| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two possible trajectories through point |
| Difficulty | Standard +0.3 Part (i) is a standard 'show that' derivation of the trajectory equation using basic kinematic equations—routine bookwork. Part (ii) applies this to find angles given numerical values, requiring solving a quadratic but following a standard method. This is a typical M2 projectiles question with no novel insight required, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = vt\cos\theta\) | B1 | aef |
| \(y = vt\sin\theta - \frac{1}{2}gt^2\) | B1 | aef; may see this with \(t\) already eliminated |
| M1 | Eliminate \(t\) | |
| \(y = x\tan\theta - \dfrac{4.9x^2}{v^2\cos^2\theta}\) | A1 | www; AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-100 = 40\tan\theta - 4.9 \times 40^2/(16^2 \times \cos^2\theta)\) | M1 | Attempt to substitute values into trajectory equation |
| A1 | ||
| \((30.625\tan^2\theta - 40\tan\theta - 69.375 = 0)\) | A1 | aef; obtain correct quadratic in \(\tan\theta\), may be unsimplified |
| \((\tan\theta = 2.2937\ldots \text{ or } -0.9876\ldots)\) | M1 | Attempt to solve quadratic in \(\tan\theta\) |
| \(\theta = 66.4°\) | A1 | |
| \(\theta = -44.6°\) | A1 | Allow 44.6 below the horizontal |
## Question 6:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = vt\cos\theta$ | B1 | aef |
| $y = vt\sin\theta - \frac{1}{2}gt^2$ | B1 | aef; may see this with $t$ already eliminated |
| | M1 | Eliminate $t$ |
| $y = x\tan\theta - \dfrac{4.9x^2}{v^2\cos^2\theta}$ | A1 | www; **AG** |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-100 = 40\tan\theta - 4.9 \times 40^2/(16^2 \times \cos^2\theta)$ | M1 | Attempt to substitute values into trajectory equation |
| | A1 | |
| $(30.625\tan^2\theta - 40\tan\theta - 69.375 = 0)$ | A1 | aef; obtain correct quadratic in $\tan\theta$, may be unsimplified |
| $(\tan\theta = 2.2937\ldots \text{ or } -0.9876\ldots)$ | M1 | Attempt to solve quadratic in $\tan\theta$ |
| $\theta = 66.4°$ | A1 | |
| $\theta = -44.6°$ | A1 | Allow 44.6 below the horizontal |
---6 A particle is projected with speed $v \mathrm {~ms} ^ { - 1 }$ from a point $O$ on horizontal ground. The angle of projection is $\theta ^ { \circ }$ above the horizontal. At time $t$ seconds after the instant of projection the horizontal displacement of the particle from $O$ is $x \mathrm {~m}$ and the upward vertical displacement from $O$ is $y \mathrm {~m}$.\\
(i) Show that
$$y = x \tan \theta - \frac { 4.9 x ^ { 2 } } { v ^ { 2 } \cos ^ { 2 } \theta } .$$
A stone is thrown from the top of a vertical cliff 100 m high. The initial speed of the stone is $16 \mathrm {~ms} ^ { - 1 }$ and the angle of projection is $\theta ^ { \circ }$ to the horizontal. The stone hits the sea 40 m from the foot of the cliff.\\
(ii) Find the two possible values of $\theta$.\\
\includegraphics[max width=\textwidth, alt={}, center]{8492ec9b-3327-4d89-aaa4-bf98cdf0ebdc-3_623_995_1475_536}
A uniform ladder $A B$ of weight $W \mathrm {~N}$ and length 4 m rests with its end $A$ on rough horizontal ground and its end $B$ against a smooth vertical wall. The ladder is inclined at an angle $\theta$ to the horizontal where $\tan \theta = \frac { 1 } { 2 }$ (see diagram). A small object $S$ of weight $2 W \mathrm {~N}$ is placed on the ladder at a point $C$, which is 1 m from $A$. The coefficient of friction between the ladder and the ground is $\mu$ and the system is in limiting equilibrium.\\
\hfill \mbox{\textit{OCR M2 2015 Q6 [10]}}