# Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $F = \mu R$ | B1 | At least once |
| Resolve horizontally | M1 | Allow with their horizontal friction |
| $S = \frac{4}{5}R\ \ (S = F_A)$ | A1 | Correct unsimplified equation |
| Resolve vertically | M1 | Allow with their vertical friction |
| $\frac{3}{5}S + R = 25g\ \ (F_B + R = 25g)$, $\left(\frac{3}{5}S + \frac{5}{4}S = 25g,\ \ S = \frac{500}{37}g\right)$ | A1 | Correct unsimplified equation |
| Moments equation | M1 | Any moments equation. Need all terms and dimensionally correct |
| $M(A): 25g \times 1.5\cos\theta = S \times 3\sin\theta + \frac{3}{5}S \times 3\cos\theta$, $\left(25g\cos\theta - \frac{6}{5}S\cos\theta = 2S\sin\theta\right)$ | A1 | Correct unsimplified equation |
| $M(B): R \times 3\cos\theta = 25g \times 1.5\cos\theta + \frac{4}{5}R \times 3\sin\theta$ | | Friction acting in wrong direction scores A0 |
| $\tan\theta = \left(\frac{25g - \frac{6}{5}S}{2S}\right) = \frac{25 - \frac{600}{37}}{\frac{1000}{37}}$ | DM1 | Substitute to form equation in $\tan\theta$ only. Condone decimals. Dependent on M marks for the equations |
| $= \frac{325}{1000}\left(= \frac{13}{40}\right)$ | A1 | Or exact equivalent $(0.325)$ |

**Special Case:** Resolving horizontally/vertically and taking moments about the centre: M1A1 for correct resolution, M2A2 for complete sets of equations to solve.

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# Question 4a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios: $ABCD = 64$, $PQRV = 4$, $RSTU = 16$, $L = 44$ | B1 | Correct mass ratios for their split |
| c of m from $AD$: $ABCD = 4a$, $PQRV = 2a$, $RSTU = 5a$, $L = (d)$ | B1 | Correct distances from vertical axis for their split. Must be multiples of $a$ |
| $M(AD)$ | M1 | Moments about $AD$ or a parallel axis. Need all terms and dimensionally consistent |
| $64 \times 4a - 4 \times 2a - 16 \times 5a = 44d$ | A1 | Correct unsimplified equation. Accept as part of a vector equation |
| $\Rightarrow d = \frac{168}{44}a = \frac{42}{11}a$ * | A1* | Obtain given answer from correct working |

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# Question 4b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| C of M of $L$ lies at midpt of $AC$ | B1 | Seen or implied |
| $M(\text{Mid pt }AB)$ | M1 | Use of moments to form equation in $k$ |
| $\left(4 - \frac{42}{11}\right)aM = 4akM$ | A1 | Correct unsimplified equation. Allow with $a$ not seen |
| $k = \frac{1}{22}$ | A1 | $0.05$ or better $(0.0454545\ldots)$. Allow with $a$ not seen |

**Alternative 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| C of M of $L$ lies at midpt of $AC$ | B1 | Seen or implied by use of $\bar{x} = \bar{y}$ or $\tan 45° = 1$ |
| Find $\bar{x}$ and $\bar{y}$ for system | M1 | |
| From $AB$: $\frac{42}{11}Ma + 8akM = (1+k)M\bar{y}$ | A1 | Correct unsimplified equations in $\bar{x}$ and $\bar{y}$. Allow with $a$ not seen |
| From $BC$: $\frac{46}{11}aM = (1+k)M\bar{x}$ | | |
| $\bar{x} = \bar{y} \Rightarrow \frac{42}{11} + 8k = \frac{46}{11} \Rightarrow k = \frac{1}{22}$ | A1 | Allow with $a$ not seen |

