Variable Force

1 A particle \(P\) of mass 0.6 kg is projected horizontally with velocity \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(O\) on a smooth horizontal surface. A horizontal force of magnitude \(0.3 x \mathrm {~N}\) acts on \(P\) in the direction \(O P\), where \(x \mathrm {~m}\) is the distance of \(P\) from \(O\). Calculate the velocity of \(P\) when \(x = 8\).

7 A particle \(P\) of mass 0.25 kg moves in a straight line on a smooth horizontal surface. \(P\) starts at the point \(O\) with speed \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and moves towards a fixed point \(A\) on the line. At time \(t \mathrm {~s}\) the displacement of \(P\) from \(O\) is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A resistive force of magnitude (5-x) N acts on \(P\) in the direction towards \(O\).

1. Form a differential equation in \(v\) and \(x\). By solving this differential equation, show that \(v = 10 - 2 x\).
2. Find \(x\) in terms of \(t\), and hence show that the particle is always less than 5 m from \(O\).

A particle \(P\) moving in a straight line has displacement \(x\) m from a fixed point \(O\) on the line at time \(t\) s. The acceleration of \(P\), in m s\(^{-2}\), is given by \(\frac{200}{x^2} - \frac{100}{x^3}\) for \(x > 0\). When \(t = 0\), \(x = 1\) and \(P\) has velocity \(10\) m s\(^{-1}\) directed towards \(O\).

1. Show that the velocity \(v\) m s\(^{-1}\) of \(P\) is given by \(v = \frac{10(1-2x)}{x}\). [5]
2. Show that \(x\) and \(t\) are related by the equation \(e^{-40t} = (2x-1)e^{2x-2}\) and deduce what happens to \(x\) as \(t\) becomes large. [5]

A particle P of mass 4 kilograms moves in such a way that its position vector at time \(t\) seconds is \(\mathbf{r}\) metres, where $$\mathbf{r} = 3t\mathbf{i} + 2e^{-3t}\mathbf{j}.$$

1. Find the initial kinetic energy of P. [4]
2. Find the time when the acceleration of P is 2 metres per second squared. [3]

7 A particle of mass 0.25 kg moves in a straight line on a smooth horizontal surface. A variable resisting force acts on the particle. At time \(t \mathrm {~s}\) the displacement of the particle from a point on the line is \(x \mathrm {~m}\), and its velocity is \(( 8 - 2 x ) \mathrm { m } \mathrm { s } ^ { - 1 }\). It is given that \(x = 0\) when \(t = 0\).

1. Find the acceleration of the particle in terms of \(x\), and hence find the magnitude of the resisting force when \(x = 1\).
2. Find an expression for \(x\) in terms of \(t\).
3. Show that the particle is always less than 4 m from its initial position.

1. In this question you must show all stages in your working. Solutions relying entirely on calculator technology are not acceptable.

A particle \(P\) is moving along the \(x\)-axis.
At time \(t\) seconds, where \(0 \leqslant t \leqslant \frac { 2 } { 3 } , P\) is \(x\) metres from the origin 0 and is moving with velocity \(\mathrm { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) in the positive x direction where $$v = ( 2 x + 1 ) ^ { \frac { 3 } { 2 } }$$ When \(\mathrm { t } = 0 , \mathrm { P }\) passes through 0 .

1. Find the value of x when the acceleration of P is \(243 \mathrm {~m} \mathrm {~s} ^ { - 2 }\)
2. Find v in terms of t .

1. Find the magnitude of \(F\) when \(t = 3\). [3]

1. Find the magnitude of \(F\) when \(t = 3\). [3]

5. A car of mass 800 kg moves along a horizontal straight road. At time \(t\) seconds, the resultant force acting on the car has magnitude \(\frac { 48000 } { ( t + 2 ) ^ { 2 } }\) newtons, acting in the direction of the motion of the car. When \(t = 0\), the car is at rest.

1. Show that the speed of the car approaches a limiting value as \(t\) increases and find this value.
2. Find the distance moved by the car in the first 6 seconds of its motion.

