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Mechanics, EXT2 M1 2025 HSC 12e

A particle of mass \(m\) kg moves along a horizontal line with an initial velocity of \(V_0 \ \text{ms}^{-1}\).

The motion of the particle is resisted by a constant force of \(m k\) newtons and a variable force of \(m v^2\) newtons, where \(k\) is a positive constant and \(v \ \text{ms}^{-1}\) is the velocity of the particle at \(t\) seconds.

Show that the distance travelled when the particle is brought to rest is  \(\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k}\right)\) metres.   (3 marks)

Show Answers Only
\(ma\) \(-m k-m v^2\)
\(a\) \(=-k-v^2\)
\(v \cdot \dfrac{d v}{d x}\) \(=-\left(k+\nu^2\right)\)
\(\dfrac{d x}{d v}\) \(=-\dfrac{v}{k+v^2}\)
\(x\) \(=-\displaystyle\frac{1}{2} \int \frac{2 v}{k+v^2}\, d v=-\frac{1}{2} \ln \left(k+v^2\right)+c\)

\(\text{At} \ \ x=0, v=v_0 \ \ \Rightarrow\ \ c=\dfrac{1}{2} \ln \left(k+V_0^2\right)\)

\(x=\dfrac{1}{2} \ln \left(k+V_0^2\right)-\dfrac{1}{2} \ln \left(k+v^2\right)=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k+v^2}\right)\)

\(\text{Find \(x\) when \(\ v=0\ \) (distance travelled):}\)

\(x=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k}\right)\ \text{metres}\)

Show Worked Solution
\(ma\) \(-m k-m v^2\)
\(a\) \(=-k-v^2\)
\(v \cdot \dfrac{d v}{d x}\) \(=-\left(k+\nu^2\right)\)
\(\dfrac{d x}{d v}\) \(=-\dfrac{v}{k+v^2}\)
\(x\) \(=-\displaystyle\frac{1}{2} \int \frac{2 v}{k+v^2}\, d v=-\frac{1}{2} \ln \left(k+v^2\right)+c\)

\(\text{At} \ \ x=0, v=v_0 \ \ \Rightarrow\ \ c=\dfrac{1}{2} \ln \left(k+V_0^2\right)\)

\(x=\dfrac{1}{2} \ln \left(k+V_0^2\right)-\dfrac{1}{2} \ln \left(k+v^2\right)=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k+v^2}\right)\)

\(\text{Find \(x\) when \(\ v=0\ \) (distance travelled):}\)

\(x=\dfrac{1}{2} \ln \left(\dfrac{k+V_0^2}{k}\right)\ \text{metres}\)

Mechanics, EXT2 M1 2024 HSC 13c

A particle of unit mass moves horizontally in a straight line. It experiences a resistive force proportional to \(v^2\), where \(v\) m s\(^{-1}\) is the speed of the particle, so that the acceleration is given by  \(-k v^2\).

Initially the particle is at the origin and has a velocity of 40 m s\(^{-1}\) to the right. After the particle has moved 15 m to the right, its velocity is 10 m s\(^{-1}\) (to the right).

  1. Show that  \(v=40 e^{-k x}\).   (3 marks)

  2. Show that  \(k=\dfrac{\ln 4}{15}\).   (1 mark)

  3. At what time will the particle's velocity be 30 m s\(^{-1}\) to the right?   (3 marks)

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i.    \(\ddot{x}=v \cdot \dfrac{d v}{d x}=-k v^2\)

  \(\dfrac{d v}{d x}\) \(=-k v\)
  \(\dfrac{d x}{d v}\) \(=-\dfrac{1}{k v}\)
  \(\displaystyle\int \frac{1}{v}\, d v\) \(=\displaystyle -\int k\, d x\)
  \(\ln \abs{v}\) \(=-k x+c\)
  \(v\) \(=e^{-k x+c}\)
    \(=A e^{-k x}\ \ (\text{where } A=e^c)\)

 
\(\text {When } x=0, v=40:\)

\(40=A e^{\circ} \ \Rightarrow \ A=40\)

\(\therefore V=40 \, e^{-k x}\)
 

ii.   \(\text {Show}\ \ k=\dfrac{\ln 4}{15}\)

\(\text {When } x=15, v=10:\)

  \(10\) \(=40 e^{-15 k}\)
  \(e^{-15 k}\) \(=\dfrac{1}{4}\)
  \(-15 k\) \(=\ln \left(\dfrac{1}{4}\right)\)
  \(15 k\) \(=\ln 4\)
  \(k\) \(=\dfrac{\ln 4}{15}\)

iii.  \(\dfrac{1}{8\,\ln 4}\  \text{seconds}\)

Show Worked Solution

i.    \(\ddot{x}=v \cdot \dfrac{d v}{d x}=-k v^2\)

  \(\dfrac{d v}{d x}\) \(=-k v\)
  \(\dfrac{d x}{d v}\) \(=-\dfrac{1}{k v}\)
  \(\displaystyle\int \frac{1}{v}\, d v\) \(=\displaystyle -\int k\, d x\)
  \(\ln \abs{v}\) \(=-k x+c\)
  \(v\) \(=e^{-k x+c}\)
    \(=A e^{-k x}\ \ (\text{where } A=e^c)\)

 
\(\text {When } x=0, v=40:\)

\(40=A e^{\circ} \ \Rightarrow \ A=40\)

\(\therefore V=40 \, e^{-k x}\)
 

ii.   \(\text {Show}\ \ k=\dfrac{\ln 4}{15}\)

\(\text {When } x=15, v=10:\)

  \(10\) \(=40 e^{-15 k}\)
  \(e^{-15 k}\) \(=\dfrac{1}{4}\)
  \(-15 k\) \(=\ln \left(\dfrac{1}{4}\right)\)
  \(15 k\) \(=\ln 4\)
  \(k\) \(=\dfrac{\ln 4}{15}\)

 

iii.   \(\text {Find}\ t\ \text {when}\ \ v=30:\)

  \(\dfrac{d v}{d t}\) \(=-k v^2\)
  \(\dfrac{d t}{d v}\) \(=-\dfrac{1}{k v^2}\)
  \(t\) \(=\displaystyle -\int \dfrac{1}{k v^2}\, d v=\dfrac{1}{k v}+c\)

 
\(\text {When}\ \ t=0, v=40:\)

\(0=\dfrac{1}{40 k}+c \ \Rightarrow \ c=-\dfrac{1}{40 k}\)
 

\(\text{Find \(t\) when  \(v=30\):}\)

  \(t\) \(=\dfrac{1}{30k}-\dfrac{1}{40k}\)
    \(=\dfrac{1}{120k}\)
    \(=\dfrac{15}{120\, \ln 4}\)
    \(=\dfrac{1}{8\,\ln 4}\  \text{seconds}\)

Mechanics, EXT2 M1 2023 HSC 14c

A projectile of mass \(M\) kg is launched vertically upwards from the origin with an initial speed \(v_0\) m s\(^{-1}\). The acceleration due to gravity is \( {g}\) ms\(^{-2}\).

