| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Easy -1.2 This is a straightforward vertical projectile motion question requiring only standard SUVAT equations. Part (i) is a routine 'show that' using v²=u²+2as with v=0 at maximum height. Part (ii) applies s=ut+½at² to find time and v=u+at for speed—both are direct substitutions with no problem-solving insight required. The calculations are simple and the question structure is entirely standard for AS-level mechanics. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| 2(i) | [0 = 302 +2(–g)s] | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| s = maximum height = 900/20 = 45 m | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 2(ii) | [33.75 = 30t – ½ gt2] | M1 |
| [5t2 – 30t + 33.75 = 0 or 4t2 – 24t + 27 = 0] | M1 | Solve a 3-term quadratic for t |
| t = 1.5 (reject t = 4.5) | A1 | |
| v = 30 – 1.5g = 15 | B1ft | Use v = u + at with u = 30 and |
| Answer | Marks | Guidance |
|---|---|---|
| v2 = 302 – 2g(33.75) = 225 → v = 15 | B1 | Use v2 = u2 + 2as with u = 30, |
| Answer | Marks | Guidance |
|---|---|---|
| or [15 = 30 – 10t] | M1 | Use s = ½ (u + v) × t with s = 33.75, u = 30 and v as found. |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Solve for t | |
| t = 1.5 | A1ft | ft on v value found |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 2:
--- 2(i) ---
2(i) | [0 = 302 +2(–g)s] | M1 | Using v2 = u2 + 2as with v = 0,
u = 30 and a = –g
For any complete method for finding maximum height s
s = maximum height = 900/20 = 45 m | A1 | AG
2
Question | Answer | Marks | Guidance
--- 2(ii) ---
2(ii) | [33.75 = 30t – ½ gt2] | M1 | Applying s = ut + ½at2 with s = 33.75, u = 30 and a = –g
[5t2 – 30t + 33.75 = 0 or 4t2 – 24t + 27 = 0] | M1 | Solve a 3-term quadratic for t
t = 1.5 (reject t = 4.5) | A1
v = 30 – 1.5g = 15 | B1ft | Use v = u + at with u = 30 and
t = 1.5
ft on t value found
Alternative method for question 2(ii)
v2 = 302 – 2g(33.75) = 225 → v = 15 | B1 | Use v2 = u2 + 2as with u = 30,
a = –g and s = 33.75 to find v
[33.75 = ½ (30 + 15) × t]
or [15 = 30 – 10t] | M1 | Use s = ½ (u + v) × t with s = 33.75, u = 30 and v as found.
or Use v = u – gt with u = 30 and v as found
M1 | Solve for t
t = 1.5 | A1ft | ft on v value found
4
Question | Answer | Marks | GuidanceA particle is projected vertically upwards with speed $30$ m s$^{-1}$ from a point on horizontal ground.
\begin{enumerate}[label=(\roman*)]
\item Show that the maximum height above the ground reached by the particle is $45$ m. [2]
\item Find the time that it takes for the particle to reach a height of $33.75$ m above the ground for the first time. Find also the speed of the particle at this time. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q2 [6]}}