A small ball of mass \(m\) is projected vertically upwards from a point \(O\) with speed \(U\). The ball is subject to air resistance of magnitude \(mkv\), where \(v\) is the speed of the ball and \(k\) is a positive constant. Find, in terms of \(U\), \(g\) and \(k\), the maximum height above \(O\) reached by the ball. (8)

4 An object of mass 0.4 kg is projected vertically upwards from the ground, with an initial speed of \(16 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A resisting force of magnitude \(0.1 v\) newtons acts on the object during its ascent, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of the object at time \(t \mathrm {~s}\) after it starts to move.

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = - 0.25 ( v + 40 )\).
2. Find the value of \(t\) at the instant that the object reaches its maximum height.

3. At time \(t = 0\), a particle of mass \(m\) is projected vertically downwards with speed \(U\) from a point above the ground. At time \(t\) the speed of the particle is \(v\) and the magnitude of the air resistance is modelled as being \(m k v\), where \(k\) is a constant. Given that \(U < \frac { \boldsymbol { g } } { \mathbf { 2 } \boldsymbol { k } }\), find, in terms of \(k , U\) and \(g\), the time taken for the particle to double its speed.
(8)

1. A particle \(P\) is moving along the positive \(x\)-axis. When the displacement of \(P\) from the origin is \(x\) metres, the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the acceleration of \(P\) is \(9 x \mathrm {~m} \mathrm {~s} ^ { - 2 }\).

When \(x = 2 , v = 6\) Show that \(v ^ { 2 } = 9 x ^ { 2 }\).
(4)

1. A particle \(P\) moves along the \(x\)-axis. At time \(t = 0 , P\) passes through the origin \(O\), moving in the positive \(x\)-direction. At time \(t\) seconds, the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(O P = x\) metres. The acceleration of \(P\) is \(\frac { 1 } { 12 } ( 30 - x ) \mathrm { m } \mathrm { s } ^ { - 2 }\), measured in the positive \(x\)-direction.
   1. Give a reason why the maximum speed of \(P\) occurs when \(x = 30\).
   Given that the maximum speed of \(P\) is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\),
2. find an expression for \(v ^ { 2 }\) in terms of \(x\).

2 On a building site, a pulley system is used for moving small amounts of material up to roof level. A light pulley, which can rotate freely, is attached with its axis horizontal to the top of some scaffolding. A light inextensible rope hangs over the pulley with a counterweight of mass \(m _ { 1 } \mathrm {~kg}\) attached to one end. Attached to the other end of the rope is a bag of negligible mass into which \(m _ { 2 } \mathrm {~kg}\) of roof tiles are placed, where \(m _ { 2 } < m _ { 1 }\). This situation is shown in Fig. 2. \begin{figure}[h] \end{figure} Initially the system is held at rest with the rope taut, the counterweight at the top of the scaffolding and the bag of tiles on the ground. When the counterweight is released, the bag ascends towards the top of the scaffolding. At time \(t \mathrm {~s}\) the velocity of the counterweight is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) downwards. The counterweight is made from a bag of negligible mass filled with sand. At the moment the counterweight is released, this bag is accidentally ripped and after this time the sand drops out at a constant rate of \(\lambda \mathrm { kg } \mathrm { s } ^ { - 1 }\).

1. Find the equation of motion for the counterweight while it still contains sand, and hence show that $$v = g t + \frac { 2 g m _ { 2 } } { \lambda } \ln \left( 1 - \frac { \lambda t } { m _ { 1 } + m _ { 2 } } \right) .$$
2. Given that the sand would run out after 10 seconds and that \(m _ { 2 } = \frac { 4 } { 5 } m _ { 1 }\), find the maximum velocity attained by the counterweight towards the ground. You may assume that the scaffolding is sufficiently high that the counterweight does not hit the ground before this velocity is reached.

A small pebble of mass \(m\) is placed in a viscous liquid and sinks vertically from rest through the liquid. When the speed of the pebble is \(v\) the magnitude of the resistance due to the liquid is modelled as \(mkv^2\), where \(k\) is a positive constant. Find the speed of the pebble after it has fallen a distance \(D\) through the liquid. [11]

3 A particle \(P\) of mass \(m \mathrm {~kg}\) is released from rest and falls vertically. When \(P\) has fallen a distance of \(x \mathrm {~m}\) it has a speed of \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The only forces acting on \(P\) are its weight and air resistance of magnitude \(\frac { 1 } { 400 } m v ^ { 2 } \mathrm {~N}\).