The projectile experiences a resistive force of magnitude \(kMv^2\) newtons, where \(k\) is a positive constant and \(v\) is the speed of the projectile at time \(t\) seconds.

  1. The maximum height reached by the particle is \(H\) metres.
  2. Show that  \(H=\dfrac{1}{2 k} \ln \left(\dfrac{k v_0{ }^2+g}{g}\right)\).  (3 marks)

  3. When the projectile lands on the ground, its speed is \(v_1 \text{m} \ \text{s}^{-1}\), where \(v_1\) is less than the magnitude of the terminal velocity.
  4. Show that  \(g\left(v_0^2-v_1^2\right)=k v_0^2 v_1^2\).  (3 marks)

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  1. \(\text{See Worked Solution}\)
  2. \(\text{See Worked Solution}\)
Show Worked Solution

i.    \(\text{Taking up as positive:}\)

\(M\ddot x\) \(=-Mg-kMv^2\)  
\(\ddot x\) \(=-g-kv^2\)  
\(v \cdot \dfrac{dv}{dx}\) \(=-(g+kv^2) \)  
\(\dfrac{dv}{dx}\) \(=-\dfrac{g+kv^2}{v} \)  
\(\dfrac{dx}{dv}\) \(=-\dfrac{v}{g+kv^2} \)  
\(x\) \(=-\dfrac{1}{2k} \displaystyle \int \dfrac{2kv}{g+kv^2}\, dv \)  
  \(=-\dfrac{1}{2k} \ln |g+kv^2|+c \)  

 
\(\text{When}\ \ x=o, \ v=v_0: \)

\(c=\dfrac{1}{2k} \ln |g+kv_0^2| \)

\(x\) \(=\dfrac{1}{2k} \ln |g+kv_0^2|-\dfrac{1}{2k} \ln |g+kv^2| \)  
  \(=\dfrac{1}{2k} \ln \Bigg{|} \dfrac{g+kv_0^2}{g+kv^2} \Bigg{|} \)  

 
\(\text{When}\ \ v=0, x=H: \)

\(H=\dfrac{1}{2k} \ln \Bigg{(} \dfrac{g+kv_0^2}{g} \Bigg{)},\ \ \ \ (k>0) \)
  

ii.   \(\text{When projectile travels downward:} \)

\(M \ddot x\) \(=Mg-kMv^2\)  
\(\ddot x\) \(=g-kv^2\)  
\(v \cdot \dfrac{dv}{dx}\) \(=g-kv^2\)  
\(\dfrac{dx}{dv}\) \(=\dfrac{v}{g-kv^2}\)  
\(x\) \(=-\dfrac{1}{2k} \displaystyle \int \dfrac{-2kv}{g-kv^2}\,dv \)  
  \(=-\dfrac{1}{2k}  \ln|g-kv^2|+c \)  

 
\(\text{When}\ \ x=0, \ v=0: \)

\(c=\dfrac{1}{2k} \ln g \)

\(x=\dfrac{1}{2k} \ln \Bigg{|} \dfrac{g}{g-kv^2} \Bigg{|} \)
 

\(\text{When}\ \ x=H, \ v=v_1: \)

\(\dfrac{1}{2k} \ln \Bigg{(} \dfrac{g+kv_0^2}{g} \Bigg{)}\) \(=\dfrac{1}{2k} \ln \Bigg{|} \dfrac{g}{g-kv_1^2} \Bigg{|} \)  
\(\dfrac{g+kv_0^2}{g} \) \(=\dfrac{g}{g-kv_1^2} \)  
\(g^2\) \(=(g+kv_0^2)(g-kv_1^2) \)  
\(g^2\) \(=g^2-gkv_1^2+gkv_0^2-k^2v_0^2v_1^2 \)  
\(gkv_0^2-gkv_1^2 \) \(=k^2v_0^2v_1^2 \)  
\(gk(v_0^2-v_1^2) \) \(=k^2v_0^2 v_1^2 \)  
\(g(v_0^2-v_1^2) \) \(=kv_0^2v_1^2 \)  
♦♦ Mean mark (ii) 31%.

Mechanics, EXT2 M1 2022 HSC 12c

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

  1. Show that  `(dv)/(dx)=-(1+3v)`  (1 mark)

  2. Hence, or otherwise, find `x` as a function of `v`.  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `x=1/3 ln((1+3u)/(1+3v))`
Show Worked Solution

i.   `F=m ddotx=ddotx\ \ (m=1)`

`ddotx` `=-(v+3v^2)`  
`v*(dv)/(dx)` `=-(v+3v^2)`  
`:. (dv)/(dx)` `=-(1+3v)\ \ text{… as required}`  

  

ii.   `(dx)/(dv)` `=- 1/(1+3v)`
  `x` `=-int1/(1+3v)\ dv`
    `=-1/3 ln(1+3v)+c`

 
`text{When}\ \ t=0, \ v=u\ \ =>\ \ c=1/3 ln(1+3u)`

`:.x` `=1/3 ln(1+3u)-1/3 ln(1+3v)`  
  `=1/3 ln((1+3u)/(1+3v))`  

Mechanics, EXT2 M1 2022 HSC 8 MC

As a projectile of mass `m` kilograms travels through air, it experiences a frictional force. The magnitude of this force is proportional to the square of the speed `v` of the projectile. The constant of proportionality is the positive number `k`. The position of the particle at time `t` is denoted by `([x],[y])`. The acceleration due to gravity is `g \ text{m s}^(-2)`.

Based on Newton's laws of motion, which equation models the motion of this projectile?

