6.06a Variable force: dv/dt or v*dv/dx methods

5. A particle of mass 0.8 kg is moving along the positive \(x\)-axis at a speed of \(5 \mathrm {~ms} ^ { - 1 }\) away from the origin \(O\). When the particle is 2 metres from \(O\) it becomes subject to a single force directed towards \(O\). The magnitude of the force is \(\frac { k } { x ^ { 2 } } \mathrm {~N}\) when the particle is \(x\) metres from \(O\). Given that when the particle is 4 m from \(O\) its speed has been reduced to \(3 \mathrm {~ms} ^ { - 1 }\),

1. show that \(k = \frac { 128 } { 5 }\),
2. find the distance of the particle from \(O\) when it comes to instantaneous rest. (4 marks)

5. When a particle of mass \(M\) is at a distance of \(x\) metres from the centre of the moon, the gravitational force, \(F\) N, acting on it and directed towards the centre of the moon is given by $$F = \frac { \left( 4.90 \times 10 ^ { 12 } \right) M } { x ^ { 2 } }$$ A rocket is projected vertically into space from a point on the surface of the moon with initial speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Given that the radius of the moon is \(\left( 1.74 \times 10 ^ { 6 } \right) \mathrm { m }\),

1. show that the speed of the rocket, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), when it is \(x\) metres from the centre of the moon is given by $$v ^ { 2 } = u ^ { 2 } + \frac { a } { x } - b$$ where \(a\) and \(b\) are constants which should be found correct to 3 significant figures.
2. Find, correct to 2 significant figures, the minimum value of \(u\) needed for the rocket to escape the moon's gravitational attraction.

3. A car starts from rest at the point \(O\) and moves along a straight line. The car accelerates to a maximum velocity, \(V \mathrm {~ms} ^ { - 1 }\), before decelerating and coming to rest again at the point \(A\). The acceleration of the car during this journey, \(a \mathrm {~ms} ^ { - 2 }\), is modelled by the formula $$a = \frac { 500 - k x } { 150 }$$ where \(x\) is the distance in metres of the car from \(O\).
Using this model and given that the car is travelling at \(16 \mathrm {~ms} ^ { - 1 }\) when it is 40 m from \(O\),

1. find \(k\),
2. show that \(V = 41\), correct to 2 significant figures,
3. find the distance \(O A\).

3. The engine of a car of mass 800 kg works at a constant rate of 32 kW . The car travels along a straight horizontal road and the resistance to motion of the car is proportional to the speed of the car. The car starts from rest and \(t\) seconds later it has a speed of \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that $$800 v \frac { \mathrm {~d} v } { \mathrm {~d} t } = 32000 - k v ^ { 2 } , \text { where } k \text { is a positive constant. }$$ Given that the limiting speed of the car is \(40 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), find
2. the value of \(k\),
3. \(v\) in terms of \(t\).

A wooden ball of mass 0.01 kg falls vertically into a pond of water. The speed of the ball as it enters the water is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When the ball is \(x\) metres below the surface of the water and moving downwards with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the water provides a resistance of magnitude \(0.02 v ^ { 2 } \mathrm {~N}\) and an upward buoyancy force of magnitude 0.158 N .

1. Show that, while the ball is moving downwards, $$- 2 v ^ { 2 } - 6 = v \frac { \mathrm {~d} v } { \mathrm {~d} x }$$
2. Hence find, to 3 significant figures, the greatest distance below the surface of the water reached by the ball.
   

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. 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)

2. At time \(t = 0\), a particle \(P\) of mass \(m\) is projected vertically upwards with speed \(\sqrt { \frac { g } { k } }\), where \(k\) is a constant. At time \(t\) the speed of \(P\) is \(v\). The particle \(P\) moves against air resistance whose magnitude is modelled as being \(m k v ^ { 2 }\) when the speed of \(P\) is \(v\). Find, in terms of \(k\), the distance travelled by \(P\) until its speed first becomes half of its initial speed.
(9)

5. A particle \(Q\) of mass 6 kg is moving along the \(x\)-axis. At time \(t\) seconds the displacement of \(Q\) from the origin \(O\) is \(x\) metres and the speed of \(Q\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The particle moves under the action of a retarding force of magnitude ( \(a + b v ^ { 2 }\) ) N, where \(a\) and \(b\) are positive constants. At time \(t = 0 , Q\) is at \(O\) and moving with speed \(U \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the positive \(x\)-direction. The particle \(Q\) comes to instantaneous rest at the point \(X\).

