Force depends on velocity v

7 An engineer is modelling the motion of a particle \(P\) of mass 0.5 kg in a wind tunnel. \(P\) is modelled as travelling in a straight line. The point \(O\) is a fixed point within the wind tunnel. The displacement of \(P\) from \(O\) at time \(t\) seconds is \(x\) metres, for \(t \geqslant 0\). You are given that \(x \geqslant 0\) for all \(t \geqslant 0\) and that \(P\) does not reach the end of the wind tunnel.
If \(t \geqslant 0\), then \(P\) is subject to three forces which are modelled in the following way.

• The first force has a magnitude of \(5 ( t + 1 ) \cosh t \mathrm {~N}\) and acts in the positive \(x\)-direction.
• The second force has a magnitude of \(0.5 x \mathrm {~N}\) and acts towards \(O\).
• The third force has a magnitude of \(\left| \frac { d x } { d t } \right| \mathrm { N }\) and acts in the direction of motion of the particle.
  1. 1. Show that the Maclaurin series for \(\mathrm { f } ( t )\) up to and including the term in \(t\) is \(6 - 5 t\).
     2. Use your answer to part (a)(ii) to show that the term in \(t ^ { 2 }\) in the Maclaurin series for \(\mathrm { f } ( t )\) is \(- 3 t ^ { 2 }\).
     3. By differentiating the differential equation in part (a) with respect to \(t\), show that the term in \(t ^ { 3 }\) in the Maclaurin series for \(\mathrm { f } ( t )\) is \(0.5 t ^ { 3 }\). You are given that the complete Maclaurin series for the function f is valid for all values of \(t \geqslant 0\).
        After 0.25 seconds \(P\) has travelled 1.43 m towards the origin.
  2. 1. By using the Maclaurin series for \(\mathrm { f } ( t )\) up to and including the term in \(t ^ { 3 }\), evaluate the suitability of the model for determining the displacement of \(P\) from \(O\) when \(t = 0.25\).
     2. Explain why it might not be sensible to use the Maclaurin series for \(\mathrm { f } ( t )\) up to and including the term in \(t ^ { 3 }\) to evaluate the suitability of the model for determining the displacement of \(P\) from \(O\) when \(t = 10\).

6 A particle \(P\) of mass 2.5 kg is free to move along the \(x\)-axis. When its displacement from the origin is \(x \mathrm {~m}\) its velocity is \(v \mathrm {~ms} ^ { - 1 }\). At time \(t = 0\) seconds, \(P\) is at the point where \(x = 1\) and is travelling in the negative \(x\)-direction with speed \(5 \mathrm {~ms} ^ { - 1 }\). At this time an impulse of \(I\) Ns is applied to \(P\) in the positive \(x\)-direction so that \(P\) moves in the positive \(x\)-direction with speed \(18 \mathrm {~ms} ^ { - 1 }\).

1. Find the value of \(I\). Subsequently, whenever \(P\) is in motion, two forces act on it. The first force acts in the positive \(x\)-direction and has magnitude \(\frac { 5 v ^ { 2 } } { x } N\). The second force acts in the negative \(x\)-direction and has magnitude 60 vN .
2. Show that the motion of \(P\) can be modelled by the differential equation \(\frac { \mathrm { dV } } { \mathrm { dx } } = \frac { \mathrm { aV } } { \mathrm { x } } + \mathrm { b }\) where \(a\) and \(b\) are constants whose values should be determined.
3. By solving the differential equation derived in part (b) find an expression for \(v\) in terms of \(x\). You are given that \(\mathrm { x } = \frac { 4 } { 3 \mathrm { e } ^ { - 24 \mathrm { t } } + 1 }\) when \(t \geqslant 0\).
4. Describe in detail the motion of \(P\) when \(t \geqslant 0\).

6 Alice places a toy, of mass 0.4 kg , on a slope. The toy is set in motion with an initial velocity of \(1 \mathrm {~ms} ^ { - 1 }\) down the slope. The resultant force acting on the toy is \(( 2 - 4 v )\) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the toy's velocity at time \(t\) seconds after it is set in motion.

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = - 10 ( v - 0.5 )\).
2. By using \(\int \frac { 1 } { v - 0.5 } \mathrm {~d} v = - \int 10 \mathrm {~d} t\), find \(v\) in terms of \(t\).
3. Find the time taken for the toy's velocity to reduce to \(0.55 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). \(7 \quad\) A small bead, of mass \(m\), is suspended from a fixed point \(O\) by a light inextensible string of length \(a\). With the string taut, the bead is at the point \(B\), vertically below \(O\), when it is set into vertical circular motion with an initial horizontal velocity \(u\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{06c3e260-8167-4616-97d4-0f360a376a0f-5_616_613_520_733} The string does not become slack in the subsequent motion. The velocity of the bead at the point \(A\), where \(A\) is vertically above \(O\), is \(v\).

6. A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string and hangs at rest at time \(t = 0\). The other end of the string is then raised vertically by an engine which is working at a constant rate \(k m g\), where \(k > 0\). At time \(t\), the distance of \(P\) above its initial position is \(x\), and \(P\) is moving upwards with speed \(v\).

1. Show that \(v ^ { 2 } \frac { \mathrm {~d} v } { \mathrm {~d} x } = ( k - v ) g\).
2. Show that \(g x = k ^ { 2 } \ln \left( \frac { k } { k - v } \right) - k v - \frac { 1 } { 2 } v ^ { 2 }\).
3. Hence, or otherwise, find \(t\) in terms of \(k , v\) and \(g\).

1. A particle of mass 2 kg is moving in a straight line on a smooth horizontal surface under the action of a horizontal force of magnitude \(F\) newtons.

