# Question 6:

## Part 6a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5mv = 2m(v - (-x))$ | M1 | Use of $I = mv - mu$ |
| $x = \frac{3v}{2}$ | A1 | Seen or implied |
| $5mv = 3m(v - (-y))$ or $2mx - 3my = 3mv - 2mv$ | M1 | Use of $I = mv - mu$ or use of CLM |
| $y = \frac{2v}{3}$ | A1 | Seen or implied |
| $2v = e\left(\frac{3v}{2} + \frac{2v}{3}\right)$ | M1 | Correct use of impact law (not necessarily with values in terms of $v$). **Allow $v - v$ on LHS** |
| $e = \frac{12}{13}$ | A1 | 0.92 or better |
| **Total: (6)** | | |

## Part 6b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Speed of $B$ after collision with wall $= vf$ | B1 | Seen or implied |
| $2 \times \frac{1}{2} \times 3m\left(y^2 - (vf)^2\right) = \frac{1}{2} \times 2m\left(x^2 - v^2\right)$ | M1 | Use KE to form an equation in $f$. Condone use of change in KE rather than loss. **Condone 2 on wrong side** |
| $3\left(\frac{4}{9} - f^2\right) = \left(\frac{9}{4} - 1\right)$ | A1 | Correct unsimplified equation for $f$ |
| $\left(f^2 = \frac{1}{36}\right)\ \ f = \frac{1}{6}$ | A1 | cao. NB: $\frac{\sqrt{31}}{6}$ comes from inconsistent subtraction |
| **Total: (4) [10]** | | |

## Question 7a:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $\dfrac{2a \times \frac{1}{2}}{3 \times \frac{\pi}{6}} \left(= \dfrac{2a}{\pi}\right)$ | B1 | Seen or implied |
| Moments about $EC$ | M1 | Dimensionally correct. Condone use of a parallel axis |
| $ad \times \dfrac{d}{2} = \dfrac{1}{6}\pi a^2 \times \dfrac{2a}{\pi} \times \sin\dfrac{\pi}{6}$ | A1 | Correct unsimplified equation |
| $\Rightarrow \left(d^2 = \dfrac{a^2}{3}\right) \quad a = \sqrt{3}d$ | A1* | Obtain **given answer** from correct working |
| | **(4)** | |

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## Question 7b:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Mass ratios $\dfrac{a^2}{\sqrt{3}} : \dfrac{\pi a^2}{6} : \dfrac{a^2}{\sqrt{3}} + \dfrac{\pi a^2}{6}$ | B1 | Or equivalent. Seen or implied |
| Moments about $BC$ | M1 | Dimensionally correct. Condone use of a parallel axis |
| $\dfrac{1}{\sqrt{3}} \times \dfrac{a}{2} + \dfrac{\pi}{6} \times \dfrac{2a}{\pi} \times \dfrac{\sqrt{3}}{2} = \left(\dfrac{1}{\sqrt{3}} + \dfrac{\pi}{6}\right)y$ | A1ft | Correct unsimplified. Follow their $\dfrac{2a}{\pi}$ |
| Distance from $BC$ $= y = \dfrac{6a}{6 + \sqrt{3}\pi}$ | A1 | Or equivalent $\left(y = \dfrac{6d}{2\sqrt{3}+\pi}\right)$ |
| Use of trig to find a relevant angle | M1 | |
| $\tan\beta^C = \dfrac{6}{6+\sqrt{3}\pi} \times \sqrt{3} \quad \left(\dfrac{\bar{y}}{d}\right)$ | A1ft | Or equivalent correct unsimplified equation for the required angle |
| $\beta = 0.737 \quad (0.74)$ | A1 | $0.74$ or better. $42.2°$ implies correct method |
| | **(7)** | |
| | **[11]** | |

---

## Question 8a:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Conservation of energy | M1 | Need all three terms and dimensionally correct. Condone sign errors |
| $\dfrac{1}{2}m \times 10^2 + mgh = \dfrac{1}{2}m \times 18^2$ | A1 | Correct unsimplified equation |
| $h = 11.4 \quad (11)$ | A1 | 3 sf or 2 sf only $\left(\text{not } \frac{80}{7}\right)$ |
| | **(3)** | |

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## Question 8b:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Vertical distance | M1 | Complete method using suvat to find angle of projection |
| $10\sin\alpha \times 2.5 - \dfrac{1}{2}g \times 2.5^2 = -11.4$ | A1ft | Follow their $h$ |
| $\alpha = 50.2°$ or $10\sin\alpha = v_V = 7.7678...$ | A1 | $50°$ or better $(50.1618...)$. Accept $50.3°$ from $11.4$. Seen or implied. Might see $\sin\alpha = \frac{43}{56}$ or $v_V = \frac{215}{28}$ |
| Horizontal distance $= 10\cos\alpha \times 2.5$ or $\sqrt{100 - v_V^2} \times 2.5$ | M1 | |
| $= 16.0 \quad (16)$ (m) | A1 | 3 sf or 2 sf only |
| | **(5)** | |

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## Question 8c:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Using energy: $\dfrac{1}{2}m \times 64 + mgs = \dfrac{1}{2}m \times 100$ | M1 | Complete method to find height above $A$ |
| $s = 1.8367...$ | A1 | 1.8 or better |
| Use of suvat to form equation in $t$ | M1 | |
| $1.84 = 10\sin 50.2° \times t - 4.9t^2$ | A1 | Correct unsimplified equation |
| Solve for $t$ and find difference between roots | DM1 | Complete method to find the required time. Dependent on 2 previous M marks |
| $T = 0.98$ or $0.978$ | A1 | 2 sf or 3 sf |
| | **(6)** | |

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## Question 8c (alt):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of Pythagoras | M1 | Complete method to find vertical component of speed |
| Vertical speed $\sqrt{64-(10\cos\alpha)^2} = 4.8...$ | A1 | Awrt 4.8 or better |
| Use of $10\sin\alpha - gt = \pm v$ to find $t$ | M1 | |
| $\begin{cases}10\sin 50.2° - gt_1 = 4.8 \\ 10\sin 50.2 - gt_2 = -4.8\end{cases}$ | A1 | Correct unsimplified equations. Could also find time to top |
| $T = t_2 - t_1 = 1.27... - 0.29...$ | DM1 | Complete method to find the required time. Dependent on 2 previous M marks |
| $= 0.98$ or $0.978$ | A1 | Final answer. 2 sf or 3 sf |
| | **(6)** | |

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## Question 8c (alt2):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of Pythagoras to form quadratic in $t$ | M1 | |
| $(10\sin\theta - gt)^2 + (10\cos\theta)^2 = 64$ | A1 | |
| Simplify and substitute for trig | M1 | |
| $36 + 9.8^2t^2 - 150.5t = 0$ | A1 | |
| $T = t_2 - t_1 = 1.27... - 0.29...$ | DM1 | Complete method to find the required time. Dependent on 2 previous M marks |
| $= 0.98$ or $0.978$ | A1 | Final answer. 2 sf or 3 sf |
| | **[14]** | |