# Question 1:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Impulse-momentum principle: $(7\mathbf{i}-5\mathbf{j})=4\mathbf{v}-4(2\mathbf{i}+3\mathbf{j})$ | M1A1 | M1 for use of impulse-momentum principle, dimensions correct, correct no. of terms, must be a *difference* of momenta. A1 for correct equation |
| $\mathbf{v}=\frac{15}{4}\mathbf{i}+\frac{7}{4}\mathbf{j}$ | A1 | A1 for correct velocity vector |
| $|\mathbf{v}|=\frac{1}{4}\sqrt{15^2+7^2}$ | M1 | M1 for attempt to find magnitude of their $\mathbf{v}$ |
| $=\frac{1}{4}\sqrt{274}=4.1\ \text{(m s}^{-1}\text{)}$ (or better) | A1 cso | A1 for exact answer or 4.1 or better |

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# Question 2a:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Use of $P=Fv$: $280=F\times 2$ | M1 | M1 for $280=F\times 2$ |
| Equation of motion: $F-75g\sin\theta=R$ | M1A1 | M1 for resolving parallel to plane with $a=0$, usual rules. A1 for correct equation |
| $140-75\times9.8\times\frac{1}{21}=R$ | | |
| $R=105$ (or 110) | A1 | A1 for 105 or 110 |

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# Question 2b:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Equation of motion: $75g\sin\theta+\frac{280}{3.5}-60=75a$ or $-75a$ | M1A2 | M1 for resolving parallel to plane with $a\neq0$. A1A0 or A0A0 for each incorrect term. Use of $280/2$ is an A error |
| $a=0.73\ \text{(m s}^{-2}\text{)}$ (0.733) or $-0.73$ ($-0.733$) | A1 | Allow negative answers |

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# Question 3a:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Integrate: $v=\int(4t-8)\ dt=2t^2-8t\ (+C)$ | M1 | M1 for attempt to integrate, at least one power increasing |
| Use $t=0, v=6$: $v=2t^2-8t+6$ | M1A1 | M1 for using initial conditions. A1 for correct expression for $v$ |
| Use factor theorem or factorise: $v=2(t-1)(t-3)\Rightarrow$ at rest for $t=1$ | M1 | M1 for showing $v=0$ when $t=1$ |
| Second value $t=3$ | A1 | A1 for $t=3$ (B1 mark) — must come from correct $v$ |

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# Question 3b:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Integrate to find distance: $s=\int v\ dt=\frac{2}{3}t^3-4t^2+Ct$ | M1A1 ft | M1 for attempt to integrate their $v$. A1 ft on their $v$ (must include non-zero $C$) |
| Correct strategy: $\left|\left[\frac{2}{3}t^3-4t^2+Ct\right]_1^3\right|+\left|\left[\frac{2}{3}t^3-4t^2+Ct\right]_3^4\right|$ | M1 | M1 (independent) for complete method to find total distance |
| $-\left(0-\frac{8}{3}\right)+\left(\frac{8}{3}-0\right)=\frac{16}{3}\ \text{(m)}$ (5.33) | A1 | A1 for $16/3$ or 5.3 or better. If $0<t<4$ used instead of $1<t<4$, treat as MR |

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# Question 4a:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Moments about $A$: $0.5\times2g+2\times5g\ (=11g)=T\cos\theta\times4=T\times\frac{3}{5}\times4$ | M1A2 | M1 for $M(A)$ with usual rules. A1A1 for correct equation in $T$ only using correct angle. Deduct 1 mark for each incorrect term |
| $T=11g\times\frac{5}{12}=\frac{55}{12}g=44.9\ (45)\ \text{(N)}$ | A1 | A1 for 45 or 44.9 (N). A0 for 45.0 |

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# Question 4b:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Resolving $\leftrightarrow$: $H=T\sin\theta$ **OR** $M(D)$: $H\times3=2g\times0.5+5g\times2$ | M1 | M1 for resolving horizontally or $M(D)$, equation in $T$ only |
| $\updownarrow$: $T\cos\theta+V=7g$ **OR** $M(B)$: $V\times4=2g\times3.5+5g\times2$ | M1A1 | M1 for resolving vertically or $M(B)$. A1 for correct equation in $T$ only |
| Pythagoras: $|R|=\sqrt{41.65^2+35.93^2}=55.0\ (55)\ \text{(N)}$ | M1A1 | M1 (independent) for squaring, adding and rooting 2 components. A1 for 55 or 55.0 |

