# Question 13:

## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. Neglect air resistance / Constant gravity / Projectile is a particle | B1 (1.2) [1] | One sensible statement. Do not accept level ground |

## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_x = 35\cos\theta$ giving $x = (35\cos\theta)t$ | B1 (3.3) | Award seen in any form |
| $u_y = 35\sin\theta$ giving $y = (35\sin\theta)t - \frac{1}{2}gt^2$ | B1 (3.3) | soi |
| Substitute for $t$: $y = (35\sin\theta)\left(\frac{x}{35\cos\theta}\right) - \frac{1}{2}g\left(\frac{x}{35\cos\theta}\right)^2$ | M1 (3.3) | Substituting for $t$ in their equation for $y$ |
| $\left[y = x\tan\theta - \frac{x^2}{250\cos^2\theta}\right]$ | A1 (1.1) [4] | Award in any form ISW |

## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER:** Using $s=22.5$, $u=35\sin\theta$, $v=0$, $a=-9.8$: $v^2=u^2+2as$ | M1 (3.1b) | Using any *suvat* in $y$-direction with $v_y=0$ |
| $0 = (35\sin\theta)^2 - 2\times9.8\times22.5$ | A1 (1.1a) | Correct equation for $\theta$ only. Either $\theta = 37°$ or $\cos\theta=\frac{4}{5}$, $\tan\theta=\frac{3}{4}$ may be used |
| $\sin\theta = 0.6$ [giving $\theta = 36.9°$] | A1 (1.1) | Allow $37°$ |
| Use trajectory with $x=110$: $y = 110\times\tan\theta° - \frac{1}{250\cos^2\theta°}\times110^2$ | M1 (3.1b) | Allow in terms of $\theta$. FT their value for $\theta$ if used. Allow this M mark for $t = \frac{110}{35\cos\theta} = \frac{55}{14}$ used in suitable equation for $y$ |
| $= 6.875$, so it goes over the wall | A1 (1.1) | Correct $y$ value |
| | E1 (3.2a) [6] | Conclusion in context from correct values |

# Mark Scheme Extraction - H640/01 November 2020

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## Question (Projectile - Wall problem):

**OR method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $s = 22.5$, $u = 35\sin\theta$, $v = 0$, $a = -9.8$ | M1 | Using any suvat in y-direction with $v_y = 0$ |
| $v^2 = u^2 + 2as$; $0 = (35\sin\theta)^2 - 2\times9.8\times22.5$ | A1 | Correct equation for $\theta$ only |
| $\sin\theta = 0.6$ giving $\theta = 36.9°$ | A1 | Allow 37°; Either $\theta = 37°$ or $\cos\theta = \frac{4}{5}$, $\tan\theta = \frac{3}{4}$ may be seen |
| Use trajectory with $y = 5$: $5 = x\tan\theta° - \frac{1}{2450}g\sec^2\theta°\, x^2$ | M1 | Allow in terms of $\theta$; FT their value for $\theta$ if used; Allow this M mark for roots of their $y=5$ used in a suitable equation for $x$ |
| $\frac{1}{160}x^2 - \frac{3}{4}x + 5 = 0$ | A1 | Both correct $x$-values |
| $x = 7.089$, $112.9$ | | |
| So particle is above height of wall between 7 m and 112.9 m away, so when $x = 110$ m it does not hit the wall | E1 [6] | Conclusion in context from correct values |

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