| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Speed at specific time or position |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question with straightforward application of SUVAT equations and impulse-momentum theorem. Part (i) requires finding maximum height (routine), part (ii) applies I=mΔv directly, parts (iii-iv) involve standard projectile motion after the impulse. All steps are textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 = (14\sin30°)^2 - 2gh\) | M1 | \(h = (14\sin30°)x1/1.4 - g(1/1.4)^2/2\) or use \((u^2\sin^2θ)/2g\) |
| \(h = 2.5\) m | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.4x15 = 0.4(14\cos30°) + I\) | M1 | Impulse = change in momentum. Not 14 or 0 for horizontal speed before impulse. aef |
| \(I = 1.15\) | A1 | |
| A1 | ||
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = (14\sin30°)^2 + 15^2\) | M1 | Not \((14\sin30°)^2 + (14\cos30°)^2\). Allow \(\sqrt{274}\). Correct trig to find an appropriate angle; not \(14\cos30\) for 15 |
| \(v = 16.6\) m s\(^{-1}\) | A1 | |
| \(\tan θ = 14\sin30°/15\) OR \(\tan ψ = 15/14\sin30°\) | M1 | |
| \(θ = 25(.)°\) OR \(ψ = 65(.)°\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t = 14\sin30°/g\) (= 1/1.4 = 0.7142...) | M1 | Rise or fall time (not to be given in (i)). Accept 10/7. \((14^2\sin(2x30) + 16.6^2\sin(2x25))/2g\). 14 resolved, 15 not |
| \(T = 1.43\) s | A1 | |
| \(R = 14\cos30°/1.4 + 15/1.4\) | M1A1 | |
| \(R = 19.4\) m | A1 | |
| [5] |
**Part (i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 = (14\sin30°)^2 - 2gh$ | M1 | $h = (14\sin30°)x1/1.4 - g(1/1.4)^2/2$ or use $(u^2\sin^2θ)/2g$ |
| $h = 2.5$ m | A1 | |
| | **[2]** | |
**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4x15 = 0.4(14\cos30°) + I$ | M1 | Impulse = change in momentum. Not 14 or 0 for horizontal speed before impulse. aef |
| $I = 1.15$ | A1 | |
| | A1 | |
| | **[3]** | |
**Part (iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = (14\sin30°)^2 + 15^2$ | M1 | Not $(14\sin30°)^2 + (14\cos30°)^2$. Allow $\sqrt{274}$. Correct trig to find an appropriate angle; not $14\cos30$ for 15 |
| $v = 16.6$ m s$^{-1}$ | A1 | |
| $\tan θ = 14\sin30°/15$ OR $\tan ψ = 15/14\sin30°$ | M1 | |
| $θ = 25(.)°$ OR $ψ = 65(.)°$ | A1 | |
| | **[4]** | |
**Part (iv)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t = 14\sin30°/g$ (= 1/1.4 = 0.7142...) | M1 | Rise or fall time (not to be given in (i)). Accept 10/7. $(14^2\sin(2x30) + 16.6^2\sin(2x25))/2g$. 14 resolved, 15 not |
| $T = 1.43$ s | A1 | |
| $R = 14\cos30°/1.4 + 15/1.4$ | M1A1 | |
| $R = 19.4$ m | A1 | |
| | **[5]** | |6 A small ball $B$ is projected with speed $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of elevation $30 ^ { \circ }$ from a point $O$ on a horizontal plane, and moves freely under gravity.\\
(i) Calculate the height of $B$ above the plane when moving horizontally.\\
$B$ has mass 0.4 kg . At the instant when $B$ is moving horizontally it receives an impulse of magnitude $I \mathrm { Ns }$ in its direction of motion which immediately increases the speed of $B$ to $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Calculate $I$.
For the instant when $B$ returns to the plane, calculate\\
(iii) the speed and direction of motion of $B$,\\
(iv) the time of flight, and the distance of $B$ from $O$.
\hfill \mbox{\textit{OCR M2 2011 Q6 [14]}}