1. Show that the distance \(O X\) is $$\frac { 3 } { b } \ln \left( 1 + \frac { b U ^ { 2 } } { a } \right) \mathrm { m }$$ Given that \(a = 12\) and \(b = 3\),
2. find, in terms of \(U\), the time taken to move from \(O\) to \(X\).

A particle \(P\) of mass 4 kg moves along a horizontal straight line under the action of a force directed towards a fixed point \(O\) on the line. At time \(t\) seconds, \(P\) is \(x\) metres from \(O\) and the force towards \(O\) has magnitude \(9 x\) newtons. The particle \(P\) is also subject to air resistance, which has magnitude \(12 v\) newtons when \(P\) is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that the equation of motion of \(P\) is $$4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 12 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 x = 0$$ It is given that the solution of this differential equation is of the form $$x = \mathrm { e } ^ { - \lambda t } ( A t + B )$$ When \(t = 0\) the particle is released from rest at the point \(R\), where \(O R = 4 \mathrm {~m}\). Find,
2. the values of the constants \(\lambda , A\) and \(B\),
3. the greatest speed of \(P\) in the subsequent motion.
   

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 $$\frac { 4 m g } { 3 } \leqslant T < \frac { 3 m g } { 2 }$$

5. A van of mass 1200 kg travels along a straight horizontal road against a resistance to motion which is proportional to the speed of the van. The engine of the van is working at a constant rate of 40 kW . The van starts from rest at time \(t = 0\). At time \(t\) seconds, the speed of the van is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When the speed of the van is \(40 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the acceleration of the van is \(0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).

1. Show that $$75 v \frac { \mathrm {~d} v } { \mathrm {~d} t } = 2500 - v ^ { 2 }$$
2. Find \(v\) in terms of \(t\).

3. A cyclist and her bicycle have a combined mass of 75 kg . The cyclist travels along a straight horizontal road. The cyclist produces a constant driving force of magnitude 150 N . At time \(t\) seconds, the speed of the cyclist is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), where \(v < \sqrt { 50 }\). As the cyclist moves, the total resistance to motion of the cyclist and her bicycle has magnitude \(3 v ^ { 2 }\) newtons. The cyclist starts from rest. At time \(t\) seconds, she has travelled a distance \(x\) metres from her starting point. Find

1. \(v\) in terms of \(x\),
2. \(t\) in terms of \(v\).

4. A particle \(P\) of mass 0.5 kg moves in a horizontal straight line. At time \(t\) seconds \(( t \geqslant 0 )\), the displacement of \(P\) from a fixed point \(O\) of the line is \(x\) metres, the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(P\) is moving in the direction of \(x\) increasing. A force of magnitude \(k x\) newtons acts on \(P\) in the direction \(P O\). The motion of \(P\) is also subject to a resistance of magnitude \(\lambda v\) newtons. Given that $$x = ( 1.5 + 10 t ) \mathrm { e } ^ { - 4 t }$$ find

1. the value of \(k\) and the value of \(\lambda\),
2. the distance from \(P\) to \(O\) when \(P\) is instantaneously at rest.

6. A particle of mass \(m\) is projected vertically upwards in a resisting medium. As the particle moves upwards, the speed \(v\) of the particle is given by $$v ^ { 2 } = k g \left( 5 \mathrm { e } ^ { - \frac { x } { 2 k } } - 4 \right)$$ where \(x\) is the distance of the particle above the point of projection and \(k\) is a positive constant.

1. Show that the magnitude of the resistance to the motion of the particle is \(\frac { m v ^ { 2 } } { 4 k }\).
   (4)
2. Find, in terms of \(k\), the greatest height reached by the particle above the point of projection.
3. Show that the time taken by the particle to reach its greatest height above the point of projection is \(\sqrt { \frac { 4 k } { g } } \arctan \left( \frac { 1 } { 2 } \right)\)

4. A body falls vertically from rest and is subject to air resistance of a magnitude which is proportional to its speed. Given that its terminal speed is \(100 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), find the time it takes for the body to attain a speed of \(60 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
(10 marks)

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 \(t\).

3 An aeroplane is taking off from a runway. It starts from rest. The resultant force in the direction of motion has power, \(P\) watts, modelled by $$P = 0.0004 m \left( 10000 v + v ^ { 3 } \right) ,$$ where \(m \mathrm {~kg}\) is the mass of the aeroplane and \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity at time \(t\) seconds. The displacement of the aeroplane from its starting point is \(x \mathrm {~m}\). To take off successfully the aeroplane must reach a speed of \(80 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) before it has travelled 900 m .

