| $\mu R = N$ or $W = R + \frac{1}{3}N$ | A1 | With coefficient(s) of friction. Condone $Wg$ |
| Take moments about $A$ or $B$ | M1 | All terms required but condone sign errors and sin/cos confusion. Terms must be resolved. |
| $M(A): 2lN\sin\theta + 2l\frac{N}{3}\cos\theta = Wl\cos\theta$ | A2 | $-1$ each error. Could be in terms of $Fs$. $-1$ if $Wg$ in place of $W$. Any friction force used should be acting in the right direction. Mark the equation, not what they have called it. |
| $M(B): 2l\cos\theta R = Wl\cos\theta + \mu R 2l\sin\theta$ | | |
| $\frac{10}{3}N + \frac{2}{3}N = W$ or $2R = W + 2\mu R \times \frac{5}{3}$ | M1 | Use $\tan\theta = \frac{5}{3}$ (substitute values for the trig ratios) |
| $\Rightarrow 4N = W \Rightarrow 4N - R = \frac{1}{3}N$ | DM1 | Equation in $N$ and $R$ (eliminate one unknown). Dependent on the moments equation |
| $\frac{11}{3}\mu R = R$ | DM1 | Solve for $\mu$. Dependent on the moments equation |
| $\mu = \frac{3}{11} (\approx 0.273)$ | A1 | $0.27$ or better |

**NB:** If $\mu$ and $\frac{1}{3}$ are used the wrong way round the candidate loses the first A1 and the final A1.

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## Question 4 Alt 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve horizontally or vertically | M1 | Allow without friction $= \mu R$ |
| $\mu R = N$ or $W = R + \frac{1}{3}N$ | A1 | With coefficient(s) of friction |
| Take moments about $A$ or $B$ | M1 | All terms required but condone sign errors and sin/cos confusion. Terms must be resolved. |
| $M(A): 2lN\sin\theta + 2l\frac{N}{3}\cos\theta = Wl\cos\theta$ | A2 | $-1$ each error. Could be in terms of $Fs$. $-1$ if $Wg$ used. Mark the equation, not what they have called it. Any friction force used should be acting in the right direction. For this method they need two moments equations – allows the marks for their best equation. |
| $M(B): 2l\cos\theta R = Wl\cos\theta + \mu R 2l\sin\theta$ | | |
| $2lN\sin\theta + 2l\frac{N}{3}\cos\theta = 2l\cos\theta R - \mu R 2l\sin\theta$ | DM1 | Use two moments equations to eliminate $W$. Dependent on the moments equation |
| Use of $\tan\theta$: $2\mu \times \frac{5}{3} + \frac{2}{3}\mu = 2 - 2\mu \times \frac{5}{3}$ | M1 | Substitute for the trig ratios |
| Solve for $\mu$: $\left(\frac{20}{3} + \frac{2}{3}\right)\mu = 2$ | DM1 | Dependent on the moments equation |
| $\mu = \frac{3}{11} (\approx 0.273)$ | A1 | $0.27$ or better |

---

## Question 4 Alt 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolving horizontally or vertically | M1 | Allow without friction $= \mu R$ |
| $\mu R = N$ or $W = R + \frac{1}{3}N$ | A1 | With coefficient(s) of friction (condone $Wg$) |
| $l\cos\theta \times R = l\cos\theta \times \frac{1}{3}N + l\sin\theta \times N + l\sin\theta \times \mu R$ | M1 | Moments about centre of rod. All terms required. Terms must be resolved. Condone sign errors and sin/cos confusion. Allow without friction $= \frac{1}{3}N$. Any friction force used should be acting in the right direction. |
| | A2 | $-1$ each error. Could be in terms of $Fs$. $-1$ if $Wg$ used. |
| $l\cos\theta \times R = l\cos\theta \times \frac{1}{3}\mu R + l\sin\theta \times \mu R + l\sin\theta \times \mu R$ | DM1 | Obtain an equation in $\mu$ and $\theta$: $\left(\cos\theta = \cos\theta \times \frac{1}{3}\mu + \sin\theta \times \mu + \sin\theta \times \mu\right)$. Dependent on the moments equation |
| $\cos\theta\left(1 - \frac{1}{3}\mu\right) = 2\mu\sin\theta \Rightarrow \tan\theta = \frac{1-\frac{1}{3}\mu}{2\mu} = \frac{5}{3}$ | M1 | Use of $\tan\theta$ (substitute values for the trig ratios) |
| Solve for $\mu$: $10\mu = 3 - \mu$ | DM1 | Dependent on the moments equation |
| $\mu = \frac{3}{11} (\approx 0.273)$ | A1 | $0.27$ or better |

