Mathematics Extension 2

A particle of unit mass is thrown vertically downwards with an initial velocity of $v_0$. It experiences a resistive force of magnitude $kv^2$, where $v$ is its velocity. Taking downwards as the positive direction, the equation of motion of the particle is given by

Let $V$ be the terminal velocity of the particle.

Show that

A particle of unit mass is thrown vertically downwards with an initial velocity of $v_0$. It experiences a resistive force of magnitude $kv^2$, where $v$ is its velocity. Taking downwards as the positive direction, the equation of motion of the particle is given by

Let $V$ be the terminal velocity of the particle.

Explain why

A particle of unit mass is moving in horizontal motion, subject to a resistance force of $v^2+v^3$, where $v$ is the object’s velocity. The particle has initial velocity $v_0$, where $v_0>0$.

Show that if the particle starts at the origin, then

satisfies the equation of motion.

A particle of unit mass is moving in horizontal motion, subject to a resistance force of $v^2+v^3$, where $v$ is the object’s velocity. The particle has initial velocity $v_0$, where $v_0>0$.

Show that the time $T$ taken to travel the distance $s$ is

A particle of unit mass is moving in horizontal motion, subject to a resistance force of $v^2+v^3$, where $v$ is the object’s velocity. The particle has initial velocity $v_0$, where $v_0>0$.

Find the distance $s$ travelled by the particle when its velocity is

A particle is moving vertically in a resistive medium under the influence of gravity. The resistive force is proportional to the velocity of the particle.

The initial speed of the particle is NOT zero.

Which of the following statements about the motion of the particle is always true?

A. If the particle is initially moving downwards, then its speed will increase.
B. If the particle is initially moving downwards, then its speed will decrease.
C. If the particle is initially moving upwards, then its speed will eventually approach a terminal speed.
D. If the particle is initially moving upwards, then its speed will not eventually approach a terminal speed.

A particle with mass $1\,\mathrm{kg}$ is moving along the $x$-axis. Initially, the particle is at the origin and has speed $u\,\mathrm{m\,s^{-1}}$ to the right. The particle experiences a resistive force of magnitude $v+3v^2$ newtons, where $v\,\mathrm{m\,s^{-1}}$ is the speed of the particle after $t$ seconds. The particle is never at rest.

Show that
$\frac{dv}{dx}=-(1+3v).$

Hence, or otherwise, find $x$ as a function of $v$.

A $20\,\mathrm{kg}$ projectile is launched upwards from the ground at $500\,\mathrm{m/s}$. It experiences a gravitational force of $200\,\mathrm N$ and air resistance, opposite to its motion, of $\frac{v^2}{18}\,\mathrm N$. Letting the ground be $x=0$, show that the acceleration is
$\ddot{x}=-10-\frac{v^2}{360}.$

Show that
$x=180\ln\left(\frac{253600}{3600+v^2}\right)$
and find the maximum height the projectile reaches.

Show that the time taken for the projectile to reach velocity $v\,\mathrm{m/s}$ is
$t=6\tan^{-1}\left(\frac{25}{3}\right)-6\tan^{-1}\left(\frac{v}{60}\right),$
and find the time taken for the projectile to reach its maximum height.

A projectile of mass $M\,\mathrm{kg}$ is launched vertically upwards from a horizontal plane with initial speed $v_0\,\mathrm{m\,s^{-1}}$, where $v_0<100\,\mathrm{m\,s^{-1}}$. The projectile experiences a resistive force of magnitude $0.1Mv$ newtons, where $v\,\mathrm{m\,s^{-1}}$ is the speed of the projectile. The acceleration due to gravity is $10\,\mathrm{m\,s^{-2}}$.

The projectile lands on the horizontal plane $7$ seconds after launch.

Find the value of $v_0$, correct to $1$ decimal place.

For the resisted vertical motion described in part (i), find an expression for the maximum height reached by the object, in terms of $k$, $g$ and $u$.

An object of mass $1\,\mathrm{kg}$ is projected vertically upwards with initial velocity $u\,\mathrm{m\,s^{-1}}$. It experiences air resistance of magnitude $kv^2$ newtons, where $v$ is velocity in $\mathrm{m\,s^{-1}}$ and $k>0$. The height above its starting point is $x$ metres, time is $t$ seconds, and acceleration due to gravity is $g\,\mathrm{m\,s^{-2}}$. Show that the time to reach maximum height is
$\frac1{\sqrt{gk}}\arctan\left(u\sqrt{\frac{k}{g}}\right)\text{ seconds}.$

A particle starts from rest and falls through a resisting medium so that its acceleration, in $\mathrm{m/s^2}$, is modelled by
$a=10\bigl(1-(kv)^2\bigr),$
where $v$ is the velocity of the particle in $\mathrm{m/s}$ and $k=0.01$. Find the velocity of the particle after $5$ seconds.

Given that $v<\dfrac{gm}{k}$, show that when
$t=\frac{3m}{k}\ln2,$
the velocity of the larger mass is
$\frac{gm}{2k}.$

Two masses, $2m$ kg and $4m$ kg, are attached by a light string. The string is placed over a smooth pulley as shown. The two masses are at rest before being released and $v$ is the velocity of the larger mass at time $t$ seconds after they are released. The force due to air resistance on each mass has magnitude $kv$, where $k$ is a positive constant.

(i) Show that
$\frac{dv}{dt}=\frac{gm-kv}{3m}.$