CAIE M1 2024 March — Question 2 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionMarch
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeSpeed at specific time or position
DifficultyModerate -0.8 This is a straightforward vertical projectile motion question requiring only standard kinematic equations. Part (a) uses v = u - gt with given values, and part (b) requires finding two times when speed equals 10 m/s then calculating distance. Both parts are routine applications of SUVAT equations with no problem-solving insight needed, making it easier than average but not trivial due to the two-part structure and need for careful sign consideration.
Spec3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form
A particle is projected vertically upwards from horizontal ground. The speed of the particle 2 seconds after it is projected is \(5\) m s\(^{-1}\) and it is travelling downwards.
  1. Find the speed of projection of the particle. [2]
  2. Find the distance travelled by the particle between the two times at which its speed is \(10\) m s\(^{-1}\). [2]
Question 2:
AnswerMarks
2(a)−5=u−g2
or 5=u+g2
or 5=0+gt t=0.5,so time to greatest height is T =2−0.5=1.5.
AnswerMarks Guidance
Hence 0=u+(−g)1.5M1 For use of constant acceleration to get an equation in
u only.
Using v=−5, t=2 and a=g.
AnswerMarks Guidance
Speed = 15 ms–1A1 Must be positive.
2
AnswerMarks
2(b)02 =102−2gs s=5 
or 102 =02 +2gs s=5 
 
1
or 10=0+gt t=1, so s=101− g12 s=5 
2
1
or (m)102 =(m)gh h=5 
AnswerMarks Guidance
2M1 For use of constant acceleration formula(e) to find
distance travelled from height with speed 10 ms–1 to
maximum height with speed 0 ms–1, a=g.
Must be a complete method, e.g.
02 =(their1 5)2−2gs s =11.25 
1 1
and 102 =(their1 5)2 −2gs s =6.25 
2 2
or
1 0=(their1 5)−gt t =0.5  ,
1
s = (their1 5+10)(their 0.5)
2
2
and with an attempt at s −s =5  .
1 2
Energy method with 2 terms, dimensionally correct.
AnswerMarks Guidance
Total distance = 10mA1FT FT their 15 ms–1 if used.
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
--- 2(a) ---
2(a) | −5=u−g2
or 5=u+g2
or 5=0+gt t=0.5,so time to greatest height is T =2−0.5=1.5.
Hence 0=u+(−g)1.5 | M1 | For use of constant acceleration to get an equation in
u only.
Using v=−5, t=2 and a=g.
Speed = 15 ms–1 | A1 | Must be positive.
2
--- 2(b) ---
2(b) | 02 =102−2gs s=5 
or 102 =02 +2gs s=5 
 
1
or 10=0+gt t=1, so s=101− g12 s=5 
2
1
or (m)102 =(m)gh h=5 
2 | M1 | For use of constant acceleration formula(e) to find
distance travelled from height with speed 10 ms–1 to
maximum height with speed 0 ms–1, a=g.
Must be a complete method, e.g.
02 =(their1 5)2−2gs s =11.25 
1 1
and 102 =(their1 5)2 −2gs s =6.25 
2 2
or
1 0=(their1 5)−gt t =0.5  ,
1
s = (their1 5+10)(their 0.5)
2
2
and with an attempt at s −s =5  .
1 2
Energy method with 2 terms, dimensionally correct.
Total distance = 10m | A1FT | FT their 15 ms–1 if used.
2
Question | Answer | Marks | Guidance
A particle is projected vertically upwards from horizontal ground. The speed of the particle 2 seconds after it is projected is $5$ m s$^{-1}$ and it is travelling downwards.

\begin{enumerate}[label=(\alph*)]
\item Find the speed of projection of the particle. [2]
\item Find the distance travelled by the particle between the two times at which its speed is $10$ m s$^{-1}$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q2 [4]}}
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