  1. `([0],[-mg])+kv([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
  2. `([0],[-mg])-kv([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
  3. `([0],[-mg])+kv^(2)([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
  4. `([0],[-mg])-kv^(2)([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
Show Answers Only

`B`

Show Worked Solution

`text{Friction}\ (F)\ text{works against velocity}`

`:.\ F prop v^2\ \ =>\ \ F=-kv^2\ \ (k>0)`

`text{→ Eliminate A and C}`
 


♦♦♦ Mean mark 26%.

`text{S}text{ince}\ \ v=abs(((dotx),(doty))):`

`- abs(kv((dotx),(doty))) = -kv abs(((dotx),(doty)))=-kv^2`

`=>B`

Mechanics, EXT2 M1 2021 HSC 14b

An object of mass 5 kg is on a slope that is inclined at an angle of 60° to the horizontal. The acceleration due to gravity is `g\ text(m s)^(-2)`  and the velocity of the object down the slope is `v\ text(m s)^(-1)`.

As well as the force due to gravity, the object is acted on by two forces, one of magnitude `2v` newtons and one of magnitude `2v^2` newtons, both acting up the slope.

  1. Show that the resultant force down the slope is  `(5sqrt3)/2 g - 2v - 2v^2`  newtons.  (2 marks)

  2. There is one value of `v` such that the object will slide down the slope at a constant speed.
  3. Find this value of `v` in `text(m s)^(-1)`, correct to 1 decimal place, given that  `g = 10`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `4.2\ \ text(ms)^-1`
Show Worked Solution

i.   

`F_s` `=\ text(force down slope)`
  `= 5g cos30 – 2v^2 – 2v`
  `= 5g · sqrt3/2 – 2v^2 – 2v`
  `= (5sqrt3)/2 g – 2v^2 – 2v`

 

ii.   `text(Constant speed occurs if)\ \ a = 0`

`F_s = ma = 0`

`2v^2 + 2v – (5sqrt3)/2 xx 10` `= 0`
`2v^2 + 2v – 25sqrt3` `= 0`

 

`v` `= (-2 + sqrt(2^2 + 4 · 2 · 25sqrt3))/(2 xx 2)`
  `= (-2 + sqrt(4 + 200sqrt3))/4`
  `= (-1 + sqrt(1 + 50 sqrt3))/2`
  `= 4.179…`
  `= 4.2\ \ text(ms)^-1\ text{(1 d.p.)}`

Mechanics, EXT2 M1 2021 HSC 15c

An object of mass 1 kg is projected vertically upwards with an initial velocity of `u` m/s. It experiences air resistance of magnitude `kv^2` newtons where `v` is the velocity of the object, in m/s, and `k` is a positive constant. The height of the object above its starting point is `x` metres. The time since projection is `t` seconds and acceleration due to gravity is `g` m/s².

  1. Show that the time for the object to reach its maximum height is  `1/sqrt{gk} arctan (u sqrt{k/g})`  seconds.  (3 marks)

  2. Find an expression for the maximum height reached by the object, in terms of `k`, `g`, and `u`.  (3 marks)

Show Answers Only
  1. `text{See Worked Solution}`
  2. `x_{text{max}} = {1}/{2k} ln ({g + ku^2}/{g})`
Show Worked Solution
i.    `{dv}/{dt}` `= – g – kv^2`
  `{dt}/{dv}` `= {-1}/{g + kv^2}`
    `= {-1}/{g} * {1}/{1 + k/g v^2}`
    `= {-1}/{g} * sqrt{g/k} * {sqrt{k/g}}/{1 + k/g v^2}`
    `= – 1/sqrt{gk} * {sqrt{k/g}}/{1 + k/g v^2}`
  `t` `= – 1/sqrt{gk} * int {sqrt{k/g}}/{1 + k/g v^2} dv`
    `= -1/sqrt{gk} tan ^-1 sqrt{k/g} v +c`

 
`text{When} \ t = 0 \ , \ v = u:`

`c = 1/sqrt{gk} tan^-1 sqrt{k/g} u`
  

`v = 0 \ text{at max height}`

`t = 1/sqrt{gk} tan^-1 (u sqrt{k/g})\ text{seconds}`
 

ii.   `text{Find max height:}`

Mean mark part (ii) 54%.
`v * {dv}/{dx}` `= -g – kv^2`
`{dv}/{dx}` `= {-g – kv^2}/{v}`
`{dx}/{dv}` `= {-v}/{g + kv^2}`
`x` `= -{1}/{2k} int {2kv}/{g + kv^2} dv`
  `= -{1}/{2k} ln (g + kv^2) + c`

 

`text{When} \ x = 0 \ , \ v = u:`

`c = {1}/{2k} ln (g + ku^2)`

`x` `= {1}/{2k} ln (g + ku^2) – {1}/{2k} ln (g + kv^2)`
  `= {1}/{2k} ln ({g + ku^2}/{g + kv^2})`
 
`text{Max height occurs when} \ v = 0:`

`:. \ x_{text{max}} = {1}/{2k} ln ({g + ku^2}/{g})`

Mechanics, EXT2 M1 2020 HSC 14b

A particle starts from rest and falls through a resisting medium so that its acceleration, in m/s2, is modelled by

`a = 10 (1 - (kv)^2)`,

where `v` is the velocity of the particle in m/s and  `k = 0.01`.

Find the velocity of the particle after 5 seconds.  (4 marks)

Show Answers Only

`46.21 \ text{ms}^-1`

Show Worked Solution
`a` `= 10 (1 – (kv)^2)`
`frac{dv}{dt}` `= 10 – 10 xx 0.01^2 xx v^2`
  `= 10 – 0.001 v^2`
`frac{dt}{dv}` `= frac{1}{10 – 0.001 \ v^2}`
  `= frac{1000}{10 \ 000 – v^2}`
`t` `= int frac{1000}{100^2 – v^2}\ dv`

 

`text{Using partial fractions}:`

`frac{1}{100^2 – v^2}` ` = frac{A}{100 + v} + frac{B}{100 – v}`
`1` `= A (100 – v) + B(100 + v)`

