OCR MEI Further Mechanics Major 2024 June — Question 1 4 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2024
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeString becomes taut problem
DifficultyModerate -0.3 This is a straightforward application of conservation of momentum for an inelastic collision, followed by a basic impulse calculation using the change in momentum. Both parts require only direct substitution into standard formulas with no problem-solving insight needed, making it slightly easier than average but not trivial since it involves two bodies and requires careful identification of the system.
Spec6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation

A car A of mass 1200 kg is about to tow another car B of mass 800 kg in a straight line along a horizontal road by means of a tow-rope attached between A and B. The tow-rope is modelled as being light and inextensible. Just before the tow-rope tightens, A is travelling at a speed of \(1.5 \text{ m s}^{-1}\) and B is at rest. Just after the tow-rope tightens, both cars have a speed of \(v \text{ m s}^{-1}\).
  1. Find the value of \(v\). [2]
  2. Calculate the magnitude of the impulse on A when the tow-rope tightens. [2]

Question 1:

AnswerMarks
1 (a)1200(1.5)=1200v+800v
v=0.9M1
A1
AnswerMarks
[2]3.3
1.1CLM soi – correct number of terms but allow sign
errors, and allow a slip in one value only

AnswerMarks
1 (b)I =800(0.9) or −I =1200(0.9)−1200(1.5) or −I =1200(0.9−1.5)
I =720 (N s)M1
A1
AnswerMarks
[2]3.3
1.1Use of Impulse = change in momentum on either A or
B (so must be using the correct mass(es)), correct
number of terms using their v from part (a) but allow
sign errors
Final answer must be positive – but condone e.g.
‘I =1200(0.9)−1200(1.5)=−720 therefore 720’ for
full marks
Question 1:
--- 1 (a) ---
1 (a) | 1200(1.5)=1200v+800v
v=0.9 | M1
A1
[2] | 3.3
1.1 | CLM soi – correct number of terms but allow sign
errors, and allow a slip in one value only
--- 1 (b) ---
1 (b) | I =800(0.9) or −I =1200(0.9)−1200(1.5) or −I =1200(0.9−1.5)
I =720 (N s) | M1
A1
[2] | 3.3
1.1 | Use of Impulse = change in momentum on either A or
B (so must be using the correct mass(es)), correct
number of terms using their v from part (a) but allow
sign errors
Final answer must be positive – but condone e.g.
‘I =1200(0.9)−1200(1.5)=−720 therefore 720’ for
full marks
A car A of mass 1200 kg is about to tow another car B of mass 800 kg in a straight line along a horizontal road by means of a tow-rope attached between A and B. The tow-rope is modelled as being light and inextensible. Just before the tow-rope tightens, A is travelling at a speed of $1.5 \text{ m s}^{-1}$ and B is at rest. Just after the tow-rope tightens, both cars have a speed of $v \text{ m s}^{-1}$.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $v$. [2]
\item Calculate the magnitude of the impulse on A when the tow-rope tightens. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2024 Q1 [4]}}