1. Find \(v ^ { 2 }\) in terms of \(x\) and show that \(v ^ { 2 }\) must be less than 3920 .
2. Find the speed of \(P\) when it has fallen 100 m .

2 A ball of mass 2 kg is projected vertically downwards with speed \(5 \mathrm {~ms} ^ { - 1 }\) through a liquid. At time \(t \mathrm {~s}\) after projection, the velocity of the ball is \(v \mathrm {~ms} ^ { - 1 }\) and its displacement from its starting point is \(x \mathrm {~m}\). The forces acting on the ball are its weight and a resistive force of magnitude \(0.2 v ^ { 2 } \mathrm {~N}\).

1. Find an expression for \(v\) in terms of \(t\).
2. Deduce what happens to \(v\) for large values of \(t\). \includegraphics[max width=\textwidth, alt={}, center]{e7091f6c-af72-49f3-b825-cdce9fb2c06f-06_803_652_251_703} A uniform square lamina of side \(2 a\) and weight \(W\) is suspended from a light inextensible string attached to the midpoint \(E\) of the side \(A B\). The other end of the string is attached to a fixed point \(P\) on a rough vertical wall. The vertex \(B\) of the lamina is in contact with the wall. The string \(E P\) is perpendicular to the side \(A B\) and makes an angle \(\theta\) with the wall (see diagram). The string and the lamina are in a vertical plane perpendicular to the wall. The coefficient of friction between the wall and the lamina is \(\frac { 1 } { 2 }\). Given that the vertex \(B\) is about to slip up the wall, find the value of \(\tan \theta\). \includegraphics[max width=\textwidth, alt={}, center]{e7091f6c-af72-49f3-b825-cdce9fb2c06f-08_581_576_269_731} A light elastic string has natural length \(8 a\) and modulus of elasticity \(5 m g\). A particle \(P\) of mass \(m\) is attached to the midpoint of the string. The ends of the string are attached to points \(A\) and \(B\) which are a distance \(12 a\) apart on a smooth horizontal table. The particle \(P\) is held on the table so that \(A P = B P = L\) (see diagram). The particle \(P\) is released from rest. When \(P\) is at the midpoint of \(A B\) it has speed \(\sqrt { 80 a g }\).
   1. Find \(L\) in terms of \(a\).
   2. Find the initial acceleration of \(P\) in terms of \(g\).

4 A particle of mass 0.2 kg moves in a straight line on a smooth horizontal surface. When its displacement from a fixed point on the surface is \(x \mathrm {~m}\), its velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The motion is opposed by a force of magnitude \(\frac { 1 } { 3 v } \mathrm {~N}\).

1. Show that \(3 v ^ { 2 } \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 5\).
2. Find the value of \(v\) when \(x = 7.4\), given that \(v = 4\) when \(x = 0\).

2 The resistance to the motion of a car is \(k v ^ { \frac { 3 } { 2 } } \mathrm {~N}\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the car's speed and \(k\) is a constant. The power exerted by the car's engine is 15000 W , and the car has constant speed \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) along a horizontal road.

1. Show that \(k = 4.8\). With the engine operating at a much lower power, the car descends a hill of inclination \(\alpha\), where \(\sin \alpha = \frac { 1 } { 15 }\). At an instant when the speed of the car is \(16 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), its acceleration is \(0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
2. Given that the mass of the car is 700 kg , calculate the power of the engine. \includegraphics[max width=\textwidth, alt={}, center]{941c0c81-a74f-49c0-acb7-1c23266fc2c8-02_579_447_1658_849} A particle \(P\) of mass 0.4 kg is attached to one end of each of two light inextensible strings which are both taut. The other end of the longer string is attached to a fixed point \(A\), and the other end of the shorter string is attached to a fixed point \(B\), which is vertically below \(A\). The string \(A P\) makes an angle of \(30 ^ { \circ }\) with the vertical and is 0.5 m long. The string \(B P\) makes an angle of \(60 ^ { \circ }\) with the vertical. \(P\) moves with constant angular speed in a horizontal circle with centre vertically below \(B\) (see diagram). The tension in the string \(A P\) is twice the tension in the string \(B P\). Calculate

3 A particle \(P\) of mass 0.3 kg is projected horizontally with speed \(u \mathrm {~ms} ^ { - 1 }\) 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 with speed \(v \mathrm {~ms} ^ { - 1 }\). There is a force of magnitude \(1.2 v ^ { 3 } \mathrm {~N}\) 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 \(v = \frac { u } { 4 u x + 1 }\).
2. Given that \(x = 2\) when \(t = 9\) find the value of \(u\).

