| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time at height |
| Difficulty | Moderate -0.5 This is a straightforward SUVAT question requiring standard application of kinematic equations with gravity. Parts (i) and (ii) involve direct substitution into familiar formulas, while part (iii) requires solving a quadratic equation but follows a predictable pattern. The multi-part structure adds some length but no conceptual challenge beyond routine mechanics problems. |
| Spec | 3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| \([0 = 25 - 10t]\) | M1 | Use of \(v = u + at\) with \(u = 25\), \(v = 0\) and \(a = -g\), or other complete method for finding \(t\) to highest point |
| \(t = 2.5\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \([20 = 25t - \frac{1}{2}gt^2]\) | M1 | Applying \(s = ut + \frac{1}{2}at^2\) with \(s = 20\), \(u = 25\) |
| \([t = 1\) and \(t = 4]\) | M1 | Solve a 3-term quadratic for \(t\), factorising or formula |
| Required time \(= 4 - 1 = 3\) seconds | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \([v^2 = 25^2 + 2 \times (-10) \times 20 \rightarrow v = \pm 15]\) | M1 | Using \(v^2 = u^2 + 2as\) with \(u = 25\), \(s = 20\) and \(a = -g\) |
| \([-15 = 15 - 10T]\) or equivalent | M1 | Use \(v\) at \(s = 20\) to find the time \(T\) taken to reach maximum height and to return to \(s = 20\) |
| Required time \(= 1.5 + 1.5 = 3\) seconds | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Max height reached at \(2.5\) s, hence reaches \(h\) after \(2\) s: \(h - 3 = 25 \times 2 - 5 \times 2^2\) | M1 | Using their \(t\) from 2(i) \(- 0.5\) in \(s = ut + \frac{1}{2}at^2\); allow finding \(h\) without taking note of the additional 3 m |
| \(h = 33\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Maximum height \(= \frac{1}{2} \times (25 + 0) \times 2.5 = 31.25\) m; In \(0.5\) s it falls distance \(\frac{1}{2} \times 10 \times 0.5^2 = 1.25\) m | M1 | For attempting to find both the maximum height and the distance fallen in \(0.5\) seconds |
| \(h = 31.25 - 1.25 + 3 = 33\) m | A1 | |
| [2] |
**Question 2(i):**
$[0 = 25 - 10t]$ | **M1** | Use of $v = u + at$ with $u = 25$, $v = 0$ and $a = -g$, or other complete method for finding $t$ to highest point
$t = 2.5$ | **A1** |
| **[2]** |
---
**Question 2(ii):**
$[20 = 25t - \frac{1}{2}gt^2]$ | **M1** | Applying $s = ut + \frac{1}{2}at^2$ with $s = 20$, $u = 25$
$[t = 1$ and $t = 4]$ | **M1** | Solve a 3-term quadratic for $t$, factorising or formula
Required time $= 4 - 1 = 3$ seconds | **A1** |
*Alternative method:*
$[v^2 = 25^2 + 2 \times (-10) \times 20 \rightarrow v = \pm 15]$ | **M1** | Using $v^2 = u^2 + 2as$ with $u = 25$, $s = 20$ and $a = -g$
$[-15 = 15 - 10T]$ or equivalent | **M1** | Use $v$ at $s = 20$ to find the time $T$ taken to reach maximum height and to return to $s = 20$
Required time $= 1.5 + 1.5 = 3$ seconds | **A1** |
| **[3]** |
---
**Question 2(iii):**
Max height reached at $2.5$ s, hence reaches $h$ after $2$ s: $h - 3 = 25 \times 2 - 5 \times 2^2$ | **M1** | Using their $t$ from 2(i) $- 0.5$ in $s = ut + \frac{1}{2}at^2$; allow finding $h$ without taking note of the additional 3 m
$h = 33$ m | **A1** |
*Alternative method:*
Maximum height $= \frac{1}{2} \times (25 + 0) \times 2.5 = 31.25$ m; In $0.5$ s it falls distance $\frac{1}{2} \times 10 \times 0.5^2 = 1.25$ m | **M1** | For attempting to find both the maximum height and the distance fallen in $0.5$ seconds
$h = 31.25 - 1.25 + 3 = 33$ m | **A1** |
| **[2]** |2 A particle $P$ is projected vertically upwards with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point 3 m above horizontal ground.\\
(i) Find the time taken for $P$ to reach its greatest height.\\
(ii) Find the length of time for which $P$ is higher than 23 m above the ground.\\
(iii) $P$ is higher than $h \mathrm {~m}$ above the ground for 1 second. Find $h$.\\
\hfill \mbox{\textit{CAIE M1 2019 Q2 [7]}}