| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| 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 two-particle kinematics problem requiring SUVAT equations with a time delay. Part (i) involves solving a quadratic inequality for when height exceeds 15m. Part (ii) requires setting up equations for equal heights with the time offset and finding velocities. While it involves multiple steps and careful bookkeeping of the time delay, the techniques are routine for M1 students and no novel insight is required—slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([15 = 20t - 5t^2 \Rightarrow 5(t^2 - 4t + 3) = 0]\) | M1 | For use of \(h = ut - \frac{1}{2}gt^2\) |
| \(t = 1, 3\) | A1 | |
| Duration is 2 s (accept \(1 < t < 3\)) | B1ft | [3] \(t_2 - t_1\) |
| (ii) | M1 | For using \(h_P = h_Q\) at time t after P's (or Q's) projection |
| \(20t - 5t^2 = 25(t - 0.4) - 5(t - 0.4)^2\) (or \(20(t + 0.4) - 5(t + 0.4)^2 = 25t - 5t^2\) or \((20 \times 0.4 - 5 \times 0.4^2) + 16t - 5t^2 = 25t - 5t^2\)) | A1 | |
| \(t = 1.2\) (or \(t = 0.8\)) | A1 | |
| \([v_P = 20 - 10x1.2; v_Q = 25 - 10x(1.2 - 0.4)\) (or \(v_P = 20 - 10x(0.8 + 0.4); v_Q = 25 - 10x0.8)]\) | M1 | For using \(v = u - gt\) for both \(v_P\) and \(v_Q\) |
| Velocities are 8 ms⁻¹ and 17 ms⁻¹ | A1 | [5] |
**(i)** $[15 = 20t - 5t^2 \Rightarrow 5(t^2 - 4t + 3) = 0]$ | M1 | For use of $h = ut - \frac{1}{2}gt^2$
$t = 1, 3$ | A1 |
Duration is 2 s (accept $1 < t < 3$) | B1ft | [3] $t_2 - t_1$
**(ii)** | M1 | For using $h_P = h_Q$ at time t after P's (or Q's) projection
$20t - 5t^2 = 25(t - 0.4) - 5(t - 0.4)^2$ (or $20(t + 0.4) - 5(t + 0.4)^2 = 25t - 5t^2$ or $(20 \times 0.4 - 5 \times 0.4^2) + 16t - 5t^2 = 25t - 5t^2$) | A1 |
$t = 1.2$ (or $t = 0.8$) | A1 |
$[v_P = 20 - 10x1.2; v_Q = 25 - 10x(1.2 - 0.4)$ (or $v_P = 20 - 10x(0.8 + 0.4); v_Q = 25 - 10x0.8)]$ | M1 | For using $v = u - gt$ for both $v_P$ and $v_Q$
Velocities are 8 ms⁻¹ and 17 ms⁻¹ | A1 | [5]
---5 Particles $P$ and $Q$ are projected vertically upwards, from different points on horizontal ground, with velocities of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. $Q$ is projected 0.4 s later than $P$. Find\\
(i) the time for which $P$ 's height above the ground is greater than 15 m ,\\
(ii) the velocities of $P$ and $Q$ at the instant when the particles are at the same height.
\hfill \mbox{\textit{CAIE M1 2010 Q5 [8]}}