 
`text{If} \ \ v = 100 \ , \ 1 = 200 B \ => \ B = frac{1}{200}`

`text{If} \ \ v = -100 \ , \ 1 = 200 A \ => \ A = frac{1}{200}`
 

`t` `= 1000 int frac{1}{200 (100 + v)}\ dv + 1000 int frac{1}{200(100 -v)}\ dv`
  `= 5 int frac{1}{100 + v}\ dv +  5 int frac{1}{100 – v}\ dv`
  `= 5 ln \ | 100 + v | – 5 ln  \|100 – v | + c`
  `= 5 ln \ | frac{100 + v}{100 – v} | + c`

 
`text{When} \ \ t = 0 , \ v = 0`

`0 = 5 ln 1  + c  \ => \ c = 0`
 

`:. t = 5 ln \ | frac{100 + v}{100 – v} |`
  

`text{Find} \ \ v \ \ text{when} \ \ t = 5 :`

`5` `= 5 ln | frac{100 + v}{100 – v} |`
`1` `= ln | frac{100 + v}{100 – v} | `
`e` `= frac{100 + v}{100 – v}`
`100 e – ve` `= 100 + v`
`v + ve` `= 100 e – 100`
`v(1 + e)` `= 100 e – 100`
`:. v` `= frac{100 e – 100}{1 + e}`
  `= 46.21 \ text{ms}^-1 \ \ (text{2 d.p.})`

Mechanics, EXT2 M1 EQ-Bank 4

A torpedo with a mass of 80 kilograms has a propeller system that delivers a force of `F` on the torpedo, at maximum power. The water exerts a resistance on the torpedo proportional to the square of the torpedo's velocity `v`.

  1. Explain why  `(dv)/(dt) = 1/80 (F - kv^2)`
     
    where `k` is a positive constant.  (1 mark)

  2. If the torpedo increases its velocity from  `text(10 ms)\ ^(−1)`  to  `text(20 ms)\ ^(−1)`, show that the distance it travels in this time, `d`, is given by
     
         `d = 40/k log_e((F - 100k)/(F - 400k))`  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `R ∝ v^2`

`R = −kv^2\ \ (k\ text{is positive constant})`

`text(Newton’s 2nd Law:)`

`text(Net Force)= mddotx` `= F – R`
`80ddotx` `= F – kv^2`
`ddotx` `= 1/80 (F – kv^2)`
`(dv)/(dt)` `= 1/80 (F – kv^2)`

 

ii.    `v · (dv)/(dx)` `= 1/80 (F – kv^2)`
  `(dv)/(dx)` `= (F – kv^2)/(80v)`
  `(dx)/(dv)` `= (80v)/(F – kv^2)`
  `x` `= −40 int (−2v)/(F – kv^2)\ dv`
    `= −40/k log_e (F – kv^2) + C`

 

`text(When)\ \ v = 10:`

`x_1 = −40/k log_e (F – 100k) + C`
 

`text(When)\ \ v = 20:`

`x_2 = −40/k log_e (F – 400k) + C`
 

`d` `= x_2 – x_1`
  `= −40/k log_e (F – 400k) + 40/k log_e (F – 100k)`
  `= 40/k log_e ((F – 100k)/(F – 400k))`

Mechanics, EXT2 M1 EQ-Bank 2

A particle with mass `m` moves horizontally against a resistance force `F`, equal to  `mv(1 + v^2)`  where `v` is the particle's velocity.

Initially, the particle is travelling in a positive direction from the origin at velocity `T`.

  1. Show that the particle's displacement from the origin, `x`, can be expressed as
     
         `x = tan^(-1)((T - v)/(1 + Tv))`  (2 marks)

     

  2. Show that the time, `t`, when the particle is travelling at velocity `v`, is given by
     
         `t = 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`  (4 marks)

     

  3. Express `v^2` as a function of `t`, and hence find the limiting values of `x` and `v`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)“
  3. `v -> 0, \ x -> tan^(−1)T`
Show Worked Solution

i.   `text(Newton’s 2nd law:)`

`F = mddotx = −mv(1 + v^2)`

`v · (dv)/(dx)` `= −v(1 + v^2)`
`(dv)/(dx)` `= −(1 + v^2)`
`(dx)/(dv)` `= −1/(1 + v^2)`
`x` `= −int 1/(1 + v^2)\ dv`
  `= −tan^(−1) v + C`

 

`text(When)\ \ x = 0, v = T:`

`C = tan^(−1)T`

`x = tan^(−1)T – tan^(−1)v`
 

`text(Let)\ \ x = A – B`

`A = tan^(−1) T \ => \ T = tan A`

`B = tan^(−1)v \ => \ v = tan B`

`tan x` `= tan(A – B)`
  `= (tan A – tan B)/(1 + tan A tan B)`
  `= (T – v)/(1 + Tv)`
`:. x` `= tan^(−1)((T – v)/(1 + Tv))`

 

ii.   `text(Using)\ \ ddotx = (dv)/(dt):`

`(dv)/(dt) = −1/(v(1 + v^2))`

`t = −int 1/(v(1 + v^2)) dv`
 

`text(Using Partial Fractions):`

`1/(v(1 + v^2)) = A/v + (Bv + C)/(1 + v^2)`

`A(1 + v^2) + (Bv + C)v = 1`

`A = 1`

`(A + B)v^2` `= 0`   `=> `    `B` `= −1`
`Cv` `= 0`   `=> `    `C` `= 0`

 

`t` `= −int 1/v\ dv + int v/(1 + v^2)\ dv`
  `= −log_e v + 1/2 int(2v)/(1 + v^2)\ dv`
  `= −log_e v + 1/2 log_e (1 + v^2) + C`
  `= −1/2 log_e v^2 + 1/2 log_e (1 + v^2) + C`
  `= 1/2 log_e ((1 + v^2)/(v^2)) + C`

 

`text(When)\ \ t = 0, v = T:`

`C = −1/2 log_e ((1 + T^2)/(T^2))`
 

`:. t` `= 1/2 log_e ((1 + v^2)/(v^2)) – 1/2 log_e((1 + T^2)/(T^2))`
  `= 1/2 log_e (((1 + v^2)/(v^2))/((1 + T^2)/(T^2)))`
  `= 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`

 

iii. `t` `= 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`
  `e^(2t)` `= (T^2(1 + v^2))/(v^2(1 + T^2))`
  `1 + v^2` `= (e^(2t)v^2(1 + T^2))/(T^2)`
  `1` `= v^2((e^(2t)(1 + T^2))/(T^2) – 1)`
  `1` `= v^2((e^(2t)(1 + T^2) – T^2)/(T^2))`
  `:. v^2` `= (T^2)/(e^(2t)(1 + T^2) – T^2)`

 
`text(As)\ \ t -> ∞:`

`v^2 -> 0, \ v -> 0`

`x = tan^(−1)T – tan^(−1)v`

`x -> tan^(−1)T`

Mechanics, EXT2 M1 EQ-Bank 1

A canon ball of mass 9 kilograms is dropped from the top of a castle at a height of `h` metres above the ground.