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]

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 \(g\).

3 An aircraft of mass 80000 kg travelling at \(90 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) touches down on a straight horizontal runway. It is brought to rest by braking and resistive forces which together are modelled by a horizontal force of magnitude ( \(27000 + 50 v ^ { 2 }\) ) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of the aircraft. Find the distance travelled by the aircraft between touching down and coming to rest.

4 A particle \(P\) of mass 0.4 kg is projected horizontally with velocity \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(O\) on a smooth horizontal surface. The motion of \(P\) is opposed by a resisting force of magnitude \(0.2 v ^ { 2 } \mathrm {~N}\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity of \(P\) at time \(t \mathrm {~s}\) after projection.

1. Show that \(v = \frac { 8 } { 1 + 4 t }\).
2. Calculate the distance \(O P\) when \(t = 1.5\).

7 A particle \(P\) of mass \(m \mathrm {~kg}\) moves in a horizontal straight line against a resistive force of magnitude \(\mathrm { mkv } ^ { 2 } \mathrm {~N}\), where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of \(P\) after it has moved a distance \(x \mathrm {~m}\) and \(k\) is a positive constant. The initial speed of \(P\) is \(u \mathrm {~ms} ^ { - 1 }\).

1. Show that \(\mathrm { x } = \frac { 1 } { \mathrm { k } } \ln 2\) when \(\mathrm { v } = \frac { 1 } { 2 } \mathrm { u }\).
   Beginning at the instant when the speed of \(P\) is \(\frac { 1 } { 2 } u\), an additional force acts on \(P\). This force has magnitude \(\frac { 5 \mathrm {~m} } { \mathrm { v } } \mathrm { N }\) and acts in the direction of increasing \(x\).
2. Show that when the speed of \(P\) has increased again to \(u \mathrm {~ms} ^ { - 1 }\), the total distance travelled by \(P\) is given by an expression of the form $$\frac { 1 } { 3 k } \ln \left( \frac { A - k u ^ { 3 } } { B - k u ^ { 3 } } \right) ,$$ stating the values of the constants \(A\) and \(B\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

A particle \(P\) of mass 0.5 kg moves along a straight line. When \(P\) is at a distance \(x\) m from a fixed point \(O\) on the line, the force acting on it is directed towards \(O\) and has magnitude \(\frac{8}{x}\) N. When \(x = 2\), the speed of \(P\) is 4 ms\(^{-1}\). Find the speed of \(P\) when it is 0.5 m from \(O\). [8 marks]

3 \includegraphics[max width=\textwidth, alt={}, center]{835616aa-0b2b-4e8c-bbbf-60b72dc5ea3e-2_145_792_1656_680} A particle \(P\) of mass 0.6 kg moves in a straight line on a smooth horizontal surface. A force of magnitude \(\frac { 3 } { x ^ { 3 } }\) newtons acts on the particle in the direction from \(P\) to \(O\), where \(O\) is a fixed point of the surface and \(x \mathrm {~m}\) is the distance \(O P\) (see diagram). The particle \(P\) is released from rest at the point where \(x = 10\). Find the speed of \(P\) when \(x = 2.5\).

A particle \(P\) of mass \(m\) kg moves along a straight line under the action of a force of magnitude \(\frac{km}{x^2}\) N, where \(k\) is a constant, directed towards a fixed point \(O\) on the line, where \(OP = x\) m. \(P\) starts from rest at \(A\), at a distance \(a\) m from \(O\). When \(OP = x\) m, the speed of \(P\) is \(v\) ms\(^{-1}\).

