| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: different start times, same height |
| Difficulty | Standard +0.3 This is a standard two-particle SUVAT problem with a time delay. Students must set up equations for both particles (with the first particle at time t+2), equate heights, solve a linear equation for t, then substitute back. It requires careful bookkeeping of the time offset but involves only routine application of s = ut + ½at² with no conceptual surprises or complex algebra. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s_2 = 20t - 0.5gt^2\) | B1 | Second particle |
| M1 | Uses \(s = ut + \frac{1}{2}at^2\) for first particle | |
| \(s_1 = 12(t+2) - 0.5g(t+2)^2\) | \*A1 | |
| \(12(t+2) - 0.5g(t+2)^2 = 20t - 0.5gt^2 \rightarrow t = \ldots\) | DM1 | Solves \(s_1 = s_2\) |
| \(t = \frac{1}{7} = 0.143\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[s = 20 \times \frac{1}{7} - 5 \times \left(\frac{1}{7}\right)^2 = 2.755\ldots\right]\) Height is \(2.76\) m | B1 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_2 = 20t - 0.5gt^2$ | B1 | Second particle |
| | M1 | Uses $s = ut + \frac{1}{2}at^2$ for first particle |
| $s_1 = 12(t+2) - 0.5g(t+2)^2$ | \*A1 | |
| $12(t+2) - 0.5g(t+2)^2 = 20t - 0.5gt^2 \rightarrow t = \ldots$ | DM1 | Solves $s_1 = s_2$ |
| $t = \frac{1}{7} = 0.143$ | A1 | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[s = 20 \times \frac{1}{7} - 5 \times \left(\frac{1}{7}\right)^2 = 2.755\ldots\right]$ Height is $2.76$ m | B1 | |
---5 A particle is projected vertically upwards from a point $O$ with a speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Two seconds later a second particle is projected vertically upwards from $O$ with a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t \mathrm {~s}$ after the second particle is projected, the two particles collide.\\
(i) Find $t$.\\
\includegraphics[max width=\textwidth, alt={}, center]{3d7f53af-dbf2-499b-9966-ae85514cef02-06_65_1569_488_328}\\
(ii) Hence find the height above $O$ at which the particles collide.\\
\hfill \mbox{\textit{CAIE M1 2017 Q5 [6]}}