# Question 7:

## Part (a)

| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{1}{2}mv^2 + \frac{3mgx^2}{4a} = mg(a+x)$ | M1 A2 (1,0) | |
| Leading to $v^2 = 2g(a+x) - \frac{3gx^2}{2a}$ | A1 | cso **(4)** |

## Part (b)

| Working | Marks | Guidance |
|---------|-------|----------|
| Greatest speed when acceleration is zero | | |
| $T = \frac{\lambda x}{a} = \frac{3mgx}{2a} = mg \Rightarrow x = \frac{2a}{3}$ | M1 A1 | |
| $v^2 = 2g\left(a + \frac{2a}{3}\right) - \frac{3g}{2a} \times \left(\frac{2a}{3}\right)^2 \left(= \frac{8ag}{3}\right)$ | M1 | |
| $v = \frac{2}{3}\sqrt{(6ag)}$ | A1 | accept exact equivalents **(4)** |

## Part (c)

| Working | Marks | Guidance |
|---------|-------|----------|
| $v = 0 \Rightarrow 2g(a+x) - \frac{3gx^2}{2a} = 0$ | M1 | |
| $3x^2 - 4ax - 4a^2 = (x-2a)(3x+2a) = 0$ | M1 A1 | |
| $x = 2a$ | | |
| At $D$: $m\ddot{x} = mg - \frac{\lambda \times 2a}{a}$ | M1 A1ft | ft their $2a$ |
| $|\ddot{x}| = 2g$ | A1 | **(6)** |
| | **[14]** | |

### Alternative (b) — Differentiation method

| Working | Marks | Guidance |
|---------|-------|----------|
| $2v\frac{dv}{dx} = 2g - \frac{3gx}{a}$ | | |
| $\frac{dv}{dx} = 0 \Rightarrow x = \frac{2a}{3}$ | M1 A1 | |
| $v^2 = 2g\left(a+\frac{2a}{3}\right) - \frac{3g}{2a}\times\left(\frac{2a}{3}\right)^2 \left(=\frac{8ag}{3}\right)$ | M1 | |
| $v = \frac{2}{3}\sqrt{(6ag)}$ | A1 | accept exact equivalents **(4)** |

### Alternative approach using SHM for (b) and (c)

| Working | Marks | Guidance |
|---------|-------|----------|
| $T = \frac{\lambda x}{a} = \frac{3mge}{2a} = mg \Rightarrow e = \frac{2a}{3}$ | bM1 bA1 | Equilibrium position |
| $m\ddot{y} = mg - \frac{\frac{3}{2}mg(y+e)}{a} = -\frac{3g}{2a}y$ | bM1 bA1 | N2L using $y$ for displacement from equilibrium |
| $\omega^2 = \frac{3g}{2a}$ | | |
| Speed at end of free fall: $u^2 = 2ga$ | cM1 | |
| $u^2 = 2ga$ when $y = -\frac{2}{3}a \Rightarrow 2ga = \frac{3g}{2a}\left(A^2 - \frac{4a^2}{9}\right)$ | cM1 | |
| $A = \frac{4a}{3}$ | cA1 | |
| Maximum speed $A\omega = \frac{4a}{3}\times\sqrt{\left(\frac{3g}{2a}\right)} = \frac{2}{3}\sqrt{(6ag)}$ | cM1 cA1 | |
| Maximum acceleration $A\omega^2 = \frac{4a}{3}\times\frac{3g}{2a} = 2g$ | cA1 | |

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