## Question 3:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{v} = \frac{\mathbf{r} - \mathbf{r_0}}{2}$ to find $\mathbf{v}$ | M1 | Or equivalent, allow difference reversed. Allow 120 min |
| $\mathbf{v} = \frac{1}{2}((55\mathbf{i}+34\mathbf{j})-(25\mathbf{i}+10\mathbf{j}))$ $(= 15\mathbf{i}+12\mathbf{j})$ | A1 | Correct unsimplified expression for $\mathbf{v}$ |
| $\mathbf{r}_A = 25\mathbf{i}+10\mathbf{j}+t(15\mathbf{i}+12\mathbf{j})$ | M1 | With the correct structure. Possible use of $\mathbf{r}_A = (55\mathbf{i}+34\mathbf{j})+(t-2)(15\mathbf{i}+12\mathbf{j})$ |
| $(\mathbf{r}_A =)\ (25+15t)\mathbf{i}+(10+12t)\mathbf{j}$ | A1* | We are looking for the RHS only to be correct. Allow order of terms to be reversed in the brackets. N.B. Use of $\mathbf{i}$'s and $\mathbf{j}$'s in columns i.e. poor notation, can score Max M1A1M1A0* |
| **[4]** | | |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{12^2+15^2}$ | M1 | Correct use of Pythagoras for their $\mathbf{v}$ |
| $\sqrt{12^2+15^2}\times\frac{1000}{3600} = \frac{5\sqrt{369}}{18} = \frac{5\sqrt{41}}{6}$ | A1 | 5.3 or better $(5.3359......)$ |
| **[2]** | | |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Position of $B$ at $t = 1.5$ (allow $t = 1.30$) | M1 | Correct use of position and direction vectors with correct structure |
| $\mathbf{r}_B = (35\mathbf{i}+51\mathbf{j})+1.5(20\mathbf{i}-6\mathbf{j})$ | A1 | Correct unsimplified |
| $\Rightarrow \begin{pmatrix}65\\42\end{pmatrix} = \begin{pmatrix}25+15t\\10+12t\end{pmatrix}$ | M1 | Use $\mathbf{r}_P = \mathbf{r}_A$ and use one component to solve for $t$. N.B. If they use $\frac{65}{42} = \frac{25+15t}{10+12t}$ and solve for $t$, it's M0 unless they go on and substitute $t = \frac{8}{3}$ into $\mathbf{r}_A$ and obtain $65\mathbf{i}+42\mathbf{j}$ |
| Obtain $t = \frac{8}{3}$ for one component | A1 | N.B. Allow $t = 2.7$ or better for A1 but not for the second A1* |
| Obtain $t = \frac{8}{3}$ for both components and hence confirm $A$ passes through $P$ **OR** sub $t = \frac{8}{3}$ into $\mathbf{r}_A$ and obtain $65\mathbf{i}+42\mathbf{j}$ and hence confirm $A$ passes through $P$ | A1* | Obtain given result from correct work. N.B. Use of $\mathbf{i}$'s and $\mathbf{j}$'s in columns i.e. poor notation, can score Max M1A1M1A1A0* but only penalise ONCE for the whole question |
| **[5]** | | |

## Question 4a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct method to form equation in $V$ only using area = distance or suvat | M1 | Condone confusion over units for time |
| $120 = \frac{1}{2}(42+33) \times V$ **OR** $\frac{1}{2} \times 9V + 33V = 120$ **OR** $42V - \frac{1}{2} \times 9V = 120$ | A1 | Correct unsimplified equation in $V$ only |
| $V = 3.2$ | A1* | Obtain given answer from correct working. N.B. correct equation required but NO intermediate lines of working |

## Question 4b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = \frac{3.2}{9}$ | M1 | Correct method |
| $= \frac{16}{45} = 0.3555... = 0.36$ or better $(\text{ms}^{-2})$ | A1 | cao |

## Question 4c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape for $Q$ | B1 | Allow crossing before $t=9$, ignore gradients but horizontal line must be above $P$ graph. B0 if starts at origin |
| $3.6$ and $6$ marked | DB1 | — |
| Correct shape for $P$ (line extended) and both ending at $t = 54$ (or equiv) marked | B1 | N.B. Allow dotted vertical lines but withhold first B1 if any continuous vertical lines |

## Question 4d:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance by $P = 120 + (12 \times 3.2)$ **OR** $\frac{1}{2}(54+45)\times 3.2$ **OR** $\frac{3.2\times 9}{2}+(45\times 3.2)$ | B1 | Seen or implied by $158.4$ used |
| Distance by $Q$: clear attempt to use area or suvat to form expression in $T$ only, using $t=54$ extended graph with triangle + rectangle or trapezium | M1 | Any complete method |
| $\frac{(48+(48-T))}{2}\times 3.6$ **OR** $\frac{1}{2}\times T\times 3.6+(48-T)\times 3.6$ | A1 | Correct unsimplified expression in $T$ only |
| $\frac{(48+(48-T))}{2}\times 3.6 = 158.4\left(=\frac{1}{2}(54+45)\times 3.2\right)$ | DM1 | Dependent on previous M; form and solve equation in $T$ only using total distances |
| $T = 8$ | A1 | Correct only |

## Question 5a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion for $P$ | M1 | Require all terms. Condone sign errors |
| $T - 3g = 3a$ | A1 | Correct unsimplified equation |

## Question 5b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion for $Q$ | M1 | Require all terms. Condone sign errors |
| $5g - T = 5a$ | A1 | Second correct equation. Condone combined equation $5g-3g=(5+3)a$ in place of one of the above |
| Solve for $T$ | DM1 | Dependent on both previous M marks. N.B. may solve using calculator |
| $T = 36.8$ | A1 | Accept $\frac{15g}{4}$ or $37$. Must be numerical value |