The canon ball experiences a resistance force due to air resistance equivalent to  `(v^2)/500`, where `v` is the speed of the canon ball in metres per second. Let  `g=9.8\ text(ms)^-2`  and the displacement, `x` metres at time `t` seconds, be measured in a downward direction.

  1. Show the equation of motion is given by
     
         `ddotx = g - (v^2)/4500`  (1 mark)

  2. Show, by integrating using partial fractions, that
     
         `v = 210((e^(7/75 t) - 1)/(e^(7/75 t) + 1))`  (5 marks)

     

  3. If the canon hits the ground after 4 seconds, calculate the height of the castle, to the nearest metre.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `78\ text(metres)`
Show Worked Solution

i.   `text(Newton’s 2nd Law:)`

`text(Net Force)` `= mddotx`
`mddotx` `= mg – (v^2)/500`
`9ddotx` `= 9g – (v^2)/500`
`ddotx` `= g – (v^2)/4500`

 

ii.    `(dv)/(dt)` `= g – (v^2)/4500`
    `= (44\ 100 – v^2)/4500`

 
`(dt)/(dv) = 4500/(44\ 100 – v^2)`
 

`text(Using Partial Fractions:)`

`1/(44\ 100 – v^2) = A/(210- v) + B/(210 – v)`

`A(210 – v) + B(210 + v) = 1`
 

`text(If)\ \ v = 210,`

`420B = 1 \ => \ B = 1/420`
 

`text(If)\ \ v = −210,`

`420A = 1 \ => \ A = 1/420`
 

`t` `= int 4500/(210^2 – v^2)\ dv`
  `= 4500/420 int 1/(210 + v) + 1/(210 – v)\ dv`
  `= 75/7 [ln(210 + v) – ln(210 – v)] + c`
  `= 75/7 ln((210 + v)/(210 – v)) + c`

 

`text(When)\ \ t = 0, v = 0:`

`0` `= 75/7 ln(210/210) + c`
`:. c` `= 0`

 

` t` `= 75/7 ln((210 + v)/(210 – v))`
`7/75 t` `= ln((210 + v)/(210 – v))`
`e^(7/75 t)` `= (210 + v)/(210 – v)`
`e^(7/75 t) (210 – v)` `= 210 + v`
`210e^(7/75 t) – 210` `= ve^(7/75 t) + v`
`210(e^(7/75 t) – 1)` `= v(e^(7/75 t) + 1)`
`:. v` `= 210((e^(7/75 t) – 1)/(e^(7/75 t) + 1))`

 

iii.    `v · (dv)/(dx)` `= (44\ 100 – v^2)/4500`
  `(dx)/(dv)` `= (4500v)/(44\ 100 – v^2)`
  `int (dx)/(dv)\ dv` `= −4500/2 int (−2v)/(44\ 100 – v^2)\ dv`
  `x` `= −2550 log_e(44\ 100 – v^2) + c`

 
`text(When)\ \ x = 0, v = 0:`

`0` `= −2250 log_e(44\ 100) + c`
`c` `= 2250 log_e(44\ 100)`
`x` `= −2250 log_e(44\ 100 – v^2) + 2250 log_e44\ 100`
  `= 2250 log_e((44\ 100)/(44\ 100 – v^2))`

 
`text(When)\ \ t = 4:`

`v` `= 210((e^(28/75) – 1)/(e^(28/75) + 1))`
  `= 38.7509…`

 

`:. h` `= 2250 log_e((44\ 100)/(44\ 100 – 38.751^2))`
  `= 77.94…`
  `= 78\ text(metres)`

Mechanics, EXT2 M1 2018 HSC 14b

A falling particle experiences forces due to gravity and air resistance. The acceleration of the particle is  `g - kv^2`, where `g` and `k` are positive constants and `v` is the speed of the particle. (Do NOT prove this.)

Prove that, after falling from rest through a distance, `h`, the speed of the particle will be  `sqrt(g/k (1 - e^(−2kh)))`.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`a = v · (dv)/(dx) = g – kv^2`

`(dv)/(dx)` `= (g – kv^2)/v`
`(dx)/(dv)` `= v/(g – kv^2)`
`x` `= int v/(g – kv^2)\ dv`
  `= −1/(2k) log_e(g – kv^2) + c`

 

`text(When)\ \ x = 0, v = 0`

`=> c = 1/(2k) log_e (g)`

`x` `= −1/(2k)log_e(g – kv^2) + 1/(2k) log_e g`
  `= −1/(2k) log_e ((g – kv^2)/g)`

 

`text(Find)\ \ v\ \ text(when)\ \ x = h:`

`log_e ((g – kv^2)/g)` `= −2kh`
`(g – kv^2)/g` `= e^(−2kh)`
`kv^2` `= g – g e^(−2kh)`
`v^2` `= g/k (1 – e^(−2kh))`
`:. v` `= sqrt(g/k (1 – e^(−2kh)))`

Mechanics, EXT2 M1 2017 HSC 13c

A particle is projected upwards from ground level with initial velocity  `1/2 sqrt(g/k)\ text(ms)^(-1)`, where `g` is the acceleration due to gravity and `k` is a positive constant. The particle moves through the air with speed  `v\ text(ms)^(-1)`  and experiences a resistive force.

The acceleration of the particle is given by  `ddot x = -g - kv^2\ text(ms)^(-2)`. Do NOT prove this.

The particle reaches a maximum height, `H`, before returning to the ground.