1. Show that \(v = \sqrt{\frac{2k(a-x)}{ax}}\). [6 marks]

\(B\) is the point half-way between \(O\) and \(A\). When \(k = \frac{1}{2}\) and \(a = 1\), the time taken by \(P\) to travel from \(A\) to \(B\) is \(T\) seconds Assuming the result that, for \(0 \leq x \leq 1\), \(\int \sqrt{\frac{x}{1-x}} dx = \arcsin(\sqrt{x}) - \sqrt{x(1-x^2)} + \text{constant}\),

1. find the value of \(T\). [5 marks]

1 A particle \(P\) of mass 0.2 kg is released from rest at a point \(O\) on a smooth horizontal surface. A horizontal force of magnitude \(t \mathrm { e } ^ { - v } \mathrm {~N}\) directed away from \(O\) acts on \(P\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity of \(P\) at time \(t \mathrm {~s}\) after release. Find the velocity of \(P\) when \(t = 2\).

3 A particle \(P\) of mass 0.8 kg moves along the \(x\)-axis on a horizontal surface. When the displacement of \(P\) from the origin \(O\) is \(x \mathrm {~m}\) the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the positive \(x\)-direction. Two horizontal forces act on \(P\). One force has magnitude \(4 \mathrm { e } ^ { - x } \mathrm {~N}\) and acts in the positive \(x\)-direction. The other force has magnitude \(2.4 x ^ { 2 } \mathrm {~N}\) and acts in the negative \(x\)-direction.

1. Show that \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 5 \mathrm { e } ^ { - x } - 3 x ^ { 2 }\).
2. The velocity of \(P\) as it passes through \(O\) is \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the velocity of \(P\) when \(x = 2\).

5 A particle \(P\) of mass 4 kg is moving in a horizontal straight line. At time \(t\) s the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\) and the displacement of \(P\) from a fixed point \(O\) on the line is \(x \mathrm {~m}\). The only force acting on \(P\) is a resistive force of magnitude \(\left( 4 \mathrm { e } ^ { - x } + 12 \right) \mathrm { e } ^ { - x } \mathrm {~N}\). When \(\mathrm { t } = 0 , \mathrm { x } = 0\) and \(v = 4\).

1. Show by integration that \(\mathrm { v } = \frac { 1 + 3 \mathrm { e } ^ { \mathrm { x } } } { \mathrm { e } ^ { \mathrm { x } } }\).
2. Find an expression for \(x\) in terms of \(t\). \includegraphics[max width=\textwidth, alt={}, center]{ad8b126c-d739-4e2a-8ce3-7811a61f5876-10_510_889_269_580} \(A B\) and \(B C\) are two fixed smooth vertical barriers on a smooth horizontal surface, with angle \(\mathrm { ABC } = 60 ^ { \circ }\). A particle of mass \(m\) is moving with speed \(u\) on the surface. The particle strikes \(A B\) at an angle \(\theta\) with \(A B\). It then strikes \(B C\) and rebounds at an angle \(\beta\) with \(B C\) (see diagram). The coefficient of restitution between the particle and each barrier is \(e\) and \(\tan \theta = 2\). The kinetic energy of the particle after the first collision is \(40 \%\) of its kinetic energy before the first collision.
   1. Find the value of \(e\).
   2. Find the size of angle \(\beta\). \includegraphics[max width=\textwidth, alt={}, center]{ad8b126c-d739-4e2a-8ce3-7811a61f5876-12_965_1059_267_502} A uniform cylinder with a rough surface and of radius \(a\) is fixed with its axis horizontal. Two identical uniform rods \(A B\) and \(B C\), each of weight \(W\) and length \(2 a\), are rigidly joined at \(B\) with \(A B\) perpendicular to \(B C\). The rods rest on the cylinder in a vertical plane perpendicular to the axis of the cylinder with \(A B\) at an angle \(\theta\) to the horizontal. \(D\) and \(E\) are the midpoints of \(A B\) and \(B C\) respectively and also the points of contact of the rods with the cylinder (see diagram). The rods are about to slip in a clockwise direction. The coefficient of friction between each rod and the cylinder is \(\mu\). The normal reaction between \(A B\) and the cylinder is \(R\) and the normal reaction between \(B C\) and the cylinder is \(N\).
   3. Find the ratio \(R : N\) in terms of \(\mu\).
   4. Given that \(\mu = \frac { 1 } { 3 }\), find the value of \(\tan \theta\).
      If you use the following page to complete the answer to any question, the question number must be clearly shown.

