# Question 8:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = 2mg\cos\alpha$ | M1A1 | |
| $F = \frac{11}{36}\times2mg\times\frac{12}{13} = \frac{22mg}{39}$ | A1* (3) | |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3mg - T = 3ma$ | M1A1 (2) | |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $T - \frac{22mg}{39} - 2mg\sin\alpha = 2ma$ $\left(T - \frac{4mg}{3} = 2ma\right)$ | M1A1 | |
| **OR:** $3mg - \frac{22mg}{39} - 2mg\sin\alpha = 5ma$ | M1A1 | |
| Solve for $a$ in terms of $g$; must reach $a = kg$ from their equations | M1 | |
| $a = \frac{1}{3}g$ | A1* (4) | |

## Part (d):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $v^2 = \frac{2gh}{3}$ | B1 | |
| $-\frac{22mg}{39} - 2mg\sin\alpha = \pm2ma$ **OR** PE Gain $= 2mgd\sin\alpha$ | M1 | |
| $\pm\frac{2g}{3} = a$ **OR** $= \frac{10mgd}{13}$ | A1 | |
| $0 = \frac{2gh}{3} - 2\times\frac{2g}{3}\times d$ | M1 | |
| $\frac{22mgd}{39} = \frac{1}{2}\times2m\times\frac{2gh}{3} - \frac{10mgd}{13}$ | M1 | |
| $d = \frac{1}{2}h$ | A1 | |
| Total distance $= \frac{1}{2}h + h = \frac{3}{2}h$ | A1ft (6) | |

## Question 8:

### Part 8(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve perpendicular to the plane | M1 | Correct terms, condone cos/sin confusion and sign errors. Allow if they use $\cos\left(\frac{12}{13}\right)$ or similar |
| Correct equation | A1 | |
| Correct given answer correctly obtained | A1* | Must see $\frac{12}{13}$ |

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### Part 8(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion for $B$ | M1 | Correct terms, condone sign errors |
| Correct equation | A1 | |

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### Part 8(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion for $A$ **OR** whole system | M1 | Correct terms but allow $F$, condone sign errors and sin/cos confusion. Allow if they use $\sin\left(\frac{5}{13}\right)$ or similar |
| Correct equation | A1 | |
| Solve for $a$ in terms of $g$, substituting trig values, solving 2 equations in $T$ and $a$ **OR** using whole system equation | M1 | Need to see trig substituted, correct terms |
| Correct given answer correctly obtained | A1* | Must see $\frac{5}{13}$. N.B. Allow full verification using equations of motion for $A$ and $B$ **OR** whole system equation |

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### Part 8(d):

**Using Newton's Second Law:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Seen or implied | B1 | |
| Equation of motion for $A$ | M1 | Correct terms, condone sign errors and sin/cos confusion |
| Correct acceleration or deceleration of $A$ | A1 | |
| Complete method to find equation in $d$, $g$ and $h$ only, using new calculated $a$ | M1 | |
| Correct answer | A1 cao | |
| Their $d$ (which must be a multiple of $h$) $+ h$ | A1ft | N.B. This mark is only dependent on the previous M |

**OR Using Work-Energy:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Seen or implied | B1 cao | |
| PE gain of $A$, condone sign errors and sin/cos confusion | M1 | |
| Correct PE gain | A1 | |
| Use work-energy principle to obtain equation in $m$, $d$, $g$ and $h$ only, using their PE expression | M1 | |
| Correct answer | A1 cao | |
| Their $d$ (which must be a multiple of $h$) $+ h$, with final answer of the form $kh$ | A1ft | |