OCR Further Mechanics 2021 November — Question 4 8 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2021
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeParticle on cone surface – no string (normal reaction only)
DifficultyStandard +0.8 This is a Further Mechanics question requiring understanding of circular motion on a cone, energy considerations, and modeling assumptions. Part (a) requires setting up force equations for both particles (including resolving forces on the cone surface and applying circular motion principles), finding the relationship between their heights, and comparing mechanical energies—a multi-step problem requiring careful geometric reasoning. Parts (b) and (c) test modeling understanding. While systematic, it demands more sophistication than standard A-level mechanics and involves non-trivial geometric relationships typical of Further Maths content.
Spec1.02z Models in context: use functions in modelling6.02i Conservation of energy: mechanical energy principle6.04a Centre of mass: gravitational effect

4 A hollow cone is fixed with its axis vertical and its vertex downwards. A small sphere \(P\) of mass \(m \mathrm {~kg}\) is moving in a horizontal circle on the inner surface of the cone. An identical sphere \(Q\) rests in equilibrium inside the cone (see diagram). \includegraphics[max width=\textwidth, alt={}, center]{c6445493-9802-46ca-b7eb-7738a831d9ee-3_586_611_404_246} The following modelling assumptions are made.
  • \(P\) and \(Q\) are modelled as particles.
  • The cone is modelled as smooth.
  • There is no air resistance.
    1. Assuming that \(P\) moves with a constant speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), show that the total mechanical energy of \(P\) is \(\frac { 3 } { 2 } \mathrm { mv } ^ { 2 } \mathrm {~J}\) more than the total mechanical energy of \(Q\).
    2. Explain how the assumption that \(P\) and \(Q\) are both particles has been used.
In practice, \(P\) will not move indefinitely in a perfectly circular path, but will actually follow an approximately spiral path on the inside surface of the cone until eventually it collides with \(Q\).
  • Suggest an improvement that could be made to the model.

  • Question 4:
    AnswerMarks Guidance
    4(a) 1
    KE of P = mv2
    2
     Csinθ=mg
    ↔ Ccosθ=ma
    cosθ a v2
    = =
    sinθ g rg
    PE of P (exceeds that of Q by)
    r rcosθ v2
    mgh=mg =mg =mg =mv2
    tanθ sinθ g
    soi
    So total ME of P exceeds that of Q by
    1 3
    =mv2+ mv2 = mv2 J
    AnswerMarks
    2 2B1
    M1
    M1
    M1
    M1
    A1
    AnswerMarks
    [6]1.2
    3.3
    3.3
    3.4
    3.4
    AnswerMarks
    2.2aBalancing forces in the vertical.
    C must be resolved
    NII in the horizontal using a
    resolved component of C
    Eliminating C (and m) between
    the two equations and using a
    correct form for a
    Using the relationship to find the
    (excess) PE of P in terms of m
    and v (and possibly g) only
    AG. Or total ME of Q = 0 but
    some justification of excess for
    PE at least must be seen in the
    AnswerMarks
    solutionSSU – change C to R if a better
    reflection of candidate solutions
    In this solution, C is the normal
    contact force between P and the
    cone and θ is the semi-vertical angle
    of the cone
    May see v2 =gh here and used later
    h is the vertical height of P above Q
    Use R instead of C?
    AnswerMarks Guidance
    4(b) One of:
    - We have assumed that the radius of the circle
    which P moves in is the same as the radius of the
    cone at that level
    - Q is at V [neither of which is quite true if P and
    AnswerMarks Guidance
    Q do not have a negligible radius]B1
    [1]3.5b Also accept e.g.
    - CofM of P lies on the edge of
    the cone
    AnswerMarks Guidance
    - CofM of Q lies at VV is the vertex of the cone
    4(c) Resistance to the motion of P should be
    included in the model.B1
    [1]3.5c eg air resistance. Allow friction.
    Question 4:
    4 | (a) | 1
    KE of P = mv2
    2
     Csinθ=mg
    ↔ Ccosθ=ma
    cosθ a v2
    = =
    sinθ g rg
    PE of P (exceeds that of Q by)
    r rcosθ v2
    mgh=mg =mg =mg =mv2
    tanθ sinθ g
    soi
    So total ME of P exceeds that of Q by
    1 3
    =mv2+ mv2 = mv2 J
    2 2 | B1
    M1
    M1
    M1
    M1
    A1
    [6] | 1.2
    3.3
    3.3
    3.4
    3.4
    2.2a | Balancing forces in the vertical.
    C must be resolved
    NII in the horizontal using a
    resolved component of C
    Eliminating C (and m) between
    the two equations and using a
    correct form for a
    Using the relationship to find the
    (excess) PE of P in terms of m
    and v (and possibly g) only
    AG. Or total ME of Q = 0 but
    some justification of excess for
    PE at least must be seen in the
    solution | SSU – change C to R if a better
    reflection of candidate solutions
    In this solution, C is the normal
    contact force between P and the
    cone and θ is the semi-vertical angle
    of the cone
    May see v2 =gh here and used later
    h is the vertical height of P above Q
    Use R instead of C?
    4 | (b) | One of:
    - We have assumed that the radius of the circle
    which P moves in is the same as the radius of the
    cone at that level
    - Q is at V [neither of which is quite true if P and
    Q do not have a negligible radius] | B1
    [1] | 3.5b | Also accept e.g.
    - CofM of P lies on the edge of
    the cone
    - CofM of Q lies at V | V is the vertex of the cone
    4 | (c) | Resistance to the motion of P should be
    included in the model. | B1
    [1] | 3.5c | eg air resistance. Allow friction.
    4 A hollow cone is fixed with its axis vertical and its vertex downwards. A small sphere $P$ of mass $m \mathrm {~kg}$ is moving in a horizontal circle on the inner surface of the cone. An identical sphere $Q$ rests in equilibrium inside the cone (see diagram).\\
    \includegraphics[max width=\textwidth, alt={}, center]{c6445493-9802-46ca-b7eb-7738a831d9ee-3_586_611_404_246}
    
    The following modelling assumptions are made.
    
    \begin{itemize}
      \item $P$ and $Q$ are modelled as particles.
      \item The cone is modelled as smooth.
      \item There is no air resistance.
    \begin{enumerate}[label=(\alph*)]
    \item Assuming that $P$ moves with a constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, show that the total mechanical energy of $P$ is $\frac { 3 } { 2 } \mathrm { mv } ^ { 2 } \mathrm {~J}$ more than the total mechanical energy of $Q$.
    \item Explain how the assumption that $P$ and $Q$ are both particles has been used.
    \end{itemize}
    
    In practice, $P$ will not move indefinitely in a perfectly circular path, but will actually follow an approximately spiral path on the inside surface of the cone until eventually it collides with $Q$.
    \item Suggest an improvement that could be made to the model.
    \end{enumerate}
    
    \hfill \mbox{\textit{OCR Further Mechanics 2021 Q4 [8]}}