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# Question 5a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Max friction $= \mu \times 10g\cos\alpha$ | B1 | Use of $\mu R$ seen or implied $(\mu \times 90.46)$ |
| Work done against friction $= 6.5 \times 10g\mu\cos\alpha\ (= 245)$ | M1 | For $6.5 \times$ their $F$ |
| Equation in $\mu$: $6.5 \times 10g\mu \times \frac{12}{13} = 245$ | DM1 | Dependent on preceding M1 |
| | A1 | Correct substituted equation |
| $\mu = 0.417$ or $0.42$ | A1 | Do not accept $\frac{5}{12}$ (cannot have an exact value following the use of $9.8$). Use of $9.81$ is a rubric infringement, so A0 if seen. |

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# Question 5b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\times 10 \times 11.5^2 - 245 - 10g \times 6.5\sin\alpha = \frac{1}{2}\times 10 \times v^2$ or equivalent | M1 | Must be using work-energy equation. All terms required. Condone sign errors and sin/cos confusion. |
| | A2 | $-1$ each error $(661.25 - 245 - 245 = 171.25)$ |
| $v = 5.85\ (5.9)\ (\text{m s}^{-1})$ | A1 | |

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# Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| At rest when $v=0$: $(2t^2 - 9t + 4) = 0$ | M1 | |
| $= (2t-1)(t-4)$ | DM1 | Solve for $t$. Dependent on the previous M1 |
| $t = \frac{1}{2},\ 4$ | A1 | Incorrect answers with no method shown score M0A0 |

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# Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = \frac{\mathrm{d}v}{\mathrm{d}t} = 4t - 9$ | M1 | Differentiate $v$ to obtain $a$ (at least one power of $t$ going down) |
| | A1 | Correct derivative |
| $t = 5,\ a = 11\ (\text{m s}^{-2})$ | A1 | |

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# Question 6c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = \int v\,\mathrm{d}t = \frac{2}{3}t^3 - \frac{9}{2}t^2 + 4t\ (+C)$ | M1 | Integrate $v$ to obtain $s$ (at least one power of $t$ going up) |
| | A1 | |
| Use of $t=0,\ t=\frac{1}{2},\ t=4,\ t=5$ (and $t=0, s=15$) as limits in integrals | DM1 | Correct strategy for their limits – requires subtraction of the negative distance. Dependent on the previous M1 and at least one positive solution for $t$ in $(0,5)$ from (a) |
| $\left[\frac{2}{3}t^3 - \frac{9}{2}t^2 + 4t(+15)\right]_0^{\frac{1}{2}} - \left[\frac{2}{3}t^3 - \frac{9}{2}t^2 + 4t(+15)\right]_{\frac{1}{2}}^{4} + \left[\frac{2}{3}t^3 - \frac{9}{2}t^2 + 4t(+15)\right]_4^5$ | A1 | NB: $\int_0^5 v\,\mathrm{d}t$ scores M0A0A0 |
| $\left(0,\ \frac{23}{24},\ -\frac{40}{3},\ \frac{-55}{6}\right) = \frac{23}{24} + \frac{343}{24} + \frac{100}{24} = 19.4\ \text{(m)}$ | A1 | $19\frac{5}{12}\ \left(\frac{233}{12}\right)$ or better |
| $\left(15,\ 15\frac{23}{24}\left(\frac{383}{24}\right),\ \frac{5}{3},\ 5.83\left(\frac{35}{6}\right)\right)$ | | |

## Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| After 4 seconds from O, horizontal speed $= u\cos\theta$ | B1 | |
| Vertical component of speed at $A = u + at$ | M1 | Complete method using *suvat* to find $v$ |
| $= u\sin\theta - 4g$ | A1 | |
| At $A$, components are $15\cos 20$ (horizontal) and $15\sin 20$ (vertical) | B1 | |
| $u\cos\theta = 15\cos 20$ and $u\sin\theta = 15\sin 20 + 4g$ | DM1 | Form simultaneous equations in $u$ and $\theta$, attempt to solve for $u$ or $\theta$. Depends on previous M1 |
| $\theta = 72.4$ (72) | A1 | Remember - A0 for the first overspecified answer |
| $u = 46.5$ (47) | A1 | |
| **[7]** | | |