## Question 5c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Force on pulley $= 2T$ | M1 | Correct for their $T$, provided $T \neq 0$ |
| $F = 73.5$ | A1ft | Accept $\frac{15g}{2}$ or $74$. Must be numerical value |

## Question 5d:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $a = \frac{g}{4}$ | B1 | Must be used in (d) |
| Complete method to find $v$ or $v^2$ with $a \neq g$ | M1 | M0 if $s=8$ is used |
| $v^2 = 2 \times a \times 2\left(v^2 = 2\times\frac{g}{4}\times 2\right)$ | A1ft | Correct unsimplified equation in $v$ or $v^2$, ft on their $a \neq \pm g$ |
| Speed at ground | M1 | Complete method with $a = \pm g$, $s=8$, and calculated $v^2$ at $s=2$, condone sign errors |
| $w^2 = v^2 + 2\times g\times 8$, $w = 12.9(13)\ (\text{ms}^{-1})$ | A1 | 3 sf or 2 sf. N.B. only penalise overaccuracy after $g=9.8$ ONCE for whole question |

## Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| First equation e.g. resolve perpendicular to slope | M1 | Need all terms. Condone sign errors and sin/cos confusion |
| $R = 5g\cos\alpha + H\sin\alpha$ | A1 | Correct unsimplified equation |
| Second equation e.g. resolve parallel to slope | M1 | Need all terms. Condone sign errors and sin/cos confusion |
| $H\cos\alpha + F = 5g\sin\alpha$ | A1 | Correct unsimplified equation |
| $F = \frac{1}{4}R$ | B1 | Seen or implied. Allow $F \leq \frac{1}{4}R$ |
| $H = 20.6$ or $20.7$ or $21$ or $\frac{40g}{19}$ | A1 | Must be 3 sf or 2 sf or exact multiple of $g$. A0 for $\frac{392}{19}$ |

## Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $S = 5g\cos\alpha\ (=4g)$ | B1 | Seen or implied |
| $F = \frac{1}{4}S$ | B1 | Used with a **new** value for the reaction |
| Equation of motion for $P$: $5g\sin\alpha - g(=3g-g) = 5a$ | M1 | Need all terms. Condone sign errors and sin/cos confusion |
| $a = \frac{2g}{5}$ or $3.92$ | A1 | Correct value for $a$. N.B. allow $-ve$ value provided clear $a$ was up the plane |
| $1.5 = \frac{1}{2}\times\frac{2g}{5}\times T^2$ | M1 | Complete method to form equation in $T$ only. Follow their $a \neq g$ |
| $(T) = 0.875$ or $0.87$ | A1 | 3 sf or 2 sf. N.B. only penalise overaccuracy after $g=9.8$ ONCE for whole question |

## Question 7a:

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $-8 = 8 - gT_1$ **OR** $0 = 8T_1 - \frac{1}{2} \times 9.8T_1^2$ **OR** $0 = 8 - gt$ and $T_1 = 2 \times \frac{8}{g}$ | M1 | Complete method using suvat. Condone sign error |
| $T_1 = 1.63$ | A1* | Given single answer correctly obtained (Allow $T$ or $t = 1.63$ or just 1.63) |
| **Total: [2]** | | |

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## Question 7b:

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $0 = u^2 - 2g \times 2$ | M1 | Complete method to find speed immediately after 1st impact. **N.B.** Could use energy. |
| $u = 2\sqrt{g} = 6.26099...$ | A1 | Seen or implied. Do not penalise for $> 3$ sf |
| Use of $I = mv - mu$ | DM1 | Dependent on M1, condone sign errors and a recalculated '8' |
| $I = 0.1(8-(-6.3))$ | A1 | Must have 8 and 6.3 or better |
| $= 1.4$ or $1.43$ (Ns) | A1 | 2sf or 3sf |
| **Total: [5]** | | |

---

## Question 7c:

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Equal heights | M1 | Complete method using suvat to form an **equation** in $T_2$. Allow with $T_2+1$, $T_2$ with $T_2+1$ used for $A$ and $T_2$ used for $B$. Allow $t$ instead of $T_2$. **N.B.** M0 if they use the same times for both or a mixture of $t$ and $t+1$. **ALT 1:** At $t=1$, $A$ is $(8-4.9)=3.1$m above the ground and moving downwards with speed $(9.8-8) \approx 1.8$ ms$^{-1}$, then use $h - s_A = s_B$ oe to find $T_2 - 1$. Must find **both** height and speed. |
| $8T_2 - \frac{1}{2}gT_2^2 = 5(T_2-1) - \frac{1}{2}g(T_2-1)^2$ **OR** $8(T_2+1) - \frac{1}{2}g(T_2+1)^2 = 5T_2 - \frac{1}{2}gT_2^2$ | A1 A1 | One distance correct; Both distances correct |
| **ALT 1:** $3.1-(1.8t+4.9t^2)=(5t-4.9t^2)$ | A1 A1 | One side correct; Both sides correct |
| $\left(t = \frac{3.1}{6.8}\right)$ $(T_2 =)\ 1.5$ or $1.46$ | A1 | Must be rounded to 2 sf or 3 sf. A0 for $\frac{99}{68} = 1.45588...$ |
| **ALT 2:** $8(T_1-T_2) - \frac{1}{2}g(T_1-T_2)^2 = 5(T_2-1) - \frac{1}{2}g(T_2-1)^2$ | A1 A1 | |
| **Total: [4]** | | **N.B.** Only penalise overaccuracy after $g = 9.8$, ONCE for parts (b) and (c) only. |
| **Question Total: [11]** | | |