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# Question 4c:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Use of $F\leq F_{\max}=\mu R$: $V\leq\mu H$ (Must have found $H$ and $V$) | M1 | M1 for use of $V\leq\mu H$. M0 for $V=H$ or $V<H$ |
| $\mu\geq\frac{V}{H}=\frac{41.65}{35.93...}=\frac{51}{44}$, 1.2 or better | A1 | Allow fraction since $g$ cancels, or 1.2 or better |

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# Question 5a:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Energy: $\frac{1}{2}m\times4.2^2+mg\times8\sin30°=\frac{1}{2}m\times u^2$ | M1A2 | M1 for energy equation with correct no. of terms. A1A1 for correct equation; deduct 1 mark per incorrect term |
| $u^2=96.04\Rightarrow u=9.8$ or $9.80$ | A1 | $49/5$ is A0 because of rubric on question paper |

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# Question 5b(i):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Their $u$ or $9.8$ | B1 ft | B1 ft for $w=$ their $u$, or if energy used again $w=9.8$. Do not award if from $v^2=u^2+2as$ without components |

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# Question 5b(ii):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\tan\theta°=\frac{w_V}{w_H}$ or $\cos\theta°=\frac{w_H}{w}$ or $\sin\theta°=\frac{w_V}{w}$ where $w_H=4.2\cos30°$; $w_V=\sqrt{(4.2\sin30°)^2+2g\times4}$; $w=$ their $u$ | M1A1 ft | M1 for complete method for equation in $\theta$ only. Allow inverted tan or cos/sin confusion. M0 if $u=0$ used. A1 ft for correct equation in $\theta$ only |
| $\theta=68$ or $68.2$ | A1 | |

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# Question 5c:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Vertical component of velocity at $C$: $4.2\sin30°-gt=-w\sin\theta°$ (Follow their $w,\theta$) | M1A1 | M1 for complete method to find time |
| $t=1.14\ \text{(s)}$ $(8/7)$ | A1 | A1 for $t=1.14$ or better |
| **OR** Using vertical distance $B$ to ground: $-4=4.2\sin30°\times t-\frac{1}{2}gt^2$ | M1A1 | |
| $t=1.14\ \text{(s)}$ $(8/7)$ | A1 | |
| Use horizontal motion — Either: $4.2\cos30°\times t$ | M1 | M1 for horizontal motion equation |
| $=4.16\ (4.2)\ \text{(m)}$ | A1 | |
| **Or**: (their $u$ or $w$)$\cos(\text{their }\theta)\times t = 4.16\ (4.2)\ \text{(m)}$ | M1A1 | A1 for 4.2 or 4.16 (m) |

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# Question 6a:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Ratio of areas: $4:1:3$ or **OR** $2:1:3$ | B1 | B1 for any correct mass ratios |
| Distances to $AB$: $3a:5a:\bar{x}$ or $1.5a:4a:\bar{x}$ | B1 | B1 for correct distances to $AB$ |
| About $AB$: $4\times3a-1\times5a=3\bar{x}$ **OR** $2\times1.5a+1\times4a=3\bar{x}$ | M1A1 | M1 for moments equation about any line parallel to $AB$. A1 for correct equation |
| $\bar{x}=\frac{7}{3}a$ **Given Answer** | A1 | A1 for correctly obtaining given answer |

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# Question 6b:

| Working/Answer | Marks | Guidance |
|---|---|---|
| About $AE$: $4\times3a-1\times2a=3\bar{y}$ **OR** $2\times3a+1\times4a=3\bar{y}$ | M1A1 | |

