| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.3 This is a standard M2 projectiles question requiring routine application of kinematic equations to derive the trajectory equation (a bookwork result), followed by straightforward substitution and calculation. Part (i) is pure derivation of a standard formula, parts (ii-iv) involve direct substitution and basic trigonometry. The collision mechanics in the second part is also standard M2 material using conservation of momentum and Newton's law of restitution, though it requires careful bookkeeping across two impacts. Overall slightly easier than average due to the routine nature of all components. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = v\cos\theta \cdot t\) | B1 | |
| \(y = v\sin\theta \cdot t - \frac{1}{2} \times 9.8t^2\) | B1 | or g |
| substitute \(t = x/v\cos\theta\) | M1 | |
| \(y = x\tan\theta - \frac{4.9x^2}{v^2\cos^2\theta}\) | A1 4 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sub \(y = -h\), \(x = h\), \(v = 14\), \(\theta = 30\) | M1 | signs must be correct |
| \(-h = h/\sqrt{3} - h^2/30\) | A1 | aef |
| solving above | M1 | |
| \(h = 47.3\) | A1 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v_v^2 = (14\sin30°)^2 - 2\times9.8\times(-47.3)\) | M1 | \(14\cos30°\) \(t=47.3\) ft & \(v_v = 14\sin30° - 9.8t\) |
| (double negative needed) ft their \(-47.3\) | A1 ft | \(t = 3.90\) (or \(dy/dx = 1/\sqrt{3} - x/15\) etc ft) |
| \(v_v = \pm31.2\) | A1 | \(v_v = \pm31.2\) \((\tan\alpha = 1/\sqrt{3} - 47.3/15)\) |
| \(\tan^{-1}(31.2/14\cos30°)\) | M1 | \(\tan^{-1}(31.2/14\cos30°)\) |
| \(\alpha = 68.8°\) below horiz/\(21.2°\) to d'vert. | A1 5 | \(68.8°/\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}mx14^2 + mx9.8x47.3 = \frac{1}{2}mv^2\) | M1 | ft \((12.1^2 + 31.2^2)\) |
| \(v = 33.5\) | A1 2 | 33.5 [15] |
# Question 6:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = v\cos\theta \cdot t$ | B1 | |
| $y = v\sin\theta \cdot t - \frac{1}{2} \times 9.8t^2$ | B1 | or g |
| substitute $t = x/v\cos\theta$ | M1 | |
| $y = x\tan\theta - \frac{4.9x^2}{v^2\cos^2\theta}$ | A1 **4** | **AG** |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sub $y = -h$, $x = h$, $v = 14$, $\theta = 30$ | M1 | signs must be correct |
| $-h = h/\sqrt{3} - h^2/30$ | A1 | aef |
| solving above | M1 | |
| $h = 47.3$ | A1 **4** | |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $v_v^2 = (14\sin30°)^2 - 2\times9.8\times(-47.3)$ | M1 | $14\cos30°$ $t=47.3$ ft & $v_v = 14\sin30° - 9.8t$ |
| (double negative needed) ft their $-47.3$ | A1 ft | $t = 3.90$ (or $dy/dx = 1/\sqrt{3} - x/15$ etc ft) |
| $v_v = \pm31.2$ | A1 | $v_v = \pm31.2$ $(\tan\alpha = 1/\sqrt{3} - 47.3/15)$ |
| $\tan^{-1}(31.2/14\cos30°)$ | M1 | $\tan^{-1}(31.2/14\cos30°)$ |
| $\alpha = 68.8°$ below horiz/$21.2°$ to d'vert. | A1 **5** | $68.8°/\ldots$ |
## Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}mx14^2 + mx9.8x47.3 = \frac{1}{2}mv^2$ | M1 | ft $(12.1^2 + 31.2^2)$ |
| $v = 33.5$ | A1 **2** | 33.5 **[15]** |
---6 A particle is projected from a point $O$ with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of elevation $\theta$ above the horizontal and it moves freely under gravity. The horizontal and upward vertical displacements of the particle from $O$ at any subsequent time, $t$ seconds, are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.\\
(i) Express $x$ and $y$ in terms of $\theta$ and $t$, and hence show that
$$y = x \tan \theta - \frac { 4.9 x ^ { 2 } } { v ^ { 2 } \cos ^ { 2 } \theta } .$$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{dd23f4a8-f7e7-4f80-bad6-7e9ec21565fc-4_551_575_667_826}
\end{center}
The particle subsequently passes through the point $A$ with coordinates $( h , - h )$ as shown in the diagram. It is given that $v = 14$ and $\theta = 30 ^ { \circ }$.\\
(ii) Calculate $h$.\\
(iii) Calculate the direction of motion of the particle at $A$.\\
(iv) Calculate the speed of the particle at $A$.\\
\includegraphics[max width=\textwidth, alt={}, center]{dd23f4a8-f7e7-4f80-bad6-7e9ec21565fc-4_278_1061_1749_543}
Two small spheres, $P$ and $Q$, are free to move on the inside of a smooth hollow cylinder, in such a way that they remain in contact with both the curved surface and the base of the cylinder. The mass of $P$ is 0.2 kg , the mass of $Q$ is 0.3 kg and the radius of the cylinder is $0.4 \mathrm {~m} . P$ and $Q$ are stationary at opposite ends of a diameter of the base of the cylinder (see diagram). The coefficient of restitution between $P$ and $Q$ is $0.5 . P$ is given an impulse of magnitude 0.8 Ns in a tangential direction.\\
(i) Calculate the speeds of the particles after $P$ 's first impact with $Q$.\\
$Q$ subsequently catches up with $P$ and there is a second impact.\\
(ii) Calculate the speeds of the particles after this second impact.\\
(iii) Calculate the magnitude of the force exerted on $Q$ by the curved surface of the cylinder after the second impact.
\hfill \mbox{\textit{OCR M2 2009 Q6 [15]}}