CAIE M1 2023 June — Question 2 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSUVAT in 2D & Gravity
TypeVertical projection: max height
DifficultyModerate -0.3 Part (a) is a standard SUVAT calculation using v²=u²+2as requiring simple substitution. Part (b) requires an additional step of using energy loss to find the rebound speed, then applying SUVAT for time of flight, but follows a well-practiced routine with clear signposting. The multi-step nature and energy consideration elevate it slightly above the most routine questions, but it remains a standard textbook exercise with no novel insight required.
Spec6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle
A particle \(P\) of mass \(0.4\) kg is projected vertically upwards from horizontal ground with speed \(10\) m s\(^{-1}\).
  1. Find the greatest height above the ground reached by \(P\). [2]
When \(P\) reaches the ground again, it bounces vertically upwards. At the first instant that it hits the ground, \(P\) loses \(7.2\) J of energy.
  1. Find the time between the first and second instants at which \(P\) hits the ground. [4]
Question 2:
AnswerMarks Guidance
2(a)0102 2(g)ss... M1
ag or 10 and solve for s (or any other complete
SUVAT method). Or using an energy method:
1
0.4gh (0.4)(10)2and solve for h (with two terms
2
using correct given values – condone lack of masses in
conservation of energy equation).
AnswerMarks
Max. height = 5 (m)A1
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
2(b)1
KE before impact  0.4102[20]
AnswerMarks Guidance
2B1FT Or loss of PE =  0.4g5using their maximum
height from (a).
1
0.4v2 207.2[v8]
2
AnswerMarks Guidance
or 0.4gh207.2[h3.2]*M1 1
M1 for 0.4v2 KE/PE before impact7.2 (must
2
be correct method of subtracting 7.2 (OE)). Or, for
finding the maximum height after first impact. Need
not solve for v (or h) for this mark.
88(g)tt ...
1
or 08t gt2 t ...
AnswerMarks Guidance
2DM1 For use of a complete method to find t. Condone sign
errors but ag. If calculating the time to the
maximum height between the first and second impacts,
then candidates must double this answer (OE). E.g.,
08(g)T, t 2T ... or
3.200.5gT2 t 2T ....
AnswerMarks
Time = 1.6 (s)A1
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
--- 2(a) ---
2(a) | 0102 2(g)ss... | M1 | For use of v2 u2 2aswith v = 0, u10and
ag or 10 and solve for s (or any other complete
SUVAT method). Or using an energy method:
1
0.4gh (0.4)(10)2and solve for h (with two terms
2
using correct given values – condone lack of masses in
conservation of energy equation).
Max. height = 5 (m) | A1
2
Question | Answer | Marks | Guidance
--- 2(b) ---
2(b) | 1
KE before impact  0.4102[20]
2 | B1FT | Or loss of PE =  0.4g5using their maximum
height from (a).
1
0.4v2 207.2[v8]
2
or 0.4gh207.2[h3.2] | *M1 | 1
M1 for 0.4v2 KE/PE before impact7.2 (must
2
be correct method of subtracting 7.2 (OE)). Or, for
finding the maximum height after first impact. Need
not solve for v (or h) for this mark.
88(g)tt ...
1
or 08t gt2 t ...
2 | DM1 | For use of a complete method to find t. Condone sign
errors but ag. If calculating the time to the
maximum height between the first and second impacts,
then candidates must double this answer (OE). E.g.,
08(g)T, t 2T ... or
3.200.5gT2 t 2T ....
Time = 1.6 (s) | A1
4
Question | Answer | Marks | Guidance
A particle $P$ of mass $0.4$ kg is projected vertically upwards from horizontal ground with speed $10$ m s$^{-1}$.

\begin{enumerate}[label=(\alph*)]
\item Find the greatest height above the ground reached by $P$. [2]
\end{enumerate}

When $P$ reaches the ground again, it bounces vertically upwards. At the first instant that it hits the ground, $P$ loses $7.2$ J of energy.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the time between the first and second instants at which $P$ hits the ground. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q2 [6]}}
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