**Alternative 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| C of M of $L$ lies at midpt of $AC$ | B1 | Seen or implied in moments equation |
| If $G$ is c of m of $L$ then $\tan ABG = \frac{42}{46}$ and take moments about $B$ | M1 | Complete method for moments about $B$ |
| $8a\sin 45° \times kM = \frac{Ma\sqrt{46^2+42^2}}{11}\sin(45°-ABG)$ | A1 | Correct unsimplified equation in $k$. Allow with $a$ not seen |
| $\Rightarrow k = \frac{1}{22}$ | A1 | Allow with $a$ not seen |

**Alternative 3:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Take moments about the centre of $ABCD$ | M1 | |
| $M \times \frac{2\sqrt{2}}{11}a = kM \times 4\sqrt{2}a$ | A1 | Correct unsimplified equation in $k$. Allow with $a$ not seen |
| $\Rightarrow k = \frac{1}{22}$ | A1 | Allow with $a$ not seen |

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# Question 5a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{a} = \frac{d\mathbf{v}}{dt}$ | M1 | Differentiate to obtain $\mathbf{a}$ – powers going down |
| $= (6t-9)\mathbf{i} + (2t+1)\mathbf{j}$ | A1 | Differentiation correct |
| $= 9\mathbf{i} + 7\mathbf{j}\ \text{(m s}^{-2})$ | A1 | ISW if go on to find $|\mathbf{a}|$ |

---

# Question 5b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Instantaneous rest $\Rightarrow \mathbf{v} = 0\mathbf{i} + 0\mathbf{j}$, $\Rightarrow 3(t-1)(t-2) = 0$ and $(t-2)(t+3) = 0$ | M1 | Set $\mathbf{v} = 0$ and solve for $t$ (need **both** components equal to zero) |
| $\Rightarrow t = 2$ | A1 | |
| $\mathbf{r} = \int \mathbf{v}\, dt$ | M1 | Integrate to obtain $\mathbf{r}$ – powers going up. Condone if no constant of integration seen |
| $= \left(t^3 - \frac{9}{2}t^2 + 6t\right)\mathbf{i} + \left(\frac{1}{3}t^3 + \frac{1}{2}t^2 - 6t\right)\mathbf{j}$ | A1 A1 | At most one error / Correct integration. Allow column vector. Allow A1A0 for correct integration and non-zero constants of integration |
| $= 2\mathbf{i} - \frac{22}{3}\mathbf{j}$, distance $= \sqrt{2^2 + \left(\frac{22}{3}\right)^2}$ | DM1 | Correct strategy to find distance, i.e. substitute their value for $t$ and use Pythagoras. Dependent on the two preceding M marks |
| $= \frac{2\sqrt{130}}{3} = 7.60\ \text{(m)}$ | A1 | 7.6 or better from correct work |

## Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 6g\cos\alpha$ | B1 | Correct normal reaction |
| Work done $= 15 \times 0.25 \times R$ | M1 | Correct method with their $R$ |
| $= 204$ (J) | A1 | Or 200(J). Accept $21g$ or better. $(20.7692...g)$. Not $\frac{2646}{13}$ |
| **Total: (3)** | | |

## Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Initial KE $-$ GPE lost $-$ WD $=$ final KE | M1 | Use of work-energy to form equation in $v$. Dimensionally correct. Ignore sign errors. Allow WD or their WD |
| $\frac{1}{2}\times6\times14^2 - 6g\times15\times\frac{5}{13} - 6g\times15\times\frac{3}{13} = \frac{1}{2}\times6v^2$ | A1ft | Unsimplified equation with at most one error |
| $\left(3\times196 - \frac{450g}{13} - \frac{270g}{13} = 3v^2\right)$ | A1ft | Correct unsimplified equation. Follow their WD |
| $v = 3.88 \quad (3.9)$ | A1 | Max 3 sf |
| Work-energy equation | M1 | Complete method using work-energy to form equation in $w$. Dimensionally correct. Ignore sign errors |
| $\frac{1}{2}\times6\times14^2 - 6g\times15\times\frac{3}{13} = \frac{1}{2}\times6w^2$ | A1ft | Correct unsimplified equation. Follow their WD or their $v$ |
| or $\frac{1}{2}mw^2 = \frac{1}{2}mv^2 + mg\times\frac{15\times5}{13}$ | | |
| $w = 11.3 \quad (11)$ | A1 | Max 3 sf |
| **Total: (7) [10]** | | |

