Variable force (velocity v) - use v dv/dx

Force or resistance depends on velocity v (e.g. kv², kv), motion is horizontal, and v dv/dx form is used (not dv/dt) to find velocity or position by integration.

8 questions · Challenging +1.1

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OCR M3 2007 June Q3
10 marks Standard +0.8
3 A particle \(P\) of mass 0.2 kg is projected horizontally with speed \(u \mathrm {~ms} ^ { - 1 }\) from a fixed point \(O\) on a smooth horizontal surface. \(P\) moves in a straight line and, at time \(t \mathrm {~s}\) after projection, \(P\) has speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and is \(x \mathrm {~m}\) from \(O\). The only force acting on \(P\) has magnitude \(0.4 v ^ { 2 } \mathrm {~N}\) and is directed towards \(O\).
  1. Show that \(\frac { 1 } { v } \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 2\).
  2. Hence show that \(v = u \mathrm { e } ^ { - 2 x }\).
  3. Find \(u\), given that \(x = 2\) when \(t = 4\).
OCR M3 2014 June Q4
10 marks Standard +0.3
4 A particle \(P\) of mass 0.4 kg is projected horizontally with speed \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a fixed point \(O\) on a smooth horizontal surface. At time \(t \mathrm {~s}\) after projection \(P\) is \(x \mathrm {~m}\) from \(O\) and is moving away from \(O\) with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). There is a force of magnitude \(1.6 v ^ { 2 } \mathrm {~N}\) resisting the motion of \(P\).
  1. Find an expression for \(\frac { \mathrm { d } v } { \mathrm {~d} x }\) in terms of \(v\), and hence show that \(v = 2 \mathrm { e } ^ { - 4 x }\).
  2. Find the distance travelled by \(P\) in the 0.5 seconds after it leaves \(O\).
Edexcel M4 2012 June Q3
16 marks Challenging +1.2
  1. Two particles, of masses \(m\) and \(2 m\), are connected to the ends of a long light inextensible string. The string passes over a small smooth fixed pulley and hangs vertically on either side. The particles are released from rest with the string taut. Each particle is subject to air resistance of magnitude \(k v ^ { 2 }\), where \(v\) is the speed of each particle after it has moved a distance \(x\) from rest and \(k\) is a positive constant.
    1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( v ^ { 2 } \right) + \frac { 4 k } { 3 m } v ^ { 2 } = \frac { 2 g } { 3 }\)
    2. Find \(v ^ { 2 }\) in terms of \(x\).
    3. Deduce that the tension in the string, \(T\), satisfies
    $$\frac { 4 m g } { 3 } \leqslant T < \frac { 3 m g } { 2 }$$
OCR MEI M4 2007 June Q4
24 marks Challenging +1.8
4 A particle of mass 2 kg starts from rest at a point O and moves in a horizontal line with velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) under the action of a force \(F \mathrm {~N}\), where \(F = 2 - 8 v ^ { 2 }\). The displacement of the particle from O at time \(t\) seconds is \(x \mathrm {~m}\).
  1. Formulate and solve a differential equation to show that \(v ^ { 2 } = \frac { 1 } { 4 } \left( 1 - \mathrm { e } ^ { - 8 x } \right)\).
  2. Hence express \(F\) in terms of \(x\) and find, by integration, the work done in the first 2 m of the motion.
  3. Formulate and solve a differential equation to show that \(v = \frac { 1 } { 2 } \left( \frac { 1 - \mathrm { e } ^ { - 4 t } } { 1 + \mathrm { e } ^ { - 4 t } } \right)\).
  4. Calculate \(v\) when \(t = 1\) and when \(t = 2\), giving your answers to four significant figures. Hence find the impulse of the force \(F\) over the interval \(1 \leqslant t \leqslant 2\).
Edexcel FM2 2019 June Q2
10 marks Challenging +1.2