3. A raindrop falls vertically under gravity through a stationary cloud. At time \(t = 0\), the raindrop is at rest and has mass \(m _ { 0 }\). As the raindrop falls, water condenses onto it from the cloud so that the mass of the raindrop increases at a constant rate \(c\). At time \(t\), the mass of the raindrop is \(m\) and the speed of the raindrop is \(v\). The resistance to the motion of the raindrop has magnitude \(m k v\), where \(k\) is a constant. Show that $$\frac { \mathrm { d } v } { \mathrm {~d} t } + v \left( k + \frac { c } { m _ { 0 } + c t } \right) = g$$

2 A railway truck of mass \(m _ { 0 }\) travels along a horizontal track. There is no driving force and the resistances to motion are negligible. The truck is being filled with coal which falls vertically into it at a mass rate \(k\). The process starts as the truck passes a point O with speed \(u\). After time \(t\), the truck has velocity \(v\) and the displacement from O is \(x\).

1. Show that \(v = \frac { m _ { 0 } u } { m _ { 0 } + k t }\) and find \(x\) in terms of \(m _ { 0 } , u , k\) and \(t\).
2. Find the distance that the truck has travelled when its speed has been halved.

1 A spherical raindrop falls through a stationary cloud. Water condenses on the raindrop and it gains mass at a rate proportional to its surface area. At time \(t\) the radius of the raindrop is \(r\). Initially the raindrop is at rest and \(r = r _ { 0 }\). The density of the water is \(\rho\).

1. Show that \(\frac { \mathrm { d } r } { \mathrm {~d} t } = k\), where \(k\) is a constant. Hence find the mass of the raindrop in terms of \(r _ { 0 } , \rho , k\) and \(t\).
2. Assuming that air resistance is negligible, find the velocity of the raindrop in terms of \(r _ { 0 } , k\) and \(t\).

A particle \(P\) of mass 3 kg moves in a straight line on a smooth horizontal plane. When the speed of \(P\) is \(v\) m s\(^{-1}\), the resultant force acting on \(P\) is a resistance to motion of magnitude \(2v\) N. Find the distance moved by \(P\) while slowing down from 5 m s\(^{-1}\) to 2 m s\(^{-1}\). [5]

7 A stone of mass \(m\) is moving along the smooth horizontal floor of a tank which is filled with a viscous liquid. At time \(t\), the stone has speed \(v\). As the stone moves, it experiences a resistance force of magnitude \(\lambda m v\), where \(\lambda\) is a constant.

1. Show that $$\frac { \mathrm { d } v } { \mathrm {~d} t } = - \lambda v$$
2. The initial speed of the stone is \(U\). Show that $$v = U \mathrm { e } ^ { - \lambda t }$$

A particle of mass \(0.4\) kg, moving on a smooth horizontal surface, passes through a point \(O\) with velocity \(10\text{ ms}^{-1}\). At time \(t\) s after the particle passes through \(O\), the particle has a displacement \(x\) m from \(O\), has a velocity \(v\text{ ms}^{-1}\) away from \(O\), and is acted on by a force of magnitude \(\frac{1}{5}v\) N acting towards \(O\). Find

1. the time taken for the velocity of the particle to reduce from \(10\text{ ms}^{-1}\) to \(5\text{ ms}^{-1}\), [5]
2. the average velocity of the particle over this time. [6]

10 A small body of mass \(m\) is thrown vertically upwards with initial velocity \(u\). Resistance to motion is \(k v ^ { 2 }\) per unit mass, where the velocity is \(v\) and \(k\) is a positive constant. Find, in terms of \(u , g\) and \(k\),

1. the time taken to reach the greatest height,
2. the greatest height to which the body will rise.

5 A particle \(P\) of mass 0.5 kg is projected vertically upwards from a point on a horizontal surface. A resisting force of magnitude \(0.02 v ^ { 2 } \mathrm {~N}\) acts on \(P\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the upward velocity of \(P\) when it is a height of \(x \mathrm {~m}\) above the surface. The initial speed of \(P\) is \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that, while \(P\) is moving upwards, \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 10 - 0.04 v ^ { 2 }\).
2. Find the greatest height of \(P\) above the surface.
3. Find the speed of \(P\) immediately before it strikes the surface after descending.

