## Question 13(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Newton's second law for the train: $5 - 2 \times 0.8 = (0.5 + 0.4)a$ | M1 | N2L for whole train with correct mass and all forces present. Also allow for 2 equations where both have correct mass and all forces present in each |
| Alternative: $5 - 0.8 - T = 0.5a$; $T - 0.8 = 0.4a$ | | |
| giving $a = \frac{34}{9} = 3.78\ \text{m s}^{-2}$ | A1 | |
| Using $v = u + at$ with $u = 0$, $t = 1.5$ | M1 | Using suvat equation(s) with $u = 0$ and their $a \neq g$ leading to a value for $v$ |
| $v = \frac{34}{9} \times 1.5 = \frac{17}{3} = 5.67\ \text{m s}^{-1}$ (3sf) | A1 | FT their $a$. Any form |
| **[4]** | | |

## Question 13(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Diagram showing weights and normal reactions (must be distinct and not vertical) | B1 | Allow if both components of weight given instead. Allow in addition to weight only if clear they are for working purposes only |
| Tensions in string and coupling parallel to inclined plane | B1 | |
| $R$ marked for both parts of the train. No additional forces | B1 | Allow if distinct if it is clear they are equal in later work |
| **[3]** | | |

## Question 13(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Newton's second law: $P - 2R - 0.9g\sin 20° = 0.9a$ | M1 | Newton's law with $m = 0.9$. Allow for incorrect weight term(s) or $R$ used instead of $2R$ |
| (fully correct) | A1 | Fully correct, any form |
| **[2]** | | |

## Question 13(d):

**Method 1:**

When $P = 5$: $5 - 2R - 0.9g\sin 20° = 0.9a$ | M1 | **3.1b** | Establishes one equation linking $R$ and $a$. FT their (c)

When $P = 5.5$: $5.5 - 2R - 0.9g\sin 20° = 0.9 \times 2a$ | M1 | **3.1b** | Establishes another equation linking $R$ and $a$. Consistent with first equation. Method need not be seen BC

Solve simultaneously giving $R = 0.742$ $\left[a = \frac{5}{9}\right]$ | A1 | **1.1b** | Correct value for $R$ ($a$ is not required)

**Alternative method:**

When $P = 5$: $a = \frac{5 - 2R - 0.9g\sin 20°}{0.9}$ | M1 | | Finds expression for $a$ when $P = 5$ or $P = 5.5$. Soi

When $P = 5.5$: $a_1 = \frac{5.5 - 2R - 0.9g\sin 20°}{0.9}$

$\frac{5.5 - 2R - 0.9g\sin 20°}{0.9} = 2\left(\frac{5 - 2R - 0.9g\sin 20°}{0.9}\right)$ | M1 | | Links corresponding acceleration for the other value of $P$. Do not allow factor of 2 on the wrong side

Giving $R = 0.742$ $\left[a = \frac{5}{9}\right]$ | A1 | | Correct value for $R$ ($a$ is not required)

**[3]**

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