| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.8 This is a straightforward projectile motion problem requiring students to find the angle of projection given initial speed and horizontal range. It involves standard SUVAT equations and resolving velocity components - routine mechanics calculations with no novel problem-solving required, making it easier than average. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(u_H = 14\cos60 = 7\) | B1 | States or uses \(14\cos60\) for the horizontal component |
| \(u_V = 14\sin60 = 7\sqrt{3}\) | B1 | States or uses \(14\sin60\) for the vertical component |
| \(s = 7\sqrt{3}\,t - \dfrac{g}{2}t^2\) | M1 | Uses \(s = ut + \dfrac{1}{2}at^2\) with \(u\) = vertical component of velocity, \(a = -g\) and \(s = \pm1.5\); PI by \(t\) = AWFW [2.54, 2.60] |
| \(-1.5 = 7\sqrt{3}\,t - 4.9t^2\) | ||
| \(4.9t^2 - 12.124t - 1.5 = 0\) | ||
| \(t = 2.592\) seconds | A1 | Obtains \(t = 2.592\); AWFW [2.54, 2.60]; Exact value is \(\dfrac{3\sqrt{10}+5\sqrt{3}}{7}\) |
| \(7t = 18.144\) | ||
| \(ut + \dfrac{1}{2}at^2 = 0.5a(2.392)^2\) | M1 | Multiplies their \(t\) value by their horizontal component provided their \(t > 0.2\) |
| \(2.860832a = 18.144\) | M1 | Substitutes \(u = 0\), their \(t - 0.2\) into \(ut + \dfrac{1}{2}at^2\) to obtain an expression for the horizontal distance travelled by the dog |
| \(a = 6.3\) | A1 | Obtains 6.3; CAO |
## Question 20:
| Answer | Mark | Guidance |
|--------|------|----------|
| $u_H = 14\cos60 = 7$ | B1 | States or uses $14\cos60$ for the horizontal component |
| $u_V = 14\sin60 = 7\sqrt{3}$ | B1 | States or uses $14\sin60$ for the vertical component |
| $s = 7\sqrt{3}\,t - \dfrac{g}{2}t^2$ | M1 | Uses $s = ut + \dfrac{1}{2}at^2$ with $u$ = vertical component of velocity, $a = -g$ and $s = \pm1.5$; PI by $t$ = AWFW [2.54, 2.60] |
| $-1.5 = 7\sqrt{3}\,t - 4.9t^2$ | | |
| $4.9t^2 - 12.124t - 1.5 = 0$ | | |
| $t = 2.592$ seconds | A1 | Obtains $t = 2.592$; AWFW [2.54, 2.60]; Exact value is $\dfrac{3\sqrt{10}+5\sqrt{3}}{7}$ |
| $7t = 18.144$ | | |
| $ut + \dfrac{1}{2}at^2 = 0.5a(2.392)^2$ | M1 | Multiplies their $t$ value by their horizontal component provided their $t > 0.2$ |
| $2.860832a = 18.144$ | M1 | Substitutes $u = 0$, their $t - 0.2$ into $ut + \dfrac{1}{2}at^2$ to obtain an expression for the horizontal distance travelled by the dog |
| $a = 6.3$ | A1 | Obtains 6.3; CAO |
**Question 20 Total: 7 marks**
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**Question Paper Total: 100 marks**20 In this question use $g = 9.8 \mathrm {~m \mathrm {~s} ^ { - 2 }$}
Find $a$\\
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\hfill \mbox{\textit{AQA Paper 2 2023 Q20 [7]}}