## Question 1:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse-momentum principle applied | M1 | Dimensionally correct, mass × velocity. Must be subtracting momenta but condone subtracting in wrong order. M0 if $g$ is included. |
| $(-6\mathbf{i}+42\mathbf{j})=2\{\mathbf{v}-(-4\mathbf{i}+3\mathbf{j})\}$ | A1 | Correct unsimplified equation |
| Find magnitude of their $\mathbf{v}$: $\sqrt{(-7)^2+24^2}$ | M1 | Correct application of Pythagoras to find magnitude of their $v$. M0 for incorrect speed if no evidence of Pythagoras being used on their velocity. |
| $25\ (\text{m s}^{-1})$ | A1 | Correct answer following the correct velocity |
| | **(4)** | |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\alpha=\dfrac{(-4\times-7)+(3\times24)}{\sqrt{(-4)^2+3^2}\times\sqrt{(-7)^2+24^2}}$ | M1 | Complete method using scalar product with their $\mathbf{v}$. Formula must be correct: $\cos\alpha=\dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$. M0 if fraction is upside down. Do not ISW. |
| $\alpha=37°$ or better | A1 | cao in degrees |
| | **(2)** | |

**Alt 1:** $\cos\alpha=\dfrac{\{(-4)^2+3^2\}+\{(-7)^2+24^2\}-(3^2+(-21)^2)}{2\times5\times\sqrt{(-7)^2+24^2}}$ | M1 | Use cosine rule in vector triangle |
| $\alpha=37°$ or better | A1 | |

**Alt 2:** $\alpha=\tan^{-1}\!\left(\dfrac{24}{7}\right)-\tan^{-1}\!\left(\dfrac{3}{4}\right)$ or $\alpha=90-\tan^{-1}\!\left(\dfrac{7}{24}\right)-\tan^{-1}\!\left(\dfrac{3}{4}\right)$ | M1 | Use of inverse tan |
| $\alpha=37°$ or better | A1 | |
| | **(2)** | |

**Total: 6 marks**

# Question 1 (part b, alt2):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method using inverse tan formulae for their **v** | M1 | Do not ISW. M0 for $\tan^{-1}\left(\frac{3}{4}\right)$ alone which gives value 36.869... |
| Correct answer in degrees | A1 | cao in degrees |

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