| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: relative motion |
| Difficulty | Standard +0.3 This is a standard M1 projectile motion question requiring SUVAT equations and understanding of when particles change direction. Part (i) needs identifying when one particle is ascending while the other descends (straightforward once you find when each reaches maximum height). Part (ii) involves setting up simultaneous equations with height and velocity relationships—routine algebraic manipulation with no novel insight required. Slightly above average due to the two-part structure and algebraic work, but well within typical M1 scope. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | For using \(\theta = u - gt\) to find times at maximum heights. |
| Times to max. height are 1.2s and 0.7s | A1 | |
| Range of values is \(0.7 < t < 1.2\) | A1 | [3] |
| (ii) | M1 | For using \(h = ut - \frac{1}{2}gt^2\) and attempting to solve \(3h_A = 8h_B\) for \(t\) |
| \(36t - 1.5gt^2 = 5t - 4gt^2\) | A1 | |
| \(t = 8/g\) | A1 | |
| M1 | For using \(v = u - gt\) | |
| Velocities are 4m \(^-1\) and \(-1\)ms \(^{-1}\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| For using \(3h_P = 8h_Q \to 3(v_P^2 - 144) \div (-20) = 8(v_Q^2 - 49) \div (-20) \to 3v_P^2 - 8v_Q^2 = 40\) | B1 | |
| For using \(v_P = 12 - 10t\) and \(v_Q = 7 - 10t\) \(\to v_P - v_Q = 5\) | B1 | |
| For eliminating \(v_Q\) (or \(v_P\)) and solving for \(v_P\) (or \(v_Q\)). | M1 | |
| \(v_P^2 - 16v_P + 48 = 0 \to v_P = 4\) (or 4, 12) | A1 | |
| Upward velocities are 4 ms\(^{-1}\) and \(-1\) ms\(^{-1}\) | A1 | [5] |
**(i)** | M1 | For using $\theta = u - gt$ to find times at maximum heights.
Times to max. height are 1.2s and 0.7s | A1 |
Range of values is $0.7 < t < 1.2$ | A1 | [3]
**(ii)** | M1 | For using $h = ut - \frac{1}{2}gt^2$ and attempting to solve $3h_A = 8h_B$ for $t$
$36t - 1.5gt^2 = 5t - 4gt^2$ | A1 |
$t = 8/g$ | A1 |
| M1 | For using $v = u - gt$
Velocities are 4m $^-1$ and $-1$ms $^{-1}$ | A1 | [5]
**Alternative for part 5(ii)**
For using $3h_P = 8h_Q \to 3(v_P^2 - 144) \div (-20) = 8(v_Q^2 - 49) \div (-20) \to 3v_P^2 - 8v_Q^2 = 40$ | B1 |
For using $v_P = 12 - 10t$ and $v_Q = 7 - 10t$ $\to v_P - v_Q = 5$ | B1 |
For eliminating $v_Q$ (or $v_P$) and solving for $v_P$ (or $v_Q$). | M1 |
$v_P^2 - 16v_P + 48 = 0 \to v_P = 4$ (or 4, 12) | A1 |
Upward velocities are 4 ms$^{-1}$ and $-1$ ms$^{-1}$ | A1 | [5]5 Two particles $P$ and $Q$ are projected vertically upwards from horizontal ground at the same instant. The speeds of projection of $P$ and $Q$ are $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively and the heights of $P$ and $Q$ above the ground, $t$ seconds after projection, are $h _ { P } \mathrm {~m}$ and $h _ { Q } \mathrm {~m}$ respectively. Each particle comes to rest on returning to the ground.\\
(i) Find the set of values of $t$ for which the particles are travelling in opposite directions.\\
(ii) At a certain instant, $P$ and $Q$ are above the ground and $3 h _ { P } = 8 h _ { Q }$. Find the velocities of $P$ and $Q$ at this instant.
\hfill \mbox{\textit{CAIE M1 2011 Q5 [8]}}