At time \(t\) seconds \(( t > 0 )\),

• the particle is moving with speed \(v \mathrm {~ms} ^ { - 1 }\)
• \(F = 2 + v\)

The time taken for the speed of the particle to increase from \(5 \mathrm {~ms} ^ { - 1 }\) to \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is \(T\) seconds.

1. Show that \(T = 2 \ln \frac { 12 } { 7 }\) The distance moved by the particle as its speed increases from \(5 \mathrm {~ms} ^ { - 1 }\) to \(10 \mathrm {~ms} ^ { - 1 }\) is \(D\) metres.
2. Find the exact value of \(D\).

8 A piston of mass 1.5 kg moves in a straight line inside a long straight horizontal cylinder. At time \(t \mathrm {~s}\) the displacement of the piston from its initial position at one end of the cylinder is \(x \mathrm {~m}\) and its velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see diagram). \includegraphics[max width=\textwidth, alt={}, center]{a8c9d007-e67f-4637-9e74-630ba9a91442-5_168_805_1726_630} The piston starts moving when \(t = 2\) and is brought to rest when it reaches the other end of the cylinder. While the piston is in motion it is acted on by a force of magnitude \(\frac { 6 } { t ^ { 2 } } \mathrm {~N}\) in the positive \(x\) direction, and also by a force of magnitude \(\frac { 3 v } { t } \mathrm {~N}\) resisting the motion.

1. Show that, while the piston is in motion, \(\frac { \mathrm { d } v } { \mathrm {~d} t } + \frac { 2 v } { t } = \frac { 4 } { t ^ { 2 } }\). The piston reaches the other end of the cylinder when \(t = 20\).
2. Find the speed of the piston immediately before it is brought to rest.
3. Show that the piston travels a distance of 5.61 m , correct to 3 significant figures. \section*{OCR} \section*{Oxford Cambridge and RSA}

4 A particle \(P\) of mass 8 kg moves in a straight line on a smooth horizontal plane. At time \(t \mathrm {~s}\) the displacement of \(P\) from a fixed point \(O\) on the line is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). Initially, \(P\) is at rest at \(O\). \(P\) is acted on by a horizontal force, directed along the line away from \(O\), with magnitude proportional to \(\sqrt { 9 + v ^ { 2 } }\). When \(v = 1.25\), the magnitude of this force is 13 N .

1. Show that \(\frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \frac { \mathrm {~d} v } { \mathrm {~d} t } = \frac { 1 } { 2 }\).
2. Find an expression for \(v\) in terms of \(t\) for \(t \geqslant 0\).
3. Find an expression for \(x\) in terms of \(t\) for \(t \geqslant 0\).

3 A particle \(P\) of mass 8 kg moves in a straight line on a smooth horizontal plane. At time \(t \mathrm {~s}\) the displacement of \(P\) from a fixed point \(O\) on the line is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). Initially, \(P\) is at rest at \(O\). \(P\) is acted on by a horizontal force, directed along the line away from \(O\), with magnitude proportional to \(\sqrt { 9 + v ^ { 2 } }\). When \(v = 1.25\), the magnitude of this force is 13 N .

1. Show that \(\frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \frac { \mathrm {~d} v } { \mathrm {~d} t } = \frac { 1 } { 2 }\).
2. Find an expression for \(v\) in terms of \(t\) for \(t \geqslant 0\).
3. Find an expression for \(x\) in terms of \(t\) for \(t \geqslant 0\).

\includegraphics{figure_6} A particle \(P\) of mass \(0.2\) kg is projected with velocity \(2\) m s\(^{-1}\) upwards along a line of greatest slope on a plane inclined at \(30°\) to the horizontal (see diagram). Air resistance of magnitude \(0.5v\) N opposes the motion of \(P\), where \(v\) m s\(^{-1}\) is the velocity of \(P\) at time \(t\) s after projection. The coefficient of friction between \(P\) and the plane is \(\frac{1}{2\sqrt{3}}\). The particle \(P\) reaches a position of instantaneous rest when \(t = T\).

1. Show that, while \(P\) is moving up the plane, \(\frac{dv}{dt} = -2.5(3 + v)\). [3]
2. Calculate \(T\). [4]
3. Calculate the speed of \(P\) when \(t = 2T\). [5]

A particle of mass \(m\) moves in a straight line. At time \(t\), its displacement from a fixed point on the line is \(s\) and its velocity is \(v\). The particle experiences a retarding force of magnitude \(mkv^2\), where \(k\) is a positive constant. Find the relationship between \(v\) and \(t\). [7]

A particle \(P\) moving in a straight line has displacement \(x\)m from a fixed point \(O\) on the line and velocity \(v\)m s\(^{-1}\) at time \(t\)s. The acceleration of \(P\), in m s\(^{-2}\), is given by \(6\sqrt{v + 9}\). When \(t = 0\), \(x = 2\) and \(v = 72\).

1. Find an expression for \(v\) in terms of \(x\). [4]
2. Find an expression for \(x\) in terms of \(t\). [5]

Suraiya, whose mass is \(m\) kg, takes a running jump into a swimming pool so that she begins to swim in a straight line with speed 0·2 ms\(^{-1}\). She continues to move in the same straight line, the only force acting on her being a resistance of magnitude \(mv^2 \sin \left(\frac{t}{100}\right)\) N, where \(v\) ms\(^{-1}\) is her speed at time \(t\) seconds after entering the pool and \(0 \leq t \leq 50\pi\).

1. Find an expression for \(v\) in terms of \(t\). [7 marks]
2. Calculate her greatest and least speeds during her motion. [3 marks]