Using  `ddot x = v (dv)/(dx)`, or otherwise, show that  `H = 1/(2k) log_e (5/4)`  metres.  (4 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`ddot x = v · (dv)/(dx) = -g – kv^2`

`(dv)/(dx)` `= – (g + kv^2)/v`
`(dx)/(dv)` `= – v/(g + kv^2)`
`:. x` `= – int v/(g + kv^2)\ dv`
  `= -1/(2k) log_e (g + kv^2) + c`

 
`text(When)\ \ x = 0,\ \ v = 1/2 sqrt(g/k)`

`0` `= -1/(2k) log_e (g + k · g/(4k)) + c`
`:. c` `= 1/(2k) log_e ((5g)/4)`

 
`:. x = 1/(2k) log_e ((5g)/4) – 1/(2k) log_e (g + kv^2)`

 
`text{Max height}\ H\ text(occurs when)\ \ v = 0:`

`H` `= 1/(2k) log_e ((5g)/4) – 1/(2k) log_e g`
  `= 1/(2k) log_e ((5g)/(4g))`
  `= 1/(2k) log_e (5/4)\ text(… as required.)`

Mechanics, EXT2 M1 2015 HSC 15a

A particle  `A` of unit mass travels horizontally through a viscous medium. When  `t = 0`, the particle is at point  `O`  with initial speed  `u`. The resistance on particle  `A`  due to the medium is  `kv^2`, where  `v`  is the velocity of the particle at time  `t`  and  `k`  is a positive constant.

When  `t = 0`, a second particle  `B`  of equal mass is projected vertically upwards from  `O`  with the same initial speed  `u`  through the same medium. It experiences both a gravitational force and a resistance due to the medium. The resistance on particle  `B`  is  `kw^2`, where  `w`  is the velocity of the particle  `B`  at time  `t`. The acceleration due to gravity is  `g`.

  1. Show that the velocity  `v`  of particle  `A`  is given by  
     
         `1/v = kt + 1/u.`  (2 marks)

  2. By considering the velocity  `w`  of particle  `B`, show that
     
         `t = 1/sqrt(gk) (tan^-1(u sqrt(k/g)) - tan^-1 (w sqrt(k/g))).`  (3 marks)

  3. Show that the velocity  `V`  of particle  `A`  when particle  `B`  is at rest is given by
     
         `1/V = 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g)).`  (1 mark)

  4. Hence, if  `u`  is very large, explain why  
     
         `V ~~ 2/pi sqrt(g/k).`  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(See Worked Solutions)`
Show Worked Solution

i.   `text(Particle)\ A:`

`ddot x` `= -kv^2`
`(dv)/(dt)` `= -kv^2`
`(dt)/(dv)` `=- 1/(kv^2)`
`t` `= -1/k int 1/v^2\ dv`
`-kt` `= -1/v + c`
`text(When)\ \ t=0,\ \ v=u\ \ \ \ =>c=1/u`
`-kt` `= -1/v + 1/u`
`:.1/v` `= kt + 1/u`

 

ii.   `text(Particle)\ B:`

`ddot x` `= -g – kw^2`
`(dw)/(dt)` `= -g – kw^2`
`(dt)/(dw)`   `=-1/(g + kw^2)`
`t`   `= – int (dw)/(g + kw^2)`
   `= -1/k int (dw)/(g/k + w^2)`
   `= -1/k xx 1/sqrt (g/k) tan^-1(w/sqrt (g/k)) + c`
  `= -1/sqrt (gk)\ tan^-1 ((sqrt k w)/sqrt g) + c`

 

`text(When)\ \ t = 0,\ \ w = u`

`=>c= 1/sqrt (gk) tan^-1 ((sqrt k u)/sqrt g)`

`:. t` `= -1/sqrt (gk) tan^-1 ((sqrt k w)/sqrt g) + 1/sqrt (gk) tan^-1 ((sqrt k u)/sqrt g)`
  `=1/sqrt (gk) (tan^-1 (u sqrt(k/g)) – 1/sqrt (gk) tan^-1 (w sqrt (k/g)))`

 

iii.   `B\ \ text(at rest when)\ \ w = 0`

`t = 1/sqrt (gk) (tan^-1 (u sqrt (k/g)))`

`:.1/V` `= k xx 1/sqrt(gk) tan^-1 (u sqrt (k/g)) + 1/u,\ \ \ \ \ text{(part (i))}`
  `= 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g))`

 

iv.   `1/V = 1/u + sqrt (k/g) tan^-1 (u sqrt (k/g))`

`text(As)\ \ u -> oo,\ \ tan^-1 (u sqrt (k/g)) -> pi/2`

`:.\ text(If)\ \ u\ \ text(is very large,)`

`1/V` `~~ 0 + sqrt (k/g) xx pi/2`
`:.V` `~~ 2/pi sqrt (g/k)`

Mechanics, EXT2 M1 2007 HSC 6b

A raindrop falls vertically from a high cloud. The distance it has fallen is given by

`x = 5 log_e ((e^(1.4 t) + e^(-1.4 t))/2)`

where `x` is in metres and `t` is the time elapsed in seconds.

  1. Show that the velocity of the raindrop, `v` metres per second, is given by
     
         `v = 7 ((e^(1.4 t) - e^(-1.4 t))/(e^(1.4 t) + e^(-1.4 t)))`  (2 marks)

  2. Hence show that  `v^2 = 49 (1 - e^(-(2x)/5)).`  (2 marks)

  3. Hence, or otherwise, show that  `ddot x = 9.8 - 0.2v^2.`  (2 marks)

  4. The physical significance of the 9.8 in part (iii) is that it represents the acceleration due to gravity.

     

    What is the physical significance of the term  `–0.2 v^2?`   (1 mark)

  5. Estimate the velocity at which the raindrop hits the ground.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `-0.2v^2\ \ text(is the air resistance)`
  5. `7\ \ text(ms)^-1`
Show Worked Solution
i.    `x` `= 5 log_e ((e^(1.4t) + e^(-1.4t))/2)`
  `v=dx/dt` `= (5(1.4e^(1.4t) – 1.4e^(-1.4t)))/(e^(1.4t) + e^(-1.4t))`
    `= 7 ((e^(1.4 t) – e^(-1.4 t))/(e^(1.4 t) + e^(-1.4 t)))`

 

ii.    `v^2` `=49 ((e^(1.4 t) – e^(-1.4 t))/(e^(1.4 t) + e^(-1.4 t)))^2`
    `=49 ((e^(2.8 t) + e^(-2.8 t)-2)/(e^(2.8 t) + e^(-2.8 t)+2))`
    `=49 ((e^(2.8 t) + e^(-2.8 t)+2-4)/(e^(2.8 t) + e^(-2.8 t)+2))`
    `=49 (((e^(1.4t) + e^(-1.4t))^2-2^2)/((e^(1.4t) + e^(-1.4t))^2))`
    `=49 (1 – (2/(e^(1.4t) + e^(-1.4t)))^2)`

