CAIE M1 2023 November — Question 5 8 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeEnergy methods in projectiles
DifficultyStandard +0.3 This is a standard multi-part projectiles and collision question requiring routine application of SUVAT equations, conservation of momentum, and basic kinematics. Part (a) uses v² = u² + 2as, part (b) applies momentum conservation with given answer, and part (c) requires finding two separate times of flight. While it has multiple steps (8 marks total), each step follows textbook procedures with no novel insight required, making it slightly easier than average.
Spec3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles
A particle \(A\) of mass 0.5 kg is projected vertically upwards from horizontal ground with speed 25 m s\(^{-1}\).
  1. Find the speed of \(A\) when it reaches a height of 20 m above the ground. [2]
When \(A\) reaches a height of 20 m, it collides with a particle \(B\) of mass 0.3 kg which is moving downwards in the same vertical line as \(A\) with speed 32.5 m s\(^{-1}\). In the collision between the two particles, \(B\) is brought to instantaneous rest.
  1. Show that the velocity of \(A\) immediately after the collision is 4.5 m s\(^{-1}\) downwards. [2]
  2. Find the time interval between \(A\) and \(B\) reaching the ground. You should assume that \(A\) does not bounce when it reaches the ground. [4]
Question 5:
AnswerMarks
5(a)v2 = 252 + 2(−g)20
1 1
OR 0.5v2 = 0.5252 − 0.5g20
AnswerMarks Guidance
2 2M1 Use of v2 = u2 + 2as with u=25,
s=20 and a=g.
OR using change in KE =  change
in PE.
AnswerMarks
Speed=15 m s-1A1
2
AnswerMarks
5(b)0.515+0.3(−32.5)=0.5v+0
Taking up as positive direction: or
0.5(−15)+0.332.5=0.5v+0
AnswerMarks Guidance
Taking down as positive direction:M1 For use of conservation of
momentum, 3 non-zero terms,
allow sign errors, using their speed
15 m s-1. Must show how 2.25 is
obtained.
[Taking up as positive direction: velocity of A = −4.5 m s−1]
[Taking down as positive direction: velocity of A = 4.5 m s−1]
AnswerMarks Guidance
Speed = 4.5 m s–1 direction downwardsA1 Any error seen in calculating v is
A0.
Must explicitly say 4.5 m s–1 and
downwards.
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
5(c)1
Downwards to be positive, for A 20=4.5t + gt 2 and solve for t
A A A
2
1
Upwards to be positive, for A −20=−4.5t − gt 2 and solve for t
A A A
AnswerMarks Guidance
2M1 Using constant acceleration
formula(e) to get a correct equation
in t and solve for t .
A A
If using quadratic formula, must be
the correct formula. If factorising,
when brackets expanded, 2 terms
correct.
1
For B 20=0+ gt 2 t=2  and solve for t
B B
AnswerMarks Guidance
2M1 Using constant acceleration
formula(e) to get a correct equation
in t and solve for t .
B B
t =1.6 or t =2
AnswerMarks
A BA1
Difference = 0.4 s onlyA1
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | v2 = 252 + 2(−g)20
1 1
OR 0.5v2 = 0.5252 − 0.5g20
2 2 | M1 | Use of v2 = u2 + 2as with u=25,
s=20 and a=g.
OR using change in KE =  change
in PE.
Speed=15 m s-1 | A1
2
--- 5(b) ---
5(b) | 0.515+0.3(−32.5)=0.5v+0
Taking up as positive direction: or
0.5(−15)+0.332.5=0.5v+0
Taking down as positive direction: | M1 | For use of conservation of
momentum, 3 non-zero terms,
allow sign errors, using their speed
15 m s-1. Must show how 2.25 is
obtained.
[Taking up as positive direction: velocity of A = −4.5 m s−1]
[Taking down as positive direction: velocity of A = 4.5 m s−1]
Speed = 4.5 m s–1 direction downwards | A1 | Any error seen in calculating v is
A0.
Must explicitly say 4.5 m s–1 and
downwards.
2
Question | Answer | Marks | Guidance
--- 5(c) ---
5(c) | 1
Downwards to be positive, for A 20=4.5t + gt 2 and solve for t
A A A
2
1
Upwards to be positive, for A −20=−4.5t − gt 2 and solve for t
A A A
2 | M1 | Using constant acceleration
formula(e) to get a correct equation
in t and solve for t .
A A
If using quadratic formula, must be
the correct formula. If factorising,
when brackets expanded, 2 terms
correct.
1
For B 20=0+ gt 2 t=2  and solve for t
B B
2 | M1 | Using constant acceleration
formula(e) to get a correct equation
in t and solve for t .
B B
t =1.6 or t =2
A B | A1
Difference = 0.4 s only | A1
4
Question | Answer | Marks | Guidance
A particle $A$ of mass 0.5 kg is projected vertically upwards from horizontal ground with speed 25 m s$^{-1}$.

\begin{enumerate}[label=(\alph*)]
\item Find the speed of $A$ when it reaches a height of 20 m above the ground. [2]
\end{enumerate}

When $A$ reaches a height of 20 m, it collides with a particle $B$ of mass 0.3 kg which is moving downwards in the same vertical line as $A$ with speed 32.5 m s$^{-1}$. In the collision between the two particles, $B$ is brought to instantaneous rest.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the velocity of $A$ immediately after the collision is 4.5 m s$^{-1}$ downwards. [2]

\item Find the time interval between $A$ and $B$ reaching the ground. You should assume that $A$ does not bounce when it reaches the ground. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q5 [8]}}
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