CAIE M1 2024 November — Question 5 10 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeTwo projectiles meeting - vertical motion only
DifficultyStandard +0.8 This is a multi-part projectiles question requiring setting up simultaneous equations for collision (part a), applying conservation of momentum for coalescence, and tracking motion in two phases (part c). While the individual techniques are standard A-level mechanics, the combination of collision timing, momentum conservation, and two-phase motion analysis with careful time-tracking makes this moderately challenging, requiring more problem-solving than routine projectile exercises.
Spec3.02h Motion under gravity: vector form6.03b Conservation of momentum: 1D two particles
A particle, \(A\), is projected vertically upwards from a point \(O\) with a speed of \(80 \text{ ms}^{-1}\). One second later a second particle, \(B\), with the same mass as \(A\), is projected vertically upwards from \(O\) with a speed of \(100 \text{ ms}^{-1}\). At time \(T\) s after the first particle is projected, the two particles collide and coalesce to form a particle \(C\).
  1. Show that \(T = 3.5\). [4]
  2. Find the height above \(O\) at which the particles collide. [1]
  3. Find the time from \(A\) being projected until \(C\) returns to \(O\). [5]
Question 5:
AnswerMarks
5(a)s =80T−1gT2
A 2
s =100(T−1)−1g(T−1)2
AnswerMarks Guidance
B 2*M1 For use of s=ut+1at2 at least once with a=gand u = 80 or
2
100 – allow t, T, t1, T1.
Two correct expressions for the displacement of both particles at
AnswerMarks Guidance
time TA1 Allow t for T.
100(T −1)−5(T −1)2 =80T −5T2DM1 Equate and attempt to solve for T or t – must not be using the
same time for both expressions (so must be using the equivalent
of T in one and T1 in the other).
AnswerMarks Guidance
Leading to T = 3.5A1 AG – no errors seen (but allow all working in terms of t).
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
5(b)s=803.5−1g3.52 =218.75m
 2 B1 OR 1002.5−1102.52.
2
1
AnswerMarks
5(c)v =80−g3.5 [=45]
A
v =100−g2.5 [=75]
AnswerMarks Guidance
B*M1 For use of v=u+atat least once to find the speed at collision
with a=g, u = 80 or 100 – with t = 2.5 or 3.5 only (but
condone 2.5 with v and 3.5 with v ).
A B
AnswerMarks Guidance
45m+75m=2mvDM1 Use of conservation of momentum, 3 non-zero terms, allow sign
errors. If total momentum before collision not correct then it
must be clear where both terms came from.
AnswerMarks
v=60A1
−218.75=60t−1gt2
AnswerMarks Guidance
2DM1 Complete method to find an equation in t using their v, their
height from part (b) and g - dependent on both previous M
marks.
AnswerMarks
t=14.9+3.5 =18.4sA1
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | s =80T−1gT2
A 2
s =100(T−1)−1g(T−1)2
B 2 | *M1 | For use of s=ut+1at2 at least once with a=gand u = 80 or
2
100 – allow t, T, t1, T1.
Two correct expressions for the displacement of both particles at
time T | A1 | Allow t for T.
100(T −1)−5(T −1)2 =80T −5T2 | DM1 | Equate and attempt to solve for T or t – must not be using the
same time for both expressions (so must be using the equivalent
of T in one and T1 in the other).
Leading to T = 3.5 | A1 | AG – no errors seen (but allow all working in terms of t).
4
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | s=803.5−1g3.52 =218.75m
 2  | B1 | OR 1002.5−1102.52.
2
1
--- 5(c) ---
5(c) | v =80−g3.5 [=45]
A
v =100−g2.5 [=75]
B | *M1 | For use of v=u+atat least once to find the speed at collision
with a=g, u = 80 or 100 – with t = 2.5 or 3.5 only (but
condone 2.5 with v and 3.5 with v ).
A B
45m+75m=2mv | DM1 | Use of conservation of momentum, 3 non-zero terms, allow sign
errors. If total momentum before collision not correct then it
must be clear where both terms came from.
v=60 | A1
−218.75=60t−1gt2
2 | DM1 | Complete method to find an equation in t using their v, their
height from part (b) and g - dependent on both previous M
marks.
t=14.9+3.5 =18.4s | A1
5
Question | Answer | Marks | Guidance
A particle, $A$, is projected vertically upwards from a point $O$ with a speed of $80 \text{ ms}^{-1}$. One second later a second particle, $B$, with the same mass as $A$, is projected vertically upwards from $O$ with a speed of $100 \text{ ms}^{-1}$. At time $T$ s after the first particle is projected, the two particles collide and coalesce to form a particle $C$.

\begin{enumerate}[label=(\alph*)]
\item Show that $T = 3.5$. [4]

\item Find the height above $O$ at which the particles collide. [1]

\item Find the time from $A$ being projected until $C$ returns to $O$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q5 [10]}}
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