 

`text(S)text(ince)\ \ x` `= 5 log_e ((e^(1.4 t) + e^(-1.4 t))/2)`
`e^(x/5)` `=(e^(1.4 t) + e^(-1.4 t))/2`

 

`:. v^2` `=49 (1 – (e^(- x/5))^2)`
  `=49(1-e^(- (2x)/5))`

 

iii.   `ddotx` `=d/(dx) (1/2 v^2)`
    `=49/2 xx  2/5 xx e^(-(2x)/5)`
    `=49/5 e^(-(2x)/5)`
    `=49/5 (1- v^2/49),\ \ \ \ text{(from part (ii))}`
    `=9.8 – 0.2v^2`

 
iv.  `-0.2v^2\ \ text(is the wind resistance acting on the rain drop.)`
 

v.   `text(Terminal velocity occurs when)\ \ ddot x=0`

`9.8 – 0.2v^2` `=0`
`v^2` `=49`
`v` `=7,\ \ \ \ (v > 0)`

 
`:.\ text(The rain drop hits the ground travelling at)\ \ 7\ \ text(ms)^-1`

Mechanics, EXT2 M1 2011 HSC 6a

Jac jumps out of an aeroplane and falls vertically. His velocity at time  `t`  after his parachute is opened is given by  `v(t)`, where  `v(0) = v_0`  and  `v(t)`  is positive in the downwards direction. The magnitude of the resistive force provided by the parachute is  `kv^2`, where  `k`  is a positive constant. Let  `m`  be Jac’s mass and  `g`  the acceleration due to gravity. Jac’s terminal velocity with the parachute open is  `v_T.`

Jac’s equation of motion with the parachute open is

`m (dv)/(dt) = mg - kv^2.`   (Do NOT prove this.)

  1. Explain why Jac’s terminal velocity  `v_T`  is given by  
     
         `sqrt ((mg)/k).`  (1 mark)

  2. By integrating the equation of motion, show that  `t`  and  `v`  are related by the equation
     
         `t = (v_T)/(2g) ln[((v_T + v)(v_T - v_0))/((v_T - v)(v_T + v_0))].`  (3 marks)

  3. Jac’s friend Gil also jumps out of the aeroplane and falls vertically. Jac and Gil have the same mass and identical parachutes.

     

    Jac opens his parachute when his speed is  `1/3 v_T.` Gil opens her parachute when her speed is  `3v_T.` Jac’s speed increases and Gil’s speed decreases, both towards  `v_T.`

     

    Show that in the time taken for Jac's speed to double, Gil's speed has halved.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `m (dv)/(dt) = mg – kv^2`

`text(As)\ \ v ->text(terminal velocity,)\ \ (dv)/(dt) -> 0`

`mg – kv_T^2` `= 0`
`v_T^2` `= (mg)/k`
`v_T` `= sqrt ((mg)/k)`

  

ii.  `m (dv)/(dt) = mg – kv^2`

`int_0^t dt` `=int_(v_0)^v m/(mg – kv^2)\ dv`
`t` `= m/k int_(v_0)^v (dv)/((mg)/k – v^2)\ \ \ \ text{(using}\ \ v_T^2= (mg)/k text{)}`
  `= (v_T^2)/g int_(v_0)^v (dv)/(v_T­^2 – v^2)`

  

♦ Mean mark part (ii) 39%.

`text(If)\ \ v_0 < v_T,\ \ text(then)\ \ v < v_T\ \ text(at all times.)`

`:. t` `=(v_T^2)/g int_(v_0)^v (dv)/(v_T­^2 – v^2)`
  `=(v_T^2)/g int_(v_0)^v (dv)/((v_T – v)(v_T + v))`
  `=(v_T^2)/(2 g v_T) int_(v_0)^v (1/((v_T – v)) + 1/((v_T + v))) dv`
  `=v_T/(2g)[-ln (v_T – v) + ln (v_T + v)]_(v_0)^v`
  `=v_T/(2g)[ln(v_T + v)-ln(v_T – v)-ln(v_T + v_0)+ln(v_T – v_0)]`
  `=v_T/(2g) ln[((v_T + v)(v_T – v_0))/((v_T – v)(v_T + v_0))]`

 

`text(If)\ \ v_0 > v_T,\ \ text(then)\ \ v > v_T\ \ text(at all times.)`

`text(Replace)\ \ v_T – v\ \ text(by) – (v – v_T)\ \ text(in the preceeding)`

`text(calculation, leading to the same result.)`

 

iii.  `text{From (ii) we have}:\ \ t = v_T/(2g) ln [((v_T + v)(v_T – v_0))/((v_T – v)(v_T + v_0))]`

`text(For Jac,)\ \ v_0 = V_T/3\ \ text(and we have to find the time)`

`text(taken for his speed to be)\ \ v = (2v_T)/3.`

`t` `=v_T/(2g) ln[((v_T + (2v_T)/3)(v_T – v_T/3))/((v_T – (2v_T)/3)(v_T + v_T/3))]`
  `=v_T/(2g) ln [(((5v_T)/3)((2v_T)/3))/((v_T/3)((4v_T)/3))]`
  `=v_T/(2g) ln[(10/9)/(4/9)]`
  `=v_T/(2g) ln (5/2)`

 

`text(For Gil),\ \ v_0 = 3v_T\ \ text(and we have to find)` 

♦♦♦ Mean mark part (iii) 16%.

`text(the time taken for her speed to be)\ \ v = (3v_T)/2.`

`t` `=v_T/(2g) ln [((v_T + (3v_T)/2)(v_T – 3v_T))/((v_T – (3v_T)/2)(v_T + 3v_T))]`
  `=v_T/(2g) ln [(((5v_T)/2)(-2v_T))/((-v_T/2)(4v_T))]`
  `=v_T/(2g) ln [(-5)/-2]`
  `=v_T/(2g) ln (5/2)`

 

`:.\ text(The time taken for Jac’s speed to double is)`

`text(the same as it takes for Gil’s speed to halve.)`

Mechanics, EXT2 M1 2012 HSC 13a

An object on the surface of a liquid is released at time  `t = 0`  and immediately sinks. Let  `x`  be its displacement in metres in a downward direction from the surface at time  `t`  seconds.

