OCR Mechanics 1 2018 December — Question 9

Exam BoardOCR
ModuleMechanics 1 (Mechanics 1)
Year2018
SessionDecember
TopicNon-constant acceleration

9 A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\), where \(v = 2 t ^ { 4 } + k t ^ { 2 } - 4\). The acceleration of \(P\) when \(t = 2\) is \(28 \mathrm {~ms} ^ { - 2 }\).
  1. Show that \(k = - 9\).
  2. Show that the velocity of \(P\) has its minimum value when \(t = 1.5\). When \(t = 1 , P\) is at the point \(( - 6.4125,0 )\).
  3. Find the distance of \(P\) from the origin \(O\) when \(P\) is moving with minimum velocity.
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    \(A\) and \(B\) are points at the upper and lower ends, respectively, of a line of greatest slope on a plane inclined at \(30 ^ { \circ }\) to the horizontal. The distance \(A B\) is \(20 \mathrm {~m} . M\) is a point on the plane between \(A\) and \(B\). The surface of the plane is smooth between \(A\) and \(M\), and rough between \(M\) and \(B\). A particle \(P\) is projected with speed \(4.2 \mathrm {~ms} ^ { - 1 }\) from \(A\) down the line of greatest slope (see diagram). \(P\) moves down the plane and reaches \(B\) with speed \(12.6 \mathrm {~ms} ^ { - 1 }\). The coefficient of friction between \(P\) and the rough part of the plane is \(\frac { \sqrt { 3 } } { 6 }\).
  4. Find the distance \(A M\).
  5. Find the angle between the contact force and the downward direction of the line of greatest slope when \(P\) is in motion between \(M\) and \(B\).