| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Standard +0.3 This is a standard projectiles question requiring extraction of angle and speed from trajectory equation (using coefficient relationships), then finding velocity components at a specific height. The methods are well-practiced M2 techniques with straightforward algebraic manipulation, making it slightly easier than average but requiring multiple connected steps. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = 3l(0)^2\) | B1 | \(\theta = \tan^{-1}0.6\) |
| \(0.017 = 10/[2(\nu\cos31)]^2\) | M1 | |
| \(v = 20\) | AG, A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Accept \(v^2 = 20^2 - 2g \times 5.2\) | |
| \(v = 17.2\) ms\(^{-1}\) | A1 | |
| \(0.017x^2 - 0.6x + 5.2 = 0\) | M1 | Solves 3 term quadratic equation |
| \(x = 20\) | A1 | Ignore smaller root if shown |
| \(dy/dx = 0.6 - 0.017(2x)\) | M1 | |
| \(\tan\alpha = 0.6 - 0.017(2 \times 20)\) | A1 | |
| \(\alpha = 4.6°\) below the horizontal | A1 [7] | 4.57° |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = 3l(0)^2\) | B1 | \(\theta = \tan^{-1}0.6\) |
| \(0.017 = 10/[2(\nu\cos31)]^2\) | M1 | |
| \(v = 20\) | AG, A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(5.2 = 20\sin31 - 10t^2/2\) | M1 | Sets up and solves a 3 term quadratic equation |
| \(t = 1.17 (t = 1.166...)\) | A1 | Ignore smaller root if shown |
| \(v_{\text{vert}} = (-)1.37(2)\) | A1 | From \(v = 20\sin31 - 10t\) |
| \(v^2 = 17.1(5)^2 + 1.37(2)^2\) | M1 | Or uses method in (ii) above 17.1(5) is horizontal velocity component |
| \(v = 17.2\) ms\(^{-1}\) | A1 | |
| \(\tan\alpha = 1.37(2)/17.1(5)\) | M1 | |
| \(\alpha = 4.6°\) below the horizontal | A1 [7] | 4.57° |
**First approach:**
**(i)**
| $\theta = 3l(0)^2$ | B1 | $\theta = \tan^{-1}0.6$ |
| $0.017 = 10/[2(\nu\cos31)]^2$ | M1 | |
| $v = 20$ | AG, A1 [3] | |
**(ii)**
| | M1 | Accept $v^2 = 20^2 - 2g \times 5.2$ |
| $v = 17.2$ ms$^{-1}$ | A1 | |
| $0.017x^2 - 0.6x + 5.2 = 0$ | M1 | Solves 3 term quadratic equation |
| $x = 20$ | A1 | Ignore smaller root if shown |
| $dy/dx = 0.6 - 0.017(2x)$ | M1 | |
| $\tan\alpha = 0.6 - 0.017(2 \times 20)$ | A1 | |
| $\alpha = 4.6°$ below the horizontal | A1 [7] | 4.57° |
**OR**
**Second approach:**
**(i)**
| $\theta = 3l(0)^2$ | B1 | $\theta = \tan^{-1}0.6$ |
| $0.017 = 10/[2(\nu\cos31)]^2$ | M1 | |
| $v = 20$ | AG, A1 [3] | |
**(ii)**
| $5.2 = 20\sin31 - 10t^2/2$ | M1 | Sets up and solves a 3 term quadratic equation |
| $t = 1.17 (t = 1.166...)$ | A1 | Ignore smaller root if shown |
| $v_{\text{vert}} = (-)1.37(2)$ | A1 | From $v = 20\sin31 - 10t$ |
| $v^2 = 17.1(5)^2 + 1.37(2)^2$ | M1 | Or uses method in (ii) above 17.1(5) is horizontal velocity component |
| $v = 17.2$ ms$^{-1}$ | A1 | |
| $\tan\alpha = 1.37(2)/17.1(5)$ | M1 | |
| $\alpha = 4.6°$ below the horizontal | A1 [7] | 4.57° | [10] |7 The equation of the trajectory of a projectile is $y = 0.6 x - 0.017 x ^ { 2 }$, referred to horizontal and vertically upward axes through the point of projection.\\
(i) Find the angle of projection of the projectile, and show that the initial speed is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the speed and direction of motion of the projectile when it is at a height of 5.2 m above the level of the point of projection for the second time.
\hfill \mbox{\textit{CAIE M2 2012 Q7 [10]}}