## Question 6c:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Use position of centre of mass of combined system | M1 | Attempt at complete method to find either coordinate of cm of combined system. **M0 if they combine masses with areas** |
| From $AB$: $\dfrac{\frac{7}{3}a \times M + 3a \times kM}{M(1+k)}$ | A1 | Correct moments equation with distances measured from a line parallel to $AB$ |
| From $AE$: $\dfrac{\frac{10}{3}a \times M}{M(1+k)}$ **OR** From $BC$: $\dfrac{\frac{8aM}{3} + 6akM}{M(1+k)}$ | A1 | Correct moments equation with distances measured from a line parallel to $AE$ |
| $AC$ vertical $\Rightarrow \dfrac{7}{3}aM + 3akM = \dfrac{10}{3}aM$ i.e. $\left(\dfrac{7}{3} + 3k = \dfrac{10}{3}\right)$ | DM1 | Dependent on first M1. Setting $\bar{x} = \bar{y}$ (or appropriate equation e.g. $\bar{y} = 6a - \bar{x}$). Must have a new $\bar{x}$ AND a new $\bar{y}$, and solving for $k$ |
| $k = \dfrac{1}{3}$ (0.33) | A1 (5) | Third A1 for $k = 1/3$ or 0.33 or better |

**Alternative: Moments about $A$**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{(\bar{x}^2 + \bar{y}^2)}\, a\sin(45-\alpha) \times M = 3a\cos 45 \times kM$ | M1A1A1 | First M1 for clear attempt at moments about $A$ – requires both terms, dimensionally correct. **N.B.** $(\bar{y}-\bar{x})\cos 45 \times M = 3a\cos 45 \times kM$ |
| Find $\alpha$ (if needed) and solve for $k$ | DM1 | Dependent on first M1. **M0 if 45 has been replaced with some other angle e.g. 30 or 60** |
| $k = \dfrac{1}{3}$ (0.33) | A1 (5) | Third A1 for $k = 1/3$ or 0.33 or better |

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## Question 7a:

| Answer/Working | Marks | Guidance |
|---|---|---|
| CLM: $6m \times 3u = 6mv + 4mw$ | M1A1 | First M1 for CLM with correct number of terms, correct masses matched with velocities. Condone sign errors and allow consistent omission of $m$'s (note $w$ is defined but $v$ is not) |
| Impact law: $\dfrac{1}{6} \times 3u = -v + w$ | M1A1 | Second M1 for NIL with $\frac{1}{6}$ on correct side of equation. Second A1 for a correct consistent equation |
| Solve for $w$: $18u = 6v + 4w$, then $3u = 6w - 6v \Rightarrow 21u = 10w$ | DM1A1 (6) | Third DM1 dependent on both M's, for solving for $w$. Third A1 for correct **given** answer ($w = 2.1u$ is A0). **N.B.** Ignore diagram if it helps the candidate |
| $w = \dfrac{21}{10}u$ | **Answer Given** | |

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## Question 7b:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(v =)\ 1.6u$ or $-1.6u$ | B1 | |

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## Question 7b (KE loss):

| Answer/Working | Marks | Guidance |
|---|---|---|
| KE loss $= \dfrac{1}{2}\times 6m \times 9u^2 - \left(\dfrac{1}{2}\times 6m \times 1.6^2u^2 + \dfrac{1}{2}\times 4m \times 2.1^2u^2\right)$ | M1A1 | First M1 for $\pm$(initial KE – sum of final KE terms), with masses and velocities correctly matched. First A1 for a correct expression for the LOSS |
| Fraction of KE lost $= \dfrac{10.5mu^2}{27mu^2} = \dfrac{7}{18}$, 0.39 or better | M1A1 (5) | Second M1 for Loss/Initial. Second A1 for $7/18$, 0.39 or better |

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## Question 7c:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $8.4mu = 4mr + ms$ | | |
| $2.1ue = -r + s$ | | |
| $r = \dfrac{2.1u}{5}(4-e)$ | M1A1 | First M1 for complete method to find $r$. First A1 for correct expression in terms of $u$ and $e$ only |
| Use $r \geq 1.6u$: $\dfrac{2.1u}{5}(4-e) \geq 1.6u$ (their $v$) | DM1 | Second M1 dependent on first, for their $r \geq 1.6u$ (their $v$) oe. **N.B. If they use $r > 1.6u$ oe, final two M marks can be scored** |
| $0 \leq e \leq \dfrac{4}{21}$ (0.190) (0.19) | DM1 A1 (5) | Third DM1 dependent on second M1, for producing $e \leq k$ where $k$ is a number. Third A1 for $0 \leq e \leq 4/21 = 0.19$ or better |