Edexcel M1 2024 June — Question 1 5 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeString becomes taut problem
DifficultyModerate -0.3 This is a standard M1 string-becomes-taut problem requiring conservation of momentum and impulse-momentum theorem. The setup is straightforward with simple mass ratio (1:3), and both parts follow directly from textbook methods with minimal algebraic manipulation. Slightly easier than average due to its routine nature and clear structure.
Spec6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation

  1. Two particles, \(A\) and \(B\), have masses \(m\) and \(3 m\) respectively. The particles are connected by a light inextensible string. Initially \(A\) and \(B\) are at rest on a smooth horizontal plane with the string slack.
Particle \(A\) is then projected along the plane away from \(B\) with speed \(U\).
Given that the common speed of the particles immediately after the string becomes taut is \(S\)
  1. find \(S\) in terms of \(U\).
  2. Find, in terms of \(m\) and \(U\), the magnitude of the impulse exerted on \(A\) immediately after the string becomes taut.

Question 1:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
CLM equation: \(mU = mS + 3mS\)M1 CLM equation with correct terms, condone sign errors and cancelled \(m\)'s or consistent extra \(g\)'s
\(S = \frac{1}{4}U\) or \(0.25U\)A1 cao (A0 if \(m\)'s not cancelled)
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
For \(A\): \(\pm m(\frac{1}{4}U - U)\)M1A1ft M1: Impulse-momentum for \(A\) or \(B\), correct terms, condone sign errors; allow \(S\) for final speed but M0 if \(m\) omitted or extra \(g\)
\(\frac{3}{4}mU\)A1 Must be positive and a multiple of \(mU\)
Alternative:
AnswerMarks
For \(B\): \(\pm 3m\frac{1}{4}U\)M1A1ft
\(\frac{3}{4}mU\)A1 (3)
# Question 1:

## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| CLM equation: $mU = mS + 3mS$ | M1 | CLM equation with correct terms, condone sign errors and cancelled $m$'s or consistent extra $g$'s |
| $S = \frac{1}{4}U$ or $0.25U$ | A1 | cao (A0 if $m$'s not cancelled) |

## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| For $A$: $\pm m(\frac{1}{4}U - U)$ | M1A1ft | M1: Impulse-momentum for $A$ or $B$, correct terms, condone sign errors; allow $S$ for final speed but M0 if $m$ omitted or extra $g$ |
| $\frac{3}{4}mU$ | A1 | Must be positive and a multiple of $mU$ |

**Alternative:**
| For $B$: $\pm 3m\frac{1}{4}U$ | M1A1ft | |
| $\frac{3}{4}mU$ | A1 (3) | |

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\begin{enumerate}
  \item Two particles, $A$ and $B$, have masses $m$ and $3 m$ respectively. The particles are connected by a light inextensible string. Initially $A$ and $B$ are at rest on a smooth horizontal plane with the string slack.
\end{enumerate}

Particle $A$ is then projected along the plane away from $B$ with speed $U$.\\
Given that the common speed of the particles immediately after the string becomes taut is $S$\\
(a) find $S$ in terms of $U$.\\
(b) Find, in terms of $m$ and $U$, the magnitude of the impulse exerted on $A$ immediately after the string becomes taut.

\hfill \mbox{\textit{Edexcel M1 2024 Q1 [5]}}