## Question 5:

### Part (i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = u + at$ | M1 | Use of a suitable constant acceleration formula |
| $5 = 0 + a \times 10 \Rightarrow a = 0.5$ | A1 | Note: The value of $a$ is not required by the question so may be implied by subsequent working |
| $F = ma \Rightarrow 120 - R = 40 \times 0.5$ | M1 | Use of Newton's 2nd Law with correct elements |
| $R = 100\text{ N}$ | E1 | |
| **[4]** | | |

### Part (ii)(A)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $F = ma \Rightarrow -100 = 40a$ | M1 | Equation to find $a$ using Newton's 2nd Law |
| $\Rightarrow a = -2.5$ | A1 | |
| When $t = 1.6$: $v = 5 + (-2.5) \times 1.6 = 1\text{ ms}^{-1}$ | A1 | CAO |
| **[3]** | | |

### Part (ii)(B)

| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t = 6$, it is stationary. $v = 0\text{ ms}^{-1}$ | B1 | |
| **[1]** | | |

### Part (iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Motion parallel to the slope: Component of weight down slope $= 40g\sin 15°\ (= 101.457...)$ | B1 | |
| $200 - 40g\sin 15° = 40a$ | M1 | Equation of motion with correct elements present. No extra forces. |
| $a = 2.463...$ | | This result is not asked for in the question |
| $v^2 - u^2 = 2as \Rightarrow 8^2 = 2 \times 2.46... \times s$ | M1 | Use of a suitable constant acceleration formula, or combination of formulae. Dependent on previous M1. |
| $\Rightarrow s = 12.989...$ rounding to $13.0\text{ m}$ | E1 | Note: If rounding is not shown for $s$, the acceleration must satisfy $2.452... < a < 2.471...$ |
| **[4]** | | |

### Part (iv)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $a$ be acceleration up the slope | | |
| $-40 \times 9.8 \times \sin 15° = 40a$ | M1 | Use of Newton's 2nd Law parallel to the slope |
| $a = -2.536...$, i.e. $2.536\text{ ms}^{-2}$ down the slope | A1 | Condone sign error |
| $s = ut + \frac{1}{2}at^2$ | | |
| $-12.989... = 8t + \frac{1}{2} \times (-2.536...)t^2$ | M1 | Dependent on previous M1. Use of a suitable constant acceleration formula (or combination) in a relevant manner. |
| $1.268...t^2 - 8t - 12.989... = 0$ | A1 | Signs must be correct |
| $t = \dfrac{8 \pm \sqrt{64 - 4 \times 1.268... \times (-12.989...)}}{2 \times 1.268...}$ | M1 | Attempt to solve a relevant three-term quadratic equation |
| $t = -1.339...$ or $7.647...$, so $t = 7.65$ seconds | A1 | |
| **[6]** | | |

**Alternative 2-stage method for Part (iv):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $-40 \times 9.8 \times \sin 15° = 40a \Rightarrow a = -2.536...$ | M1, A1 | Use of Newton's 2nd Law parallel to slope; condone sign error |
| $v = u + at \Rightarrow 0 = 8 - 2.536...t \Rightarrow t = 3.154...$ | M1, A1 | Use of suitable constant acceleration formula for either $t$ or $s$ |
| $s = 8 \times 3.154... - \frac{1}{2} \times 2.536... \times 3.154...^2 \Rightarrow s = 12.616...$ | | |
| Distance to bottom $= 12.989... + 12.616... = 25.605...$ | | |
| $s = ut + \frac{1}{2}at^2 \Rightarrow 25.605... = \frac{1}{2} \times 2.536... \times t^2$ | M1 | Use of a suitable constant acceleration formula |
| $t = 4.493...$ | | |
| Total time $= 3.154... + 4.493... = 7.647...\text{ s}$ | A1 | |