5 A particle \(P\) of mass \(m \mathrm {~kg}\) is projected vertically upwards through a liquid. Student \(A\) measures \(P\) 's initial speed as \(( 8.5 \pm 0.25 ) \mathrm { m } \mathrm { s } ^ { - 1 }\) and they also record the time for \(P\) to attain its greatest height above the initial point of projection as over 3 seconds.
In an attempt to model the motion of \(P\) student \(B\) determines that \(t\) seconds after projection the only forces acting on \(P\) are its weight and the resistance from the liquid. Student \(B\) models the resistance from the liquid to be of magnitude \(m v ^ { 2 } - 6 m v\), where \(v\) is the speed of the particle.

1. (a) Show that \(\frac { \mathrm { d } t } { \mathrm {~d} v } = - \frac { 1 } { ( v - 3 ) ^ { 2 } + 0.8 }\).
   (b) Determine whether student \(B\) 's model is consistent with the time recorded by \(A\) for \(P\) to attain its greatest height. After attaining its greatest height \(P\) now falls through the liquid. Student \(C\) claims that the time taken for \(P\) to achieve a speed of \(1 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when falling through the liquid is given by $$- \int _ { 0 } ^ { 1 } \frac { 1 } { ( v - 3 ) ^ { 2 } + 0.8 } \mathrm {~d} v$$
2. Explain why student \(C\) 's claim is incorrect and write down the integral which would give the correct time for \(P\) to achieve a speed of \(1 \mathrm {~ms} ^ { - 1 }\) when falling through the liquid.

1 One end of a light elastic rope of natural length 2.5 m and modulus of elasticity 80 N is attached to a fixed point \(A\). A stone \(S\) of mass 8 kg is attached to the other end of the rope. \(S\) is held at a point 6 m vertically below \(A\) and then released. Find the initial acceleration of \(S\).

3. A rocket is fired vertically upwards with speed \(U\) from a point on the Earth's surface. The rocket is modelled as a particle \(P\) of constant mass \(m\), and the Earth as a fixed sphere of radius \(R\). At a distance \(x\) from the centre of the Earth, the speed of \(P\) is \(v\). The only force acting on \(P\) is directed towards the centre of the Earth and has magnitude \(\frac { c m } { x ^ { 2 } }\), where \(c\) is a constant.

1. Show that \(v ^ { 2 } = U ^ { 2 } + 2 c \left( \frac { 1 } { x } - \frac { 1 } { R } \right)\). The kinetic energy of \(P\) at \(x = 2 R\) is half of its kinetic energy at \(x = R\).
2. Find \(c\) in terms of \(U\) and \(R\).
   (3)

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 \(t = 4\).

2 A car of mass 1800 kg is travelling along a straight horizontal road. The power of the car's engine is constant. There is a constant resistance to motion of 650 N .

1. Find the power of the car's engine, given that the car's acceleration is \(0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) when its speed is \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
2. Find the steady speed which the car can maintain with the engine working at this power.

1 A bullet of mass 0.2 kg is fired into a fixed vertical barrier. It enters the barrier horizontally with speed \(250 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and emerges horizontally after a time \(T\) seconds with speed \(40 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). There is a constant horizontal resisting force of magnitude 1200 N . Find \(T\).

5. A rocket is launched vertically upwards under gravity from rest at time \(t = 0\). The rocket propels itself upward by ejecting burnt fuel vertically downwards at a constant speed \(u\) relative to the rocket. The initial mass of the rocket, including fuel, is \(M\). At time \(t\), before all the fuel has been used up, the mass of the rocket, including fuel, is \(M ( 1 - k t )\) and the speed of the rocket is \(v\).

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { k u } { 1 - k t } - g\).
2. Hence find the speed of the rocket when \(t = \frac { 1 } { 3 k }\).

4 A train of mass 400000 kg is moving on a straight horizontal track. The power of the engine is constant and equal to 1500 kW and the resistance to the train's motion is 30000 N . Find

1. the acceleration of the train when its speed is \(37.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\),
2. the steady speed at which the train can move.

3 A car of mass 1250 kg travels down a straight hill with the engine working at a power of 22 kW . The hill is inclined at \(3 ^ { \circ }\) to the horizontal and the resistance to motion of the car is 1130 N . Find the speed of the car at an instant when its acceleration is \(0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).