1. Formulate and solve a differential equation for \(v\) in terms of \(x\). Hence show that the aeroplane takes off successfully.
2. Formulate a differential equation for \(v\) in terms of \(t\). Solve the differential equation to show that \(v = 100 \tan ( 0.04 t )\). What feature of this result casts doubt on the validity of the model?
3. In fact the model is only valid for \(0 \leqslant t \leqslant 11\), after which the power remains constant at the value attained at \(t = 11\). Will the aeroplane take off successfully?

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.

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 \(1 \leqslant t \leqslant 2\).

1 A rocket in deep space starts from rest and moves in a straight line. The initial mass of the rocket is \(m _ { 0 }\) and the propulsion system ejects matter at a constant mass rate \(k\) with constant speed \(u\) relative to the rocket. At time \(t\) the speed of the rocket is \(v\).

1. Show that while mass is being ejected from the rocket, \(\left( m _ { 0 } - k t \right) \frac { \mathrm { d } v } { \mathrm {~d} t } = u k\).
2. Hence find an expression for \(v\) in terms of \(t\).
3. Find the speed of the rocket when its mass is \(\frac { 1 } { 3 } m _ { 0 }\).

2 A car of mass \(m \mathrm {~kg}\) starts from rest at a point O and moves along a straight horizontal road. The resultant force in the direction of motion has power \(P\) watts, given by \(P = m \left( k ^ { 2 } - v ^ { 2 } \right)\), where \(v \mathrm {~ms} ^ { - 1 }\) is the velocity of the car and \(k\) is a positive constant. The displacement from O in the direction of motion is \(x \mathrm {~m}\).

1. Show that \(\left( \frac { k ^ { 2 } } { k ^ { 2 } - v ^ { 2 } } - 1 \right) \frac { \mathrm { d } v } { \mathrm {~d} x } = 1\), and hence find \(x\) in terms of \(v\) and \(k\).
2. How far does the car travel before reaching \(90 \%\) of its terminal velocity?

1 A raindrop increases in mass as it falls vertically from rest through a stationary cloud. At time \(t \mathrm {~s}\) the velocity of the raindrop is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and its mass is \(m \mathrm {~kg}\). The rate at which the mass increases is modelled as \(\frac { m g } { 2 ( v + 1 ) } \mathrm { kg } \mathrm { s } ^ { - 1 }\). Resistances to motion are neglected.

1. Write down the equation of motion of the raindrop. Hence show that $$\left( 1 - \frac { 1 } { v + 2 } \right) \frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 1 } { 2 } g .$$
2. Solve this differential equation to find an expression for \(t\) in terms of \(v\). Calculate the time it takes for the velocity of the raindrop to reach \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
3. Describe, with reasons, what happens to the acceleration of the raindrop for large values of \(t\).

4 A parachutist of mass 90 kg falls vertically from rest. The forces acting on her are her weight and resistance to motion \(R \mathrm {~N}\). At time \(t \mathrm {~s}\) the velocity of the parachutist is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the distance she has fallen is \(x \mathrm {~m}\). While the parachutist is in free-fall (i.e. before the parachute is opened), the resistance is modelled as \(R = k v ^ { 2 }\), where \(k\) is a constant. The terminal velocity of the parachutist in free-fall is \(60 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that \(k = \frac { g } { 40 }\).
2. Show that \(v ^ { 2 } = 3600 \left( 1 - \mathrm { e } ^ { - \frac { g x } { 1800 } } \right)\). When she has fallen 1800 m , she opens her parachute.
3. Calculate, by integration, the work done against the resistance before she opens her parachute. Verify that this is equal to the loss in mechanical energy of the parachutist. As the parachute opens, the resistance instantly changes and is now modelled as \(R = 90 v\).
4. Calculate her velocity just before opening the parachute, correct to four decimal places.
5. Formulate and solve a differential equation to calculate the time it takes after opening the parachute to reduce her velocity to \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1 At time \(t\) a rocket has mass \(m\) and is moving vertically upwards with velocity \(v\). The propulsion system ejects matter at a constant speed \(u\) relative to the rocket. The only additional force acting on the rocket is its weight.

1. Derive the differential equation \(m \frac { \mathrm {~d} v } { \mathrm {~d} t } + u \frac { \mathrm {~d} m } { \mathrm {~d} t } = - m g\). The rocket has initial mass \(m _ { 0 }\) of which \(75 \%\) is fuel. It is launched from rest. Matter is ejected at a constant mass rate \(k\).
2. Assuming that the acceleration due to gravity is constant, find the speed of the rocket at the instant when all the fuel is burnt.