## Question 16:

### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolve vertically: $R = mg - T\sin\theta$ | B1 | Must be explicit; may be seen on diagram. Allow $R + T\sin\theta = mg$ if $F = \mu(mg - T\sin\theta)$ also seen |
| $F = \mu R$, so $F = \mu(mg - T\sin\theta)$ | M1 | Allow only if $R$ seen explicitly or correct vertical equation seen; allow $F = \mu mg$ only if $R = mg$ seen explicitly |
| Resolve horizontally: $T\cos\theta - F = ma$; $ma = T\cos\theta - \mu mg + T\mu\sin\theta$ | M1 | All forces correct and no extras; allow sign errors |
| $a = \frac{T}{m}\cos\theta - \mu g + \frac{T}{m}\mu\sin\theta$ | A1 | AG — complete argument needed |

## Question 16(b):

$$\frac{da}{d\theta} = -\frac{T}{m}\sin\theta + \frac{T}{m}\mu\cos\theta$$ | **M1** | AO 3.1a | Attempt to differentiate with respect to $\theta$

$$-\frac{T}{m}\sin\alpha + \frac{T}{m}\mu\cos\alpha = 0$$ | **M1** | AO 1.1a | Equate their derivative to 0 and attempt to rearrange using a trig identity. Condone using $\theta$ not $\alpha$

$$\frac{\sin\alpha}{\cos\alpha} = \mu \text{ so } \alpha = \tan^{-1}\mu$$ | **A1** | AO 1.1 | Must be $\alpha =$ ... Also allow for $\alpha = \frac{\pi}{2} - \tan^{-1}\frac{1}{\mu}$

**Alternative solution:**

Maximum $a$ when $\frac{T}{m}(\cos\theta + \mu\sin\theta)$ is max | — | —

Acceleration is $(R\cos(\theta - \beta))$ where $\beta = \tan^{-1}\mu$ | **M1, M1** | Uses trig identity; Attempt to find the value of $\beta$

Max acceleration when $(\alpha - \beta) = 0$, giving $\alpha = \tan^{-1}\mu$ | **A1** | Must be $\alpha =$ ... Also allow for $\alpha = \frac{\pi}{2} - \tan^{-1}\frac{1}{\mu}$

**Total: [3]**

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