OCR MEI Further Mechanics A AS (Further Mechanics A AS) 2018 June

Question 1
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1 Forces of magnitude \(4 \mathrm {~N} , 3 \mathrm {~N} , 5 \mathrm {~N}\) and \(R \mathrm {~N}\) act on a particle in the directions shown in Fig. 1. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-2_697_780_443_639} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} The particle is in equilibrium. Find each of the following.
  • The value of \(R\).
  • The value of \(\theta\).
Question 2
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2 A car of mass 1350 kg travels along a straight horizontal road. Throughout this question the resistance force to the motion of the car is modelled as constant and equal to 920 N .
  1. Calculate the power, in kW , developed by the car when the car is travelling at a constant speed of \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The car is now used to tow a caravan of mass 1050 kg along the same road. When the car tows the caravan at a constant speed of \(20 \mathrm {~ms} ^ { - 1 }\) the power developed by the car is 45 kW .
  2. Find the additional resistance force due to the caravan. In the remaining parts of this question the power developed by the car is constant and equal to 68 kW and the resistance force due to the caravan is modelled as constant and equal to the value found in part (ii). When the car and caravan pass a point A on the same straight horizontal road the speed of the car and caravan is \(20 \mathrm {~ms} ^ { - 1 }\).
  3. Find the acceleration of the car and caravan at point A . The car and caravan later pass a point B on the same straight horizontal road with speed \(28 \mathrm {~ms} ^ { - 1 }\). The distance \(A B\) is \(1024 m\).
  4. Find the time taken for the car and caravan to travel from point A to point B .
  5. Suggest one way in which any of the modelling assumptions used in this question could have been improved.
Question 3
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3 Jodie is doing an experiment involving a simple pendulum. The pendulum consists of a small object tied to one end of a piece of string. The other end of the string is attached to a fixed point O and the object is allowed to swing between two fixed points A and B and back again, as shown in Fig. 3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-3_328_350_584_886} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure} Jodie thinks that \(P\), the time the pendulum takes to swing from A to B and back again, depends on the mass, \(m\), of the small object, the length, \(l\), of the piece of string, and the acceleration due to gravity \(g\). She proposes the formula \(P = k m ^ { \alpha } l ^ { \beta } g ^ { \gamma }\).
  1. What is the significance of \(k\) in Jodie's formula?
  2. Use dimensional analysis to determine the values of \(\alpha , \beta\) and \(\gamma\). Jodie finds that when the mass of the object is 1.5 kg and the length of the string is 80 cm the time taken for the pendulum to swing from A to B and back again is 1.8 seconds.
  3. Use Jodie's formula and your answers to part (ii) to find each of the following.
    (A) The value of \(k\)
    (B) The time taken for the pendulum to swing from A to B and back again when the mass of the object is 0.9 kg and the length of the string is 1.4 m
  4. Comment on the assumption made by Jodie that the formula for the time taken for the pendulum to swing from A to B and back again is dependent on \(m , l\) and \(g\).
Question 4
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4 A uniform lamina ABDE is in the shape of an equilateral triangle ABC of side 12 cm from which an equilateral triangle of side 6 cm has been removed from corner \(C\). The lamina is situated on coordinate axes as shown in Fig. 4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-4_501_753_406_646} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
  1. Explain why angle \(\mathrm { BDA } = 90 ^ { \circ }\).
  2. Find the coordinates of the centre of mass of the lamina ABDE . The lamina ABDE is now freely suspended from D and hangs in equilibrium.
  3. Calculate the angle DE makes with the downward vertical.
Question 5
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5 A small ball is held at a height of 160 cm above a horizontal surface. The ball is released from rest and rebounds from the surface. After its first bounce on the surface the ball reaches a height of 122.5 cm .
  1. Find the height reached by the ball after its second bounce on the surface. After \(n\) bounces the height reached by the ball is less than 10 cm .
  2. Find the minimum possible value of \(n\).
  3. State what would happen if the same ball is released from rest from a height of 160 cm above a different horizontal surface and
    (A) the coefficient of restitution between the ball and the new surface is 0 ,
    (B) the coefficient of restitution between the ball and the new surface is 1 .
Question 6
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6 A uniform rod AB has length \(2 a\) and weight \(W\). The rod is in equilibrium in a horizontal position. The end A rests on a smooth plane which is inclined at an angle of \(30 ^ { \circ }\) to the horizontal. The force exerted on AB by the plane is \(R\). The end B is attached to a light inextensible string inclined at an angle of \(\theta\) to AB as shown in Fig. 6. The rod and string are in the same vertical plane, which also contains the line of greatest slope of the plane on which A lies. The tension in the string is \(T\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-5_474_862_479_616} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Add the forces \(R\) and \(T\) to the copy of Fig. 6 in the Printed Answer Booklet.
  2. By taking moments about B , find an expression for \(R\) in terms of \(W\).
  3. By resolving horizontally, show that \(6 T \cos \theta = W \sqrt { 3 }\).
  4. By finding a second equation connecting \(T\) and \(\theta\), determine
    • the value of \(\theta\),
    • an expression for \(T\) in terms of \(W\).