# Question 5:

## Main Method

| Answer/Working | Mark | Guidance |
|---|---|---|
| **i** direction: $K = 0.5(15\cos\theta - 12)$ **or** $K = 0.5v_1 - 6$ | M1A1 | |
| **j** direction: $K = 0.5 \times 15\sin\theta$ **or** $K = 0.5v_2$ | M1A1 | |
| $(12 + 2K)^2 + (2K)^2 = 15^2$ | **DM1** | Dependent on both previous M marks; use $\cos^2\theta + \sin^2\theta = 1$ or magnitude |
| $K = 1.37...$ | A1 | $K = (3\sqrt{34}-12)/4 = 1.4$ or better |
| $\theta = 10.6°\ (10.550...)$ | A1 | |

## Alt (i)

| Answer/Working | Mark | Guidance |
|---|---|---|
| **i** direction: $K = 0.5(15\cos\theta - 12)$ | M1A1 | |
| **j** direction: $K = 0.5 \times 15\sin\theta$ | M1A1 | |
| $6 = 7.5(\cos\theta - \sin\theta)$ | **DM1** | Eliminate $K$ to give $15\cos\theta - 12 = 15\sin\theta$ |
| $K = 1.37...$ | A1 | |
| $\theta = 10.6°\ (10.5500...)$ | A1 | |

## Alt (ii) — Vector Triangle / Sine Rule

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sine rule: $\dfrac{\sin 135}{7.5} = \dfrac{\sin(45-\theta)}{6}$ | M1A1 | First M1 for applying sine rule |
| $\dfrac{K\sqrt{2}}{\sin\theta} = \dfrac{7.5}{\sin 135}$ | M1A1 | Second M1 for applying sine rule again |
| $K = 1.37...$ | M1A1 | Third M1 for solving for $\theta$ |
| $\sin(45-\theta) = 0.5656......,\ \theta = 10.6°\ (10.5500...)$ | A1 | |

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# Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| CLM: $3mu = 3mv + kmw$ $\quad (3u = 3v + kw)$ | M1A1 | Correct no. of terms; condone sign errors |
| Impact (NEL): $w - v = \frac{1}{9}u$ $\quad (3w - 3v = \frac{1}{3}u)$ | M1A1 | Correct way up; condone sign errors |
| $\frac{10}{3}u = w(k+3)$ | **DM1** | Eliminate $v_P$; dependent on both previous M marks |
| $w = \dfrac{10u}{3(k+3)}$ **ANSWER GIVEN** | A1 | |

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# Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = w - \frac{u}{9} = \frac{10u}{3(3+k)} - \frac{u}{9} < 0$ $\quad v_P = \frac{u(27-k)}{9(k+3)} < 0$ | M1A1ft | M1 for finding $v_P$; A1ft on direction of their $v_P$ |
| $k > 27$ | A1 | |

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# Question 6c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = 7 \Rightarrow w = \dfrac{u}{3}$ | B1 | Seen or implied |
| CLM: $7 \times \text{their}\ w = 7.5x$ | M1 | |
| $x = \dfrac{14u}{45}$ | | |
| KE lost $= \frac{1}{2} \times 7m \times \frac{u^2}{9} - \frac{1}{2} \times 7.5m \times x^2$ | M1A1 | $\pm(\text{KE loss of }Q - \text{KE gain of }R)$; allow M1 if $m$'s omitted |
| $= \dfrac{7}{270}mu^2$ | A1 | $7mu^2/270$ or $0.026\,mu^2$ or better |

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# Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Energy: $\frac{1}{2}m \cdot 4^2 + mgh = \frac{1}{2}m \cdot 7^2$ | M1A2 | |
| $h = \dfrac{49 - 16}{2g} = 1.7\ \text{m}\ (1.68\ \text{m})$ | A1 | |

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# Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal speed: $2.5 = 4\cos\alpha$ | M1A1 | |
| $\alpha = \cos^{-1}\!\left(\dfrac{5}{8}\right) = 51°,\ 51.3°$ or better | A1 | |

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# Question 7c:

## Main Method

| Answer/Working | Mark | Guidance |
|---|---|---|
| Vertical speed at B $= \sqrt{7^2 - 2.5^2}\ (= 6.54)$ | M1A1 | |
| $v = u + at$: $\quad -6.54... = 4\sin\alpha - gt$ | M1A1 | |
| $t = 0.986\ \text{s}\ \ (0.99)$ | A1 | |
| Horizontal distance $= 2.5t = 2.5$ or $2.46\ \text{m}$ $\quad (4\cos\alpha)\,t$ | M1A1 | |

## Alternative 1

| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = ut + \frac{1}{2}at^2$: $\quad -1.68... = 3.12...T - 4.9T^2$ | M1A1 | |
| Solving $T = \dfrac{-3.122 \pm \sqrt{3.12^2 - 4 \times 4.9 \times (-1.68)}}{2 \times 4.9}$ | M1A1 | |
| $T = 0.986\ \text{s}\ \ (0.99)$ | A1 | |
| Horizontal distance $= 2.5t = 2.5$ or $2.46\ \text{m}$ $\quad (4\cos\alpha)\,t$ | M1A1 | |