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## Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| KE gain $=$ final KE $-$ initial KE | M1 | KE equation for $B$. Allow for change in KE |
| $\frac{48}{25}mu^2 = \frac{1}{2}mw^2 - \frac{1}{2}mu^2$ | A1 | Correct unsimplified equation to find $w$ |
| $\left(w^2 = \frac{121}{25}u^2, \quad w = \frac{11}{5}u\right)$ | | |
| CLM: $3m\times2u + mu = 3mv + mw$ | M1 | All terms required. Condone sign errors |
| $\left(7mu = 3mv + \frac{11}{5}mu\right)\left(v = \frac{8}{5}u\right)$ | A1 | Correct unsimplified equation in $v$ and $w$ or their $w$ |
| Impact law: $w - v = e(2u - u)$ | M1 | Used correctly |
| | A1 | Correct unsimplified equation in $v$ and $w$ or their $v$ and $w$ |
| Solve for $e$ | DM1 | Dependent on the preceding M marks |
| $\frac{3}{5}u = eu, \quad e = \frac{3}{5}$ | A1 | |
| **Total: (8)** | | |

## Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Impact law: $fw = v$ | M1 | Condone sign error |
| $f = \frac{8}{11}$ | A1 | 0.73 or better. Final answer must be positive |
| **Total: (2) [10]** | | |

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## Question 8a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal component: $p = 8$ | B1 | |
| Vertical component: $-12 = q - 3g$ | M1 | Complete method to find $q$ using suvat. Condone sign errors |
| $q = 17.4$ | A1 | 17 or better |
| Speed $= \sqrt{8^2 + 17.4^2}$ | M1 | Use of Pythagoras to find speed using their velocity. Independent M mark |
| $= 19.2 \quad (19)\left(\text{ms}^{-1}\right)$ | A1 | 3 sf or 2 sf |
| **Total: (5)** | | |

## Question 8b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of Pythagoras to find vertical component | M1 | |
| vertical component $= \pm6$ | A1 | Seen or implied. Accept without $+/-$ |
| $-6 = 6 - 9.8T$ | DM1 | Complete method using suvat to find required time. Dependent on the previous M1 |
| $T = 1.22 \quad (1.2)$ | A1 | 3 sf or 2 sf. Not $\frac{60}{49}$ |
| **Total: (4)** | | |

**Alternative method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use suvat and Pythagoras to form an equation in $t$ | M1 | Or an inequality |
| $8^2 + (17.4 - gt)^2 = 100$ | A1 | Correct unsimplified equation for $t$. Accept inequality |
| Solve for $T$ | DM1 | Complete method to obtain $T$. Dependent on the previous M1 |
| $T = 1.22 \quad (1.2)$ | A1 | 3 sf or 2 sf. Not $\frac{60}{49}$ |
| **Total: (4)** | | |

## Question 8c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Velocity perpendicular $\Rightarrow$ vertical component $= \frac{2}{3}\times8$ | M1 | Complete method to find vertical component of velocity at $B$ |
| $= \frac{16}{3}$ | A1 | |
| $(-12)^2 = \left(\frac{16}{3}\right)^2 - 2g(-h)$ | DM1 | Complete method to find the required vertical distance using their vertical component of the velocity. Dependent on the previous M1 |
| $h = 5.90 \quad (5.9)$ (m) | A1 | Max 3 sf |
| **Total: (4)** | | |

**Alternative method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}8\\17.4-gt\end{pmatrix}\cdot\begin{pmatrix}8\\-12\end{pmatrix} = 0$ and time $= 3-t$ | M1 | Complete method to find the time from $B$ to $A$ |
| Time $= 3 - 1.23... = 1.768...$ | A1 | |
| $s = vt - \frac{1}{2}gt^2 = 12t - 4.9t^2$ | DM1 | Complete method to find the required vertical distance using their time. Dependent on the previous M1 |
| $s = 5.9$ (m) | A1 | Max 3 sf |
| **Total: [13]** | | |