  1. A particle, \(P\), of mass 0.4 kg is moving along the positive \(x\)-axis, in the positive \(x\) direction under the action of a single force. At time \(t\) seconds, \(t > 0 , P\) is \(x\) metres from the origin \(O\) and the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The force is acting in the direction of \(x\) increasing and has magnitude \(\frac { k } { v }\) newtons, where \(k\) is a constant.
At \(x = 3 , v = 2\) and at \(x = 6 , v = 2.5\)
  1. Show that \(v ^ { 3 } = \frac { 61 x + 9 } { 24 }\) The time taken for the speed of \(P\) to increase from \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is \(T\) seconds.
  2. Use algebraic integration to show that \(T = \frac { 81 } { 61 }\)
Edexcel FM2 2020 June Q3
10 marks Challenging +1.2
  1. A particle \(P\) of mass 0.5 kg is moving along the positive \(x\)-axis in the direction of \(x\) increasing. At time \(t\) seconds \(( t \geqslant 0 ) , P\) is \(x\) metres from the origin \(O\) and the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The resultant force acting on \(P\) is directed towards \(O\) and has magnitude \(k v ^ { 2 } \mathrm {~N}\), where \(k\) is a positive constant.
When \(x = 1 , v = 4\) and when \(x = 2 , v = 2\)
  1. Show that \(v = a b ^ { x }\), where \(a\) and \(b\) are constants to be found. The time taken for the speed of \(P\) to decrease from \(4 \mathrm {~ms} ^ { - 1 }\) to \(2 \mathrm {~ms} ^ { - 1 }\) is \(T\) seconds.
  2. Show that \(T = \frac { 1 } { 4 \ln 2 }\)
OCR Further Mechanics 2024 June Q7
14 marks Standard +0.8
  1. Show that \(B\) 's motion can be modelled by the differential equation \(\frac { 1 } { \mathrm { v } } \frac { \mathrm { dv } } { \mathrm { dx } } = - 4\).
    1. Solve the differential equation in part (a) to find the particular solution for \(v\) in terms of \(x\) and \(u\).
    2. By considering the behaviour of \(v\) as \(x \longrightarrow \infty\) describe one feature of the model that is not realistic. At the instant when \(B\) reaches the point \(A\), where \(\mathrm { x } = \mathrm { X }\), its speed is \(V \mathrm {~ms} ^ { - 1 }\). The work done by the resistance as \(B\) moves from \(O\) to \(A\) is denoted by \(W \mathrm {~J}\).
    1. Use the formula \(\mathrm { W } = \int \mathrm { F } \mathrm { dx }\) to determine an expression for \(W\) in terms of \(X\) and \(u\).
    2. Explain the relevance of the sign of your answer in part (c)(i).
    3. By writing your answer to part (c)(i) in terms of \(V\) and \(u\) show how the quantity \(W\) relates to the energy of \(B\).
OCR Further Mechanics 2023 June Q5
13 marks Challenging +1.3
A particle \(P\) of mass \(2\) kg moves along the \(x\)-axis. At time \(t = 0\), \(P\) passes through the origin \(O\) with speed \(3\) m s\(^{-1}\). At time \(t\) seconds the displacement of \(P\) from \(O\) is \(x\) m and the velocity of \(P\) is \(v\) m s\(^{-1}\), where \(t \geqslant 0\), \(x \geqslant 0\) and \(v \geqslant 0\). While \(P\) is in motion the only force acting on \(P\) is a resistive force \(F\) of magnitude \((v^2 + 1)\) N acting in the negative \(x\)-direction.
  1. Find an expression for \(v\) in terms of \(x\). [5]
  2. Determine the distance travelled by \(P\) while its speed drops from \(3\) m s\(^{-1}\) to \(2\) m s\(^{-1}\). [2]
Particle \(Q\) is identical to particle \(P\). At a different time, \(Q\) is moving along the \(x\)-axis under the influence of a single constant resistive force of magnitude \(1\) N. When \(t' = 0\), \(Q\) is at the origin and its speed is \(3\) m s\(^{-1}\).
  1. By comparing the motion of \(P\) with the motion of \(Q\) explain why \(P\) must come to rest at some finite time when \(t < 6\) with \(x < 9\). [3]
  2. Sketch the velocity-time graph for \(P\). You do not need to indicate any values on your sketch. [1]
  3. Determine the maximum displacement of \(P\) from \(O\) during \(P\)'s motion. [2]