| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two projectiles meeting - 2D flight |
| Difficulty | Challenging +1.2 This is a standard two-projectile problem requiring systematic application of SUVAT equations in both horizontal and vertical directions. Part (i) involves equating times of flight and comparing horizontal distances, while part (ii) requires equating both position components simultaneously. Though multi-step with two scenarios, the techniques are routine for M2 students with no novel geometric insight required—slightly above average due to algebraic manipulation and the two-part structure. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For \(P\): \(4.9t^2 = 60\) | M1 | Signs must be consistent |
| \(t = 3.5(0)\) | A1 | aef |
| For \(Q\): \(0 = 25\sin\theta \times t - \frac{1}{2} \times 9.8 \times t^2\) | M1 | |
| \(\theta = 43.3\) | A1 | |
| \(PQ = (25\cos\theta - 15) \times t_c\) | M1 | |
| \(= 11.2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(25\cos\theta(t) = 15(t)\) and solving for \(\theta\) | M1 | Equating horizontal components of velocity (or displacement) and solving for \(\theta\) |
| \(\theta = 53.1\) | A1 | |
| For \(Q\): \(s_{y1} = 25\sin\theta \times t - \frac{1}{2} \times 9.8 \times t^2\) | B1 | |
| For \(P\): \(s_{y2} = \pm\frac{1}{2} \times 9.8 \times t^2\) | B1 | |
| Using \(s_{y1} + s_{y2} = 60\) | *M1 | |
| Solving for \(t\) | M1dep* | |
| \(t = 3\) | A1 | |
| \(v = 25\sin\theta - 9.8 \times 3\) | M1 | Other methods include finding time to max height for Q |
| \(v = -9.4\) therefore falling | A1 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For $P$: $4.9t^2 = 60$ | M1 | Signs must be consistent |
| $t = 3.5(0)$ | A1 | aef |
| For $Q$: $0 = 25\sin\theta \times t - \frac{1}{2} \times 9.8 \times t^2$ | M1 | |
| $\theta = 43.3$ | A1 | |
| $PQ = (25\cos\theta - 15) \times t_c$ | M1 | |
| $= 11.2$ | A1 | |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $25\cos\theta(t) = 15(t)$ and solving for $\theta$ | M1 | Equating horizontal components of velocity (or displacement) and solving for $\theta$ |
| $\theta = 53.1$ | A1 | |
| For $Q$: $s_{y1} = 25\sin\theta \times t - \frac{1}{2} \times 9.8 \times t^2$ | B1 | |
| For $P$: $s_{y2} = \pm\frac{1}{2} \times 9.8 \times t^2$ | B1 | |
| Using $s_{y1} + s_{y2} = 60$ | *M1 | |
| Solving for $t$ | M1dep* | |
| $t = 3$ | A1 | |
| $v = 25\sin\theta - 9.8 \times 3$ | M1 | Other methods include finding time to max height for Q |
| $v = -9.4$ therefore falling | A1 | |7 A particle $P$ is projected horizontally with speed $15 \mathrm {~ms} ^ { - 1 }$ from the top of a vertical cliff. At the same instant a particle $Q$ is projected from the bottom of the cliff, with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\theta ^ { \circ }$ above the horizontal. $P$ and $Q$ move in the same vertical plane. The height of the cliff is 60 m and the ground at the bottom of the cliff is horizontal.\\
(i) Given that the particles hit the ground simultaneously, find the value of $\theta$ and find also the distance between the points of impact with the ground.\\
(ii) Given instead that the particles collide, find the value of $\theta$, and determine whether $Q$ is rising or falling immediately before this collision.
\hfill \mbox{\textit{OCR M2 2012 Q7 [15]}}