| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Moderate -0.3 This is a standard projectiles question requiring straightforward application of SUVAT equations in two dimensions. While it involves projection from an elevated point (slightly more complex than ground-to-ground), the calculations are routine: resolve initial velocity, find maximum height using vertical motion, then find range by solving a quadratic for time when y = 0. No novel problem-solving or geometric insight required—just methodical application of standard mechanics formulas. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0 = (19\sin11°)^2 - 2gh\). Hence max height is \(\frac{(19\sin11°)^2}{19.6} + 1.53 = 2.20\) m | M1, B1, A1, A1 (4 marks) | For use of relevant const acc equation for \(h\); For correct vertical component \(19\sin11°\); For correct expression for \(h\) (\(= 0.67\)); For correct answer 2.20 |
| (ii) EITHER: Time to top point is \(\frac{19\sin11°}{g} = 0.3699\) M1. Time to fall is \(\sqrt{\frac{2 \times 2.20}{9.8}} = 0.6701\) M1. Total time of flight is 1.04 A1. Horiz dist is \(19\cos11° \times 1.04 = 19.4\) m A1 M1 A1 | M1, M1, A1, A1, M1, A1 | For use of relevant const acc equation for \(t_{\text{up}}\); For use of relevant const acc eqn for \(t_{\text{down}}\); For a correct expression for \(t_{\text{down}}\); For correct value (or expression); For any use of \(x = (19\cos11°)t\); For correct answer 19.4 |
| OR: \(-1.53 = x\tan11° - \frac{gx^2}{2(19\cos11°)^2}\). Hence \(x = 19.4\) | M1, B1, A1, M1, A2 (6 marks) | For relevant use of trajectory equation; For \(y = -1.53\) correctly substituted; For completely correct equation for \(x\); For attempt to solve relevant quadratic; For correct answer 19.4 |
(i) $0 = (19\sin11°)^2 - 2gh$. Hence max height is $\frac{(19\sin11°)^2}{19.6} + 1.53 = 2.20$ m | M1, B1, A1, A1 (4 marks) | For use of relevant const acc equation for $h$; For correct vertical component $19\sin11°$; For correct expression for $h$ ($= 0.67$); For correct answer 2.20
(ii) EITHER: Time to top point is $\frac{19\sin11°}{g} = 0.3699$ M1. Time to fall is $\sqrt{\frac{2 \times 2.20}{9.8}} = 0.6701$ M1. Total time of flight is 1.04 A1. Horiz dist is $19\cos11° \times 1.04 = 19.4$ m A1 M1 A1 | M1, M1, A1, A1, M1, A1 | For use of relevant const acc equation for $t_{\text{up}}$; For use of relevant const acc eqn for $t_{\text{down}}$; For a correct expression for $t_{\text{down}}$; For correct value (or expression); For any use of $x = (19\cos11°)t$; For correct answer 19.4
OR: $-1.53 = x\tan11° - \frac{gx^2}{2(19\cos11°)^2}$. Hence $x = 19.4$ | M1, B1, A1, M1, A2 (6 marks) | For relevant use of trajectory equation; For $y = -1.53$ correctly substituted; For completely correct equation for $x$; For attempt to solve relevant quadratic; For correct answer 19.4
---6 An athlete 'puts the shot' with an initial speed of $19 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $11 ^ { \circ }$ above the horizontal. At the instant of release the shot is 1.53 m above the horizontal ground. By treating the shot as a particle and ignoring air resistance, find\\
(i) the maximum height, above the ground, reached by the shot,\\
(ii) the horizontal distance the shot has travelled when it hits the ground.
\hfill \mbox{\textit{OCR M2 Q6 [10]}}