| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a standard projectile motion problem requiring application of kinematic equations at a specific time, then finding when velocity makes a given angle. Part (i) involves setting up equations for position at t=0.6s with the constraint that the angle of elevation is 45°, leading to a straightforward algebraic solution. Part (ii) requires finding when tan(angle) of velocity equals 1. Both parts use routine mechanics techniques with no novel insight required, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x = (\vcos60)0.6\) and \(y = (\vsin60)0.6 - g0.6^2/2\) | M1 | Finds both coordinates in terms of \(t = 0.6\) |
| DM1 | Relates coordinates and \(45°\) angle | |
| \(\tan45 = [(\vsin60)0.6 - g0.6^2/2]/[(\vcos60)0.6]\) | A1 | \((\vsin60)0.6 - g0.6^2/2 = (\vcos60)0.6\) |
| \(v = 8.2(0) \text{ ms}^{-1}\) | AG A1 | [4] |
| (ii) \(\text{Relates velocity components and } 45°\) | M1 | |
| \(8.2\sin60 - gt = 8.2\cos60\) | A1 | \(\tan45 = (8.2\sin60 - gt)/(8.2\cos60)\) |
| \(T = 0.3(00) \text{ s}\) | A1 | [3] |
**(i)** $x = (\vcos60)0.6$ and $y = (\vsin60)0.6 - g0.6^2/2$ | M1 | Finds both coordinates in terms of $t = 0.6$
| DM1 | Relates coordinates and $45°$ angle
$\tan45 = [(\vsin60)0.6 - g0.6^2/2]/[(\vcos60)0.6]$ | A1 | $(\vsin60)0.6 - g0.6^2/2 = (\vcos60)0.6$
$v = 8.2(0) \text{ ms}^{-1}$ | AG A1 | [4]
**(ii)** $\text{Relates velocity components and } 45°$ | M1
$8.2\sin60 - gt = 8.2\cos60$ | A1 | $\tan45 = (8.2\sin60 - gt)/(8.2\cos60)$
$T = 0.3(00) \text{ s}$ | A1 | [3]
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\includegraphics[max width=\textwidth, alt={}, center]{d1f1f036-1676-443e-b733-ca1fe79972d4-2_525_913_1123_616}
A particle $P$ is projected from a point $O$ at an angle of $60 ^ { \circ }$ above horizontal ground. At an instant 0.6 s after projection, the angle of elevation of $P$ from $O$ is $45 ^ { \circ }$ (see diagram).\\
(i) Show that the speed of projection of $P$ is $8.20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, correct to 3 significant figures.\\
(ii) Calculate the time after projection when the direction of motion of $P$ is $45 ^ { \circ }$ above the horizontal.
\hfill \mbox{\textit{CAIE M2 2011 Q2 [7]}}