The displacement of \(P\) from \(O\) is \(x\) m at time \(t\) s.

1. Find an expression for \(x\) in terms of \(t\), while \(P\) is moving upwards. [2]
2. Find, correct to 3 significant figures, the greatest height above \(O\) reached by \(P\). [2]

2 A duck of mass 2 kg is travelling with horizontal speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when it lands on a lake. The duck is brought to rest by the action of resistive forces, acting in the direction opposite to the duck's motion and having total magnitude \(\left( 2 v + 3 v ^ { 2 } \right) \mathrm { N }\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of the duck. Show that the duck comes to rest after travelling approximately 1.30 m from the point of its initial contact with the surface of the lake.

11 A car of mass 800 kg has a constant power output of 32 kW while travelling on a horizontal road. At time \(t \mathrm {~s}\) the car's speed is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the resistive force has magnitude \(20 v \mathrm {~N}\).

1. Show that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 1600 - v ^ { 2 } } { 40 v }\).
2. Given that \(v = 0\) when \(t = 0\), solve this differential equation to find \(v\) in terms of \(t\). State what the solution predicts as \(t\) becomes large.

6 A particle \(P\) of mass 0.2 kg is released from rest at a point \(O\) on a plane inclined at \(30 ^ { \circ }\) to the horizontal. At time \(t \mathrm {~s}\) after its release, \(P\) has velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and displacement \(x \mathrm {~m}\) down the plane from \(O\). The coefficient of friction between \(P\) and the plane increases as \(P\) moves down the plane, and equals \(0.1 x ^ { 2 }\).

1. Show that \(2 v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 10 - ( \sqrt { } 3 ) x ^ { 2 }\).
2. Calculate the maximum speed of \(P\).
3. Find the value of \(x\) at the point where \(P\) comes to rest.

3 A particle of mass 0.2 kg lies at rest on a smooth horizontal table. A horizontal force of magnitude \(F\) newtons acts on the particle in a constant direction for 0.1 seconds. At time \(t\) seconds, $$F = 5 \times 10 ^ { 3 } t ^ { 2 } , \quad 0 \leqslant t \leqslant 0.1$$ Find the value of \(t\) when the speed of the particle is \(2 \mathrm {~ms} ^ { - 1 }\).
(4 marks)

A vertical ladder is fixed to a wall in a harbour. On a particular day the minimum depth of water in the harbour occurs at 0900 hours. The next time the water is at its minimum depth is 2115 hours on the same day. The bottom step of the ladder is 1 m above the lowest level of the water and 9 m below the highest level of the water. The rise and fall of the water level can be modelled as simple harmonic motion and the thickness of the step can be assumed to be negligible. Find

1. the speed, in metres per hour, at which the water level is moving when it reaches the bottom step of the ladder, [7]
2. the length of time, on this day, between the water reaching the bottom step of the ladder and the ladder being totally out of the water once more. [4]

7 A particle \(P\) of mass 0.2 kg is released from rest at a point \(O\) on a rough plane inclined at \(60 ^ { \circ }\) to the horizontal, and travels down a line of greatest slope. The coefficient of friction between \(P\) and the plane is 0.3 . A force of magnitude \(0.6 x \mathrm {~N}\) acts on \(P\) in the direction \(P O\), where \(x \mathrm {~m}\) is the displacement of \(P\) from \(O\).

1. Show that \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 5 \sqrt { } 3 - 1.5 - 3 x\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity of \(P\) at a displacement \(x \mathrm {~m}\) from \(O\).
2. Find the value of \(x\) for which \(P\) reaches its maximum velocity, and calculate this maximum velocity.
3. Calculate the magnitude of the acceleration of \(P\) immediately after it has first come to instantaneous rest.

3 A particle \(P\) of mass 0.4 kg is released from rest at a point \(O\) on a smooth plane inclined at \(30 ^ { \circ }\) to the horizontal. \(P\) moves down the line of greatest slope through \(O\). The velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when its displacement from \(O\) is \(x \mathrm {~m}\). A retarding force of magnitude \(0.2 v ^ { 2 } \mathrm {~N}\) acts on \(P\) in the direction \(P O\).

1. Show that \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 5 - 0.5 v ^ { 2 }\).
2. Express \(v\) in terms of \(x\).