# Question 3:

## Part (a)

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| For equilibrium: $R(\perp \text{ plane})$  $N = 1.5g\cos 30$ | M1, A1 | For resolution of forces parallel or perpendicular to the plane. Weight must be resolved. Condone sin/cos confusion. Correct equation for $N$ (12.7) |
| $R(\parallel \text{ plane})$  $F = 1.5g\cos 60$ | A1 | Correct equation for $F$ (7.35). Condone $\mu R$ |
| $\frac{F}{N} = \frac{\cos 60}{\cos 30} = 0.577... < 0.6 \therefore \text{equilibrium}$ | M1 | Use of $F_{\max} = \mu N$ and compare with $F$, or find the value of $\frac{F}{N}$ and compare with $\mu$ |
| Conclusion: equilibrium | A1 (5) | Reach **given conclusion** correctly. They must make some comment, however brief. |

**ALT for first 3 marks:**

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| Resolve vertically: $N\cos 30 + F\cos 60 = 1.5g$ | M1A1 | |
| Resolve horizontally: $N\cos 60 = F\cos 30$ | A1 | |

**ALT for last 2 marks:**

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $F_{\max} = 0.6 \times 12.73 = 7.63 > 7.35$ | M1 | |
| $\therefore P$ is at rest | A1 | |

> **Note:** If the candidate has given the equation of motion for the particle moving down the plane then A1 for $1.5g\sin 30 - \mu R = \pm 1.5a$. To score more they need to comment correctly: $a = -0.19$ impossible (M1), conclude particle cannot be moving (A1).

> Candidates who think the diagram applies to (a) will score nothing in (a) but if they carry their results forward into (b) then their work can score the marks available in (b).

## Part (b)

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $R(\perp \text{ plane})$  $N = 1.5g\cos 30 + X\cos 60$ | M1 | Requires all 3 terms. Condone sin/cos confusion and sign errors. |
| $R(\parallel \text{ plane})$  $X\cos 30 = 1.5g\cos 60 + F$ | M1 | Requires all 3 terms. Condone sin/cos confusion and sign errors. |
| (Both equations correct unsimplified) | A1 | Both equations correct unsimplified. |
| $N = 1.5g\cos 30 + \frac{\cos 60}{\cos 30}(1.5g\cos 60 + 0.6N)$ | DM1 | Use $F = 0.6N$ to form an equation in $N$ or in $X$. Dependent on the two previous M marks. OR: $0.6(X\cos 60 + 1.5g\cos 30) + 1.5g\sin 30 = X\cos 30$ |
| (i)  $N = 26$ or $26.0$ (N) | A1 | First value found correctly ($N$ or $X$) |
| (ii)  $X = (N - 1.5g\cos 30) \div \cos 60$ | DM1 | Substitute their $N$ (or $X$) to find $X$ (or $N$). Dependent on previous M mark. |
| $X = 26$ or $26.5$ | A1 (7) | Second value found correctly. |

**Alt:**

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $N\cos 30 - F\cos 60 = 1.5g$,  $N\cos 30 - 0.6N\cos 60 = 1.5g$ | M1, DM1 | Resolve vertically. Condone sin/cos confusion. Must have all terms. Use $F = 0.6N$ |
| (correct unsimplified equation) | A1 | |
| $N = \frac{1.5g}{\cos 30 - 0.6\cos 60} = 26$ or $26.0$ | A1 | |
| $X = F\cos 30 + N\cos 60 = N(0.6\cos 30 + \cos 60)$ | M1, DM1 | Resolve horizontally. Follow their $N$. Must have all terms. Condone sin/cos confusion. Substitute for $F$ and $N$. |
| $X = 26$ or $26.5$ | A1 | |

**Total: [12]**

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