OCR MEI Further Mechanics Major 2023 June — Question 7 9 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeConnected particles with pulley
DifficultyStandard +0.3 This is a standard Further Maths mechanics problem involving connected particles, friction, and the work-energy principle. Part (a) requires straightforward calculation of friction work using W = μRd. Part (b) applies work-energy principle with given final velocity—methodical but routine for FM students. The coefficient of friction simplifies nicely (√3/6), and the multi-step nature is typical for this level. Slightly above average difficulty due to the FM context and work-energy application, but follows standard problem-solving patterns without requiring novel insight.
Spec3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces6.01a Dimensions: M, L, T notation6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle

One end of a rope is attached to a block A of mass 2 kg. The other end of the rope is attached to a second block B of mass 4 kg. Block A is held at rest on a fixed rough ramp inclined at \(30°\) to the horizontal. The rope is taut and passes over a small smooth pulley P which is fixed at the top of the ramp. The part of the rope from A to P is parallel to a line of greatest slope of the ramp. Block B hangs vertically below P, at a distance \(d\) m above the ground, as shown in the diagram. \includegraphics{figure_7} Block A is more than \(d\) m from P. The blocks are released from rest and A moves up the ramp. The coefficient of friction between A and the ramp is \(\frac{1}{2\sqrt{3}}\). The blocks are modelled as particles, the rope is modelled as light and inextensible, and air resistance can be ignored.
  1. Determine, in terms of \(g\) and \(d\), the work done against friction as A moves \(d\) m up the ramp. [3]
  2. Given that the speed of B immediately before it hits the ground is \(1.75 \text{ m s}^{-1}\), use the work–energy principle to determine the value of \(d\). [5]
  3. Suggest one improvement, apart from including air resistance, that could be made to the model to make it more realistic. [1]

Question 7:
AnswerMarks Guidance
7(a) R = 2 g c o s 3 0
number of terms – allow sin/cos confusion (M0 if 2
used for the weight)
1
F = ( 2 g c o s 3 0 ) ( = 12 g )
AnswerMarks Guidance
2 3M1dep* 3.4
Work done against friction is 12 g dA1 1.1
[3]
AnswerMarks Guidance
7(b) PE (lost by) B: 4gd
PE (gained by) A: 2 g ( d s i n 3 0 ) ( = g d )B1 1.1
KE (gained by) A and B: 1(2)(1.75)2+1(4)(1.75)2
AnswerMarks Guidance
2 2B1 1.1
 3.0625 or  6.125 or  9.1875)
AnswerMarks Guidance
12 ( 2 ) (1 .7 5 ) 2 + 12 ( 4 ) (1 .7 5 ) 2 = ( 4 g d − g d ) − 12 g dM1 3.4
but must be the correct number of terms (so must be
considering KE for A and B and three terms in d) and
dimensionally correct
AnswerMarks Guidance
d = 0.375A1 2.2a
[5]
AnswerMarks Guidance
7(c) • Consider the dimensions of the pulley or block(s)
• Consider the weight/mass of the rope
• More accurate value of g
• Friction at the pulley
AnswerMarks Guidance
• Elastic ropeB1 3.5c
modelled blocks as particles’ (oe)
B0 for ‘include air resistance/resistance/wind’ or
‘friction/resistance should be proportional to speed or
speed squared’
[1]
Question 7:
7 | (a) | R = 2 g c o s 3 0 | M1* | 3.3 | Resolving perpendicular to the plane for A – correct
number of terms – allow sin/cos confusion (M0 if 2
used for the weight)
1
F = ( 2 g c o s 3 0 ) ( = 12 g )
2 3 | M1dep* | 3.4 | Use of F =R with correct and their R
Work done against friction is 12 g d | A1 | 1.1 | Must be in terms of g and d
[3]
7 | (b) | PE (lost by) B: 4gd | B1 | 1.1 | Allow  4 g d or  3 9 .2 d
PE (gained by) A: 2 g ( d s i n 3 0 ) ( = g d ) | B1 | 1.1 | Allow  g d or 9.8d
KE (gained by) A and B: 1(2)(1.75)2+1(4)(1.75)2
2 2 | B1 | 1.1 | KE (gained) for either A or B (so B1 for either
 3.0625 or  6.125 or  9.1875)
12 ( 2 ) (1 .7 5 ) 2 + 12 ( 4 ) (1 .7 5 ) 2 = ( 4 g d − g d ) − 12 g d | M1 | 3.4 | Work energy principle – condone sign errors and slips
but must be the correct number of terms (so must be
considering KE for A and B and three terms in d) and
dimensionally correct
d = 0.375 | A1 | 2.2a
[5]
7 | (c) | • Consider the dimensions of the pulley or block(s)
• Consider the weight/mass of the rope
• More accurate value of g
• Friction at the pulley
• Elastic rope | B1 | 3.5c | Must be suggesting an improvement so B0 for ‘do not
modelled blocks as particles’ (oe)
B0 for ‘include air resistance/resistance/wind’ or
‘friction/resistance should be proportional to speed or
speed squared’
[1]
One end of a rope is attached to a block A of mass 2 kg. The other end of the rope is attached to a second block B of mass 4 kg. Block A is held at rest on a fixed rough ramp inclined at $30°$ to the horizontal. The rope is taut and passes over a small smooth pulley P which is fixed at the top of the ramp. The part of the rope from A to P is parallel to a line of greatest slope of the ramp. Block B hangs vertically below P, at a distance $d$ m above the ground, as shown in the diagram.

\includegraphics{figure_7}

Block A is more than $d$ m from P. The blocks are released from rest and A moves up the ramp.

The coefficient of friction between A and the ramp is $\frac{1}{2\sqrt{3}}$.

The blocks are modelled as particles, the rope is modelled as light and inextensible, and air resistance can be ignored.

\begin{enumerate}[label=(\alph*)]
\item Determine, in terms of $g$ and $d$, the work done against friction as A moves $d$ m up the ramp. [3]
\item Given that the speed of B immediately before it hits the ground is $1.75 \text{ m s}^{-1}$, use the work–energy principle to determine the value of $d$. [5]
\item Suggest one improvement, apart from including air resistance, that could be made to the model to make it more realistic. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2023 Q7 [9]}}