## Question Alt7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| After 4 seconds from O, horizontal speed $= u\cos\theta$ | B1 | |
| At $t=4$, $s = vt - \frac{1}{2}gt^2$ | M1 | Complete method to find the vertical height at $A$ |
| $= 98.9$....... | A1 | |
| At $A$, components are $15\cos 20$ (horizontal) and $15\sin 20$ (vertical) | B1 | |
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - 2gh$ | DM1 | Conservation of energy. Equation needs to include all three terms but condone sign error(s) |
| $u = 46.5$ (47) | A1 | Remember - A0 for the first overspecified answer |
| $\theta = 72.4$ (72) | A1 | Beware inappropriate use of *suvat* |

## Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $-15\sin 20 = 15\sin 20 - gt$ or $0 = 15\sin 20t - \frac{1}{2}gt^2$ | M1 | Complete method using *suvat* or otherwise to find the time to travel from $A$ to $B$ |
| $t = 1.05$ (s) or $1.0$ (s) | A1 | |
| **[2]** | | |

## Question 7c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Total time $= 4 + (1.05) + 4$ | B1ft | Follow their $t$, or $\frac{2u\sin\theta}{g}$ for their $u, \theta$ |
| Range $= 46.5 \times \cos 72.4 \times (8+1.05)$ (or $15\cos 20 \times 9.05$) | M1 | Correct method to find $OC$ for their $t, u$ and $\theta$ |
| $= 128$ (m) or $127$ (m) (130) | A1 | |
| **[3]** | | |
| **(12)** | | |

## Question 8a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4mu - 2mu = mv + mw$ | M1 | Equation for CLM. Requires all 4 terms. Condone sign errors. Condone $m$ missing throughout |
| | A1 | |
| $w - v = 6eu$ | M1 | Impact law. $e$ must be used correctly. Condone sign errors |
| | A1 | Signs should be consistent with equation for CLM |
| $v + w = 2u$; $w - v = 6eu$; $2w = 2u + 6eu$, $(w = u + 3eu)$ | DM1 | Solve for $w$. Dependent on the two previous M marks |
| For $Q$ and $R$ to collide require $w > 3u$ | M1 | Use **inequality** to compare their $w$ with $3u$ |
| $u + 3eu > 3u$, $e > \frac{2}{3}$ | A1 | Reach **Given answer** with no errors seen |
| **[7]** | | |

## Question 8a alt:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Collision between $Q$ and $R \Rightarrow w > 3u$ | M1 | |
| Magnitude of impulse on $Q > 5mu$ | | |
| Magnitude of impulse on $P > 5mu$ | A1 | |
| $\Rightarrow v < -u$ | M1 | |
| $e = \frac{w-v}{4u+2u}$ | M1 | Impact law |
| Speed of separation after collision $> u + 3u$ | M1 | |
| $e > \frac{4u}{4u+2u}$ | A1 | |
| $e > \frac{2}{3}$ | A1 | Reach given inequality with no errors seen |

## Question 8b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = u + 3eu = \frac{13}{4}u$, $v = -\frac{5}{4}u$ | B1 | Correct values seen or implied |
| $mw + 3mu = mx + my$ and $\frac{3}{4}(w - 3u) = y - x$ | M1 | Equations for CLM and impact. Allow with $w$ or their $w$. All terms required. Condone sign errors. $e$ must be used correctly |
| | A1 | Correct equations in $w$ or their $w$ |
| $\frac{25}{4}u = x + y$, $\frac{3u}{16} = y - x$ | | |
| Solve for $x$ (or $kx$) | DM1 | Dependent on the previous M1. Need to get far enough to give a convincing concluding argument |
| $\frac{97}{16}u = 2x$, $x = \frac{97u}{32}$ ($= 3.03125u$) | A1 | Correct expression for $kx$ |
| $P$ and $Q$ moving away from each other, so no collision | A1 | Reach **given conclusion** with no errors seen |
| **[6]** | | |
| **(13)** | | |