Edexcel M3 — Question 1

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
TopicVariable Force

  1. A particle of mass \(m \mathrm {~kg}\) moves in a horizontal straight line. Its initial speed is \(u \mathrm {~ms} ^ { - 1 }\) and the only force acting on it is a variable resistance of magnitude \(m k v \mathrm {~N}\), where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of the particle after \(t\) seconds and \(k\) is a constant.
    Show that \(v = u e ^ { - k t }\).
  2. A particle \(P\) of mass \(m \mathrm {~kg}\) moves in a horizontal circle at one end of a light inextensible string of length 40 cm , as shown. The other end of the string is attached to a fixed point \(O\).
    The angular velocity of \(P\) is \(\omega \mathrm { rad } \mathrm { s } ^ { - 1 }\).
    If the angle \(\theta\) which the string makes with the vertical must not
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    (7 marks)
    exceed \(60 ^ { \circ }\), calculate the greatest possible value of \(\omega\).
  3. A particle \(P\) of mass \(m \mathrm {~kg}\) is attached to one end of a light elastic string of natural length 0.5 m and modulus of elasticity \(\frac { m g } { 2 } \mathrm {~N}\). The other end of the string is attached to a fixed point \(O\) and \(P\) hangs vertically below \(O\).
    1. Find the stretched length of the string when \(P\) rests in equilibrium.
    2. Find the elastic potential energy stored in the string in the equilibrium position. \(P\), which is still attached to the string, is now held at rest at \(O\) and then lowered gently into its equilibrium position.
    3. Find the work done by the weight of the particle as it moves from \(O\) to the equilibrium position.
    4. Explain the discrepancy between your answers to parts (b) and (c).
    5. A particle \(P\), of mass \(m \mathrm {~kg}\), is attached to two light elastic strings, each of natural length \(l \mathrm {~m}\) and modulus of elasticity 3 mg N . The other ends of the strings are attached to the fixed points \(A\) and \(B\), where \(A B\) is horizontal and \(A B = 2 l \mathrm {~m}\). If \(P\) rests in equilibrium vertically below the mid-point of \(A B\), with each string making an angle \(\theta\) with the vertical, show that
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    $$\cot \theta - \cos \theta = \frac { 1 } { 6 } .$$ \section*{MECHANICS 3 (A) TEST PAPER 2 Page 2}