October 2020 Paper 1 Q13
13 A projectile is fired from ground level at \(35\,\mathrm{m\,s^{-1}}\) at an angle of \(\theta^\circ\) above the horizontal.
The projectile travels above horizontal ground towards a wall that is 110 m away from the point of projection and 5 m high. The projectile reaches a maximum height of 22.5 m.
| Scheme | Marks | AO |
|---|---|---|
| eg. Neglect air resistance Constant gravity Projectile is a particle | B1 | 1.2 |
| [1] |
Notes
B1: One sensible statement.
Do not accept level ground
| Scheme | Marks | AO |
|---|---|---|
| \(u_x = 35\cos\theta\) giving \(x = (35\cos\theta)t\) | B1 | 3.3 |
| \(u_y = 35\sin\theta\) giving \(y = (35\sin\theta)t - \dfrac{1}{2}gt^2\) | B1 | 3.3 |
| Substitute for \(t\) | M1 | 3.3 |
| \(y = (35\sin\theta)\left(\dfrac{x}{35\cos\theta}\right) - \dfrac{1}{2}g\left(\dfrac{x}{35\cos\theta}\right)^2\) \(\left[y = x\tan\theta - \dfrac{x^2}{250\cos^2\theta}\right]\) | A1 | 1.1 |
| [4] |
Notes
B1: Award seen in any form
B1: soi
M1: Substituting for \(t\) in their equation for \(y\)
A1: Award in any form ISW
| Scheme | Marks | AO |
|---|---|---|
| EITHER Using \(s = 22.5,\quad u = 35\sin\theta,\quad v = 0,\quad a = -9.8\) \(v^2 = u^2 + 2as\) | M1 | 3.1b |
| \(0 = (35\sin\theta)^2 - 2 \times 9.8 \times 22.5\) | A1 | 1.1a |
| \(\sin\theta = 0.6\) [giving \(\theta = 36.9^\circ\)] | A1 | 1.1 |
| Use the trajectory with \(x = 110\) \(y = 110 \times \tan\theta^\circ - \dfrac{1}{250\cos^2\theta^\circ} \times 110^2\) | M1 | 3.1b |
| \(= 6.875\) | A1 | 1.1 |
| So it goes over the wall | E1 | 3.2a |
| [6] |
Notes
M1: Using any suvat in \(y\)-direction with \(v_y = 0\)
A1: Correct equation for \(\theta\) only
A1: allow \(37^\circ\)
Either \(\theta = 37^\circ\) or \(\cos\theta = \dfrac{4}{5},\ \tan\theta = \dfrac{3}{4}\) may be used.
M1: Allow in terms of \(\theta\)
FT their value for \(\theta\) if used
Allow this M mark for \(t = \dfrac{110}{35\cos\theta} = \dfrac{55}{14}\) used in a suitable equation for \(y\)
A1: Correct \(y\) value
E1: Conclusion in context from correct values
Alternative (OR)
| Scheme | Marks |
|---|---|
| Using \(s = 22.5,\quad u = 35\sin\theta,\quad v = 0,\quad a = -9.8\) \(v^2 = u^2 + 2as\) | M1 |
| \(0 = (35\sin\theta)^2 - 2 \times 9.8 \times 22.5\) | A1 |
| \(\sin\theta = 0.6\) giving \(\theta = 36.9^\circ\) | A1 |
| Use the trajectory with \(y = 5\) \(5 = x\tan\theta^\circ - \dfrac{1}{2450}g\sec^2\theta^\circ x^2\) | M1 |
| \(\dfrac{1}{160}x^2 - \dfrac{3}{4}x + 5 = 0\) \(x = 7.089,\quad 112.9\) | A1 |
| So particle is above the height of the wall between 7 m and 112.9 m away, so when \(x = 110\) m so it does not hit the wall | E1 |
M1: Using any suvat in \(y\)-direction with \(v_y = 0\)
A1: Correct equation for \(\theta\) only
A1: allow \(37^\circ\)
Either \(\theta = 37^\circ\) or \(\cos\theta = \dfrac{4}{5},\ \tan\theta = \dfrac{3}{4}\) may be seen.
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 \(x\).
A1: Both correct \(x\)-values
E1: Conclusion in context from correct values