CAIE M1 2023 June — Question 1 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeMultiple sequential collisions
DifficultyModerate -0.8 This is a straightforward two-stage momentum conservation problem requiring only direct application of the conservation formula with no conceptual complications. Part (a) involves one equation with one unknown expression, and part (b) uses that result in a second conservation equation to find m. The calculations are simple arithmetic with no algebraic manipulation challenges, making this easier than a typical A-level question.
Spec6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form
Two particles \(P\) and \(Q\), of masses \(m\) kg and \(0.3\) kg respectively, are at rest on a smooth horizontal plane. \(P\) is projected at a speed of \(5\) m s\(^{-1}\) directly towards \(Q\). After \(P\) and \(Q\) collide, \(P\) moves with a speed of \(2\) m s\(^{-1}\) in the same direction as it was originally moving.
  1. Find, in terms of \(m\), the speed of \(Q\) after the collision. [2]
After this collision, \(Q\) moves directly towards a third particle \(R\), of mass \(0.6\) kg, which is at rest on the plane. \(Q\) is brought to rest in the collision with \(R\), and \(R\) begins to move with a speed of \(1.5\) m s\(^{-1}\).
  1. Find the value of \(m\). [2]
Question 1:
AnswerMarks Guidance
1(a)m50m20.3v M1
terms (with m appearing in two terms); allow sign
errors.
AnswerMarks Guidance
Speed 10m (ms–1)A1 M1A0 if using g in momentum terms.
v10mis A0.
2
AnswerMarks Guidance
1(b)0.310m000.61.5 [3m0.9] M1
R (so must be using correct masses of 0.3 and 0.6) to
form a linear equation in m using their answer from
(a); 2 non-zero terms; allow sign errors.
AnswerMarks Guidance
m = 0.3A1FT 3
FT
their +ve coefficient of m from (a)
Condone including kg in answer.
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 1:
--- 1(a) ---
1(a) | m50m20.3v | M1 | Attempt at conservation of momentum; 3 non-zero
terms (with m appearing in two terms); allow sign
errors.
Speed 10m (ms–1) | A1 | M1A0 if using g in momentum terms.
v10mis A0.
2
--- 1(b) ---
1(b) | 0.310m000.61.5 [3m0.9] | M1 | Attempt at conservation of momentum between Q and
R (so must be using correct masses of 0.3 and 0.6) to
form a linear equation in m using their answer from
(a); 2 non-zero terms; allow sign errors.
m = 0.3 | A1FT | 3
FT
their +ve coefficient of m from (a)
Condone including kg in answer.
2
Question | Answer | Marks | Guidance
Two particles $P$ and $Q$, of masses $m$ kg and $0.3$ kg respectively, are at rest on a smooth horizontal plane. $P$ is projected at a speed of $5$ m s$^{-1}$ directly towards $Q$. After $P$ and $Q$ collide, $P$ moves with a speed of $2$ m s$^{-1}$ in the same direction as it was originally moving.

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$, the speed of $Q$ after the collision. [2]
\end{enumerate}

After this collision, $Q$ moves directly towards a third particle $R$, of mass $0.6$ kg, which is at rest on the plane. $Q$ is brought to rest in the collision with $R$, and $R$ begins to move with a speed of $1.5$ m s$^{-1}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $m$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q1 [4]}}
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