Air resistance kv² - falling from rest or projected downward

4 A small object of mass 0.4 kg is released from rest at a point 8 m above the ground. The object descends vertically and when its downwards displacement from its initial position is \(x \mathrm {~m}\) the object has velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). While the object is moving, a force of magnitude \(0.2 v ^ { 2 } \mathrm {~N}\) opposes the motion.

1. Show that \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 10 - 0.5 v ^ { 2 }\).
2. Express \(v\) in terms of \(x\).
3. Find the increase in the value of \(v\) during the final 4 m of the descent of the object.

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 \(g\).

5 The pilot of a hot air balloon keeps it at a fixed altitude by dropping sand from the balloon. Each grain of sand has mass \(m \mathrm {~kg}\) and is released from rest. When a grain has fallen a distance \(x \mathrm {~m}\), it has speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Each grain falls vertically and the only forces acting on it are its weight and air resistance of magnitude \(m k v ^ { 2 } \mathrm {~N}\), where \(k\) is a positive constant.

1. Show that \(\left( \frac { v } { g - k v ^ { 2 } } \right) \frac { \mathrm { d } v } { \mathrm {~d} x } = 1\).
2. Find \(v ^ { 2 }\) in terms of \(k , g\) and \(x\). Hence show that, as \(x\) becomes large, the limiting value of \(v\) is \(\sqrt { \frac { g } { k } }\).
3. Given that the altitude of the balloon is 300 m and that each grain strikes the ground at \(90 \%\) of its limiting velocity, find \(k\).

7 A particle \(P\) of mass 0.2 kg is released from rest at a point \(O\) and falls vertically. Air resistance of magnitude \(\frac { v ^ { 2 } } { 2000 } \mathrm {~N}\) acts upwards on \(P\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity of \(P\) when it has fallen a distance of \(x \mathrm {~m}\).

1. Show that \(\left( \frac { 400 v } { 3920 - v ^ { 2 } } \right) \frac { \mathrm { d } v } { \mathrm {~d} x } = 1\).
2. Find \(v ^ { 2 }\) in terms of \(x\) and hence show that \(v ^ { 2 } < 3920\) for all values of \(x\).
3. Find the work done against the air resistance while \(P\) is falling, from \(O\), to the point where its downward acceleration is \(5.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). }{www.ocr.org.uk}) after the live examination series.
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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 .

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 \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

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

1. Find an expression for \(v\) in terms of \(t\). [6]
2. Deduce what happens to \(v\) for large values of \(t\). [1]

A particle \(P\) of mass \(m\) kg is held at rest at a point \(O\) and released so that it moves vertically under gravity against a resistive force of magnitude \(0.1mv^2\) N, where \(v\) m s\(^{-1}\) is the velocity of \(P\) at time \(t\) s.

1. Find an expression for \(v\) in terms of \(t\). [6]
2. Find an expression for \(v^2\) in terms of \(x\). [5]

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

A particle \(P\) of mass \(mk\) falls from rest due to gravity. There is a resistance force of magnitude \(mkv^2\) N, where \(v\) ms\(^{-1}\) is the speed of \(P\) after it has fallen a distance \(x\) m and \(k\) is a positive constant.

1. By using \(v \frac{dv}{dx} = \frac{dv}{dt}\) and appropriate differential equation, show that $$v^2 = \frac{g}{k}(1 - e^{-2kx}).$$ [7] It is given that \(k = 0.01\). The speed of \(P\) when \(x = 0.2\) comes to approximately \(v\) ms\(^{-1}\).
2. 1. Find \(V\) correct to 2 decimal places. [1]
   2. Hence find how far \(P\) has fallen when its speed is \(\frac{1}{2}V\) ms\(^{-1}\). [2]

A small sphere \(S\), of mass \(m\) kg is released from rest at the surface of a liquid in a right circular cylinder whose axis is vertical. When \(S\) is moving downwards with speed \(v\) ms\(^{-1}\), the viscous resistive force acting upwards on it has magnitude \(v^2\) N.

1. Write down a differential equation for the motion of \(S\), clearly defining any symbol(s) that you introduce. [4 marks]
2. Find, in terms of \(m\), the distance \(S\) has fallen when its speed is \(\sqrt{\frac{mg}{2}}\) ms\(^{-1}\). [9 marks]

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]

A particle P of mass 5 kg is released from rest at a point O and falls vertically. A resistance of magnitude \(0.05v^2\) N acts vertically upwards on P, where \(v \text{ m s}^{-1}\) is the velocity of P when it has fallen a distance \(x\) m.

1. Show that \(\left(\frac{100v}{980-v^2}\right)\frac{dv}{dx} = 1\). [2]
2. Verify that \(v^2 = 980(1-e^{-0.02x})\). [4]
3. Determine the work done against the resistance while P is falling from O to the point where P's acceleration is \(8.36 \text{ m s}^{-2}\). [5]