Variable mass problems (mass increasing)

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 \(t\).

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 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 \(t\).

1 A raindrop of mass \(m\) falls vertically from rest under gravity. Initially the mass of the raindrop is \(m _ { 0 }\). As it falls it loses mass by evaporation at a rate \(\lambda m\), where \(\lambda\) is a constant. Its motion is modelled by assuming that the evaporation produces no resultant force on the raindrop. The velocity of the raindrop is \(v\) at time \(t\). The forces on the raindrop are its weight and a resistance force of magnitude \(k m v\), where \(k\) is a constant.

1. Find \(m\) in terms of \(m _ { 0 } , \lambda\) and \(t\).
2. Write down the equation of motion of the raindrop. Solve this differential equation and hence show that \(v = \frac { g } { \lambda - k } \left( \mathrm { e } ^ { ( \lambda - k ) t } - 1 \right)\).
3. Find the velocity of the raindrop when it has lost half of its initial mass.

4 A raindrop falls from rest through a stationary cloud. The raindrop has mass \(m\) and speed \(v\) when it has fallen a distance \(x\). You may assume that resistances to motion are negligible.

1. Derive the equation of motion $$m v \frac { \mathrm {~d} v } { \mathrm {~d} x } + v ^ { 2 } \frac { \mathrm {~d} m } { \mathrm {~d} x } = m g .$$ Initially the mass of the raindrop is \(m _ { 0 }\). Two different models for the mass of the raindrop are suggested.
   In the first model \(m = m _ { 0 } \mathrm { e } ^ { k _ { 1 } x }\), where \(k _ { 1 }\) is a positive constant.
2. Show that $$v ^ { 2 } = \frac { g } { k _ { 1 } } \left( 1 - \mathrm { e } ^ { - 2 k _ { 1 } x } \right) ,$$ and hence state, in terms of \(g\) and \(k _ { 1 }\), the terminal velocity of the raindrop according to this first model. In the second model \(m = m _ { 0 } \left( 1 + k _ { 2 } x \right)\), where \(k _ { 2 }\) is a positive constant.
3. By considering the expression obtained from differentiating \(v ^ { 2 } \left( 1 + k _ { 2 } x \right) ^ { 2 }\) with respect to \(x\), show that, for the second model, the equation of motion in part (i) may be written as $$\frac { d } { d x } \left[ v ^ { 2 } \left( 1 + k _ { 2 } x \right) ^ { 2 } \right] = 2 g \left( 1 + k _ { 2 } x \right) ^ { 2 } .$$ Hence find an expression for \(v ^ { 2 }\) in terms of \(g , k _ { 2 }\) and \(x\).
4. Suppose that the models give the same value for the speed of the raindrop at the instant when it has doubled its initial mass. Find the exact value of \(\frac { k _ { 1 } } { k _ { 2 } }\), giving your answer in the form \(\frac { a } { b }\) where \(a\) and \(b\) are integers. are integers. \section*{END OF QUESTION PAPER}

7. A motor boat of mass \(M\) is moving in a straight line, with its engine switched off, across a stretch of still water. The boat is moving with speed \(U\) when, at time \(t = 0\), it develops a leak. The water comes in at a constant rate so that at time \(t\), the mass of water in the boat is \(\lambda t\). At time \(t\) the speed of the boat is \(v\) and it experiences a total resistance to motion of magnitude \(2 \lambda v\).

1. Show that \(( M + \lambda t ) \frac { \mathrm { d } v } { \mathrm {~d} t } + 3 \lambda v = 0\).
   (6)
2. Show that the time taken for the speed of the boat to reduce to \(\frac { 1 } { 2 } U\) is \(\frac { M } { \lambda } \left( 2 ^ { \frac { 1 } { 3 } } - 1 \right)\).
   (6) The boat sinks when the mass of water inside the boat is \(M\).
3. Show that the boat does not sink before the speed of the boat is \(\frac { 1 } { 2 } U\).

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 $$\frac { \mathrm { d } v } { \mathrm {~d} t } + v \left( k + \frac { c } { m _ { 0 } + c t } \right) = g$$

7. A raindrop absorbs water as it falls vertically under gravity through a cloud. In a model of the motion the cloud is assumed to consist of stationary water particles. At time \(t\), the mass of the raindrop is \(m\) and the speed of the raindrop is \(v\). At time \(t = 0\), the raindrop is at rest. The rate of increase of the mass of the raindrop with respect to time is modelled as being \(m k v\), where \(k\) is a positive constant.

1. Ignoring air resistance, show from first principles, that $$\frac { \mathrm { d } v } { \mathrm {~d} t } = g - k v ^ { 2 }$$
2. Find the time taken for the raindrop to reach a speed of \(\frac { 1 } { 2 } \sqrt { } \left( \frac { g } { k } \right)\)

4. A particle \(P\), whose initial mass is \(m _ { 0 }\), is projected vertically upwards from the ground at time \(t = 0\) with speed \(\frac { g } { k }\), where \(k\) is a constant. As the particle moves upwards it gains mass by picking up small droplets of moisture from the atmosphere. The droplets are at rest before they are picked up. At time \(t\) the speed of \(P\) is \(v\) and its mass has increased to \(m _ { 0 } \mathrm { e } ^ { k t }\). Assuming that, during the motion, the acceleration due to gravity is constant,

1. show that, while \(P\) is moving upwards, $$k v + \frac { \mathrm { d } v } { \mathrm {~d} t } = - g$$
2. find, in terms of \(m _ { 0 }\), the mass of \(P\) when it reaches its greatest height above the ground.
   (6)

6. A small object \(P\), of mass \(m _ { 0 }\), is projected vertically upwards from the ground with speed \(U\). As \(P\) moves upwards it picks up droplets of moisture from the atmosphere. The droplets are at rest immediately before they are picked up. In a model of the motion, \(P\) is modelled as a particle, air resistance is assumed to be negligible and the acceleration due to gravity is assumed to have the constant value of \(g\). When \(P\) is at a height \(x\) above the ground, the combined mass of \(P\) and the moisture is \(m _ { 0 } ( 1 + k x )\), where \(k\) is a constant, and the speed of \(P\) is \(v\).

1. Show that, while \(P\) is moving upwards $$\frac { \mathrm { d } } { \mathrm {~d} x } \left( v ^ { 2 } \right) + \frac { 2 k v ^ { 2 } } { ( 1 + k x ) } = - 2 g$$ The general solution of this differential equation is given by \(v ^ { 2 } = \frac { A } { ( 1 + k x ) ^ { 2 } } - \frac { 2 g } { 3 k } ( 1 + k x )\),
   where \(A\) is an arbitrary constant. Given that \(U = \sqrt { 2 g h }\) and \(k = \frac { 7 } { 3 h }\)
2. find, in terms of \(h\), the height of \(P\) above the ground when \(P\) first comes to rest.