| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle below horizontal or horizontal |
| Difficulty | Standard +0.3 This is a standard projectile motion problem with downward projection requiring application of SUVAT equations in 2D. Part (i) involves calculating displacement components, (ii) requires finding time to ground then horizontal range, and (iii) needs velocity components at impact. While it has multiple parts and requires careful sign conventions for downward projection, the techniques are routine for M2 level with no novel problem-solving required—slightly above average due to the multi-step nature and downward projection setup. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x = (26\cos30°) \times 2.3\); \(y = (26\sin30°) \times 2.3 + g \times 2.3^2/2\); \(d^2 = 51.8^2 + 56.35^2\); \(d = 76.5\) m | B1, B1, M1, A1 [4] | = 51.788...; = 56.35 |
| (ii) \(80 = (26\sin30°)t + 10t^2/2\); \(t = 2.91\) s [or (42.06-13)/10]; \(x = (2.906 \times 26\cos30°) = 65.4\) m OR \(80 = x\tan30° + 10x^2/(2 \times 26^2 \times \cos^2 30°)\); \(x = 65.4\) | M1, A1, A1 [3] | or \(v^2 = (26\sin30°)^2 + 2 \times 10 \times 80\) with v = 42.06; = \(26\sin30° + 10t\) solved for t; Uses trajectory equation; Attempts to solve the quadratic equation |
| (iii) \(v^2 = (26\sin30°)^2 + 2g \times 80\); \(v^2 = (26\sin30°)^2 + 2g \times 80 + (26\cos30°)^2\); \(V = 47.7\) ms\(^{-1}\); \(\alpha = \tan^{-1}[(42.06)/(26\cos30°)] = 61.8°\) | B1, M1, A1, A1 [4] | v = 42.06. Accept v = \(26\sin30° + 10 \times 2.91\) or award correct method to find \(\alpha\); Below horizontal (1.08) |
(i) $x = (26\cos30°) \times 2.3$; $y = (26\sin30°) \times 2.3 + g \times 2.3^2/2$; $d^2 = 51.8^2 + 56.35^2$; $d = 76.5$ m | B1, B1, M1, A1 [4] | = 51.788...; = 56.35
(ii) $80 = (26\sin30°)t + 10t^2/2$; $t = 2.91$ s [or (42.06-13)/10]; $x = (2.906 \times 26\cos30°) = 65.4$ m OR $80 = x\tan30° + 10x^2/(2 \times 26^2 \times \cos^2 30°)$; $x = 65.4$ | M1, A1, A1 [3] | or $v^2 = (26\sin30°)^2 + 2 \times 10 \times 80$ with v = 42.06; = $26\sin30° + 10t$ solved for t; Uses trajectory equation; Attempts to solve the quadratic equation
(iii) $v^2 = (26\sin30°)^2 + 2g \times 80$; $v^2 = (26\sin30°)^2 + 2g \times 80 + (26\cos30°)^2$; $V = 47.7$ ms$^{-1}$; $\alpha = \tan^{-1}[(42.06)/(26\cos30°)] = 61.8°$ | B1, M1, A1, A1 [4] | v = 42.06. Accept v = $26\sin30° + 10 \times 2.91$ or award correct method to find $\alpha$; Below horizontal (1.08)6 A particle $P$ is projected with speed $26 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ below the horizontal, from a point $O$ which is 80 m above horizontal ground.\\
(i) Calculate the distance from $O$ of the particle 2.3 s after projection.\\
(ii) Find the horizontal distance travelled by $P$ before it reaches the ground.\\
(iii) Calculate the speed and direction of motion of $P$ immediately before it reaches the ground.
\hfill \mbox{\textit{CAIE M2 2011 Q6 [11]}}