The equation of motion is given by

`(dv)/(dt) = 10 − (v^2)/40`,

where  `v`  is the velocity of the object.  

  1. Show that  `v = (20(e^t − 1))/(e^t + 1)`.   (4 marks)

  2. Use  `(dv)/(dt) = v (dv)/(dx)`  to show that  
     
         `x = 20\ log_e(400/(400 − v^2))`   (2 marks)

     

  3. How far does the object sink in the first 4 seconds?   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `text(See Worked Solutions.)`
  3. `40log_e((e^4 + 1)/(2e^2))\ text(m)`
Show Worked Solution
i.   `(dv)/(dt)` `= 10 − (v^2)/40`
  `(dv)/(dt)` `= (400 − v^2)/40`
  `dt` `= 40/(400 −v^2)\ dv`
  `int dt` `=int 40/(400 −v^2)\ dv`
  `t` `= int (1/(20 + v) + 1/(20 − v))\ dv`
    `= log_e(20 + v) − log_e(20 − v) + c`

 

`text(When)\ \ t = 0, v = 0\ \ => \ c = 0`

`t=` ` log_e((20 + v)/(20 − v))`
`e^t=` ` (20 + v)/(20 − v)`
`20e^t-ve^t=` ` 20 + v`
`v+ ve^t=` ` 20e^t − 20`
`v(1+e^t)=` `20(e^t − 1)`
`v=` ` (20(e^t − 1))/(e^t + 1)\ \ \ \ text(… as required)`

 

ii.   `v (dv)/(dx)` `= 10 − (v^2)/40`
  `(40v\ dv)/(400 − v^2)` `= dx`
`int dx` `= int (40v)/(400 − v^2)\ dv`
 `x` `= -20log_e(400 − v^2) + c`

 

`text(When)\ \ x = 0, v = 0\ \ \ => c = 20log_e 400`

`:.x` `= 20log_e400 − 20log_e(400 − v^2)`
  `= 20log_e((400)/(400 − v^2))\ \ \ \ text(… as required)`

 

iii.  `text(When)\ \ t = 4,\  v = (20(e^4 − 1))/(e^4 + 1)`

`x` `= 20log_e[400/(400 −((20(e^4 − 1))/(e^4 + 1))^2)]`
  `= 20log_e[((e^4 + 1)^2)/((e^4 + 1)^2 − (e^4 − 1)^2)]`
  `= 20log_e(((e^4 + 1)^2)/((e^4+1 + e^4 − 1)(e^4 + 1 − e^4 + 1)))`
  `= 20log_e(((e^4 + 1)^2)/(4e^4))`
  `= 40log_e\ (e^4 + 1)/(2e^2)\ \ text(metres)`

Mechanics, EXT2 M1 2013 HSC 15d

A ball of mass `m` is projected vertically into the air from the ground with initial velocity  `u`. After reaching the maximum height `H` it falls back to the ground. While in the air, the ball experiences a resistive force `kv^2`, where `v` is the velocity of the ball and `k` is a constant.

The equation of motion when the ball falls can be written as

`m dot v = mg-kv^2.`      (Do NOT prove this.)

  1. Show that the terminal velocity  `v_T` of the ball when it falls is
  2.    `sqrt ((mg)/k).`  (1 mark)

  3. Show that when the ball goes up, the maximum height  `H`  is
  4.    `H = (v_T^2)/(2g) ln (1 + u^2/(v_T^2)).`  (3 marks)

  5. When the ball falls from height  `H`  it hits the ground with velocity  `w`.
  6. Show that  `1/w^2 = 1/u^2 + 1/(v_T^2).`  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `m dot v = mg-kv^2`

`t = 0,\ \ \ v = 0,\ \ \ x = 0\ \ \ text(Ball falling)`

`text{For terminal velocity}\ \(v_T),\ \ \  dot v = 0`

`v_T^2` `= (mg)/k`
`:.v_T` `= sqrt ((mg)/k)`

 

ii.  `text(When the ball rises),\ \ m dot v = -mg-kv^2`

♦ Mean mark 43%.

MARKER’S COMMENT: More than half of students incorrectly wrote the equation to solve as `m dot v=mg-kv^2!`
`text(Using)\ \ dot v` `= v (dv)/(dx)`
`mv (dv)/(dx)` `= -mg-kv^2`
`dx` `=(-mv)/(mg + kv^2) dv` 
`int_0^H dx` `= -int_u^0 (mv)/(mg + kv^2) dv`
`[x]_0^H` `= -m/(2k) [log_e (mg + kv^2)]_u^0`
`H` `= -m/(2k) (log_e (mg)-log_e (mg + ku^2))`
  `= m/(2k) log_e ((mg + ku^2)/(mg))`
  `= m/(2k) log_e (1 + (ku^2)/(mg))\ \ \ \ text{(using}\ \ k = (mg)/v_T^2 text{)}`
`:.H` `=(v_T^2)/(2g) log_e (1 + u^2/(v_T^2))`

 

 

iii.  `text(When the ball falls),\ \ m dot v = mg-kv^2`

♦♦ Mean mark 14%.
`mv (dv)/(dx)` `= mg-kv^2`
`dx` `=(mv)/(mg-kv^2) dv` 

 

`int_0^H dx` `= int_0^w (mv)/(mg-kv^2)dv`
`[x]_0^H` ` =-m/(2k)[log_e (mg-kv^2)]_0^w`
`H` `= -m/(2k) (log_e (mg-kw^2)-log_e (mg))`
  `= -m/(2k) log_e ((mg-kw^2)/(mg))`
  `= -m/(2k) log_e (1-(kw^2)/(mg))\ \ \ \ text{(using}\ \ k = (mg)/v_T^2 text{)}`
`H` `=-(v_T^2)/(2g) log_e (1-w^2/(v_T^2))`
  `=-(v_T^2)/(2g) log_e ((v_T^2-w^2)/(v_T^2))`
  `=(v_T^2)/(2g) log_e ((v_T^2)/(v_T^2-w^2))`

 

`text{Using part (ii):}`

`(v_T^2)/(2g) log_e ((v_T^2)/(v_T^2-w^2))` `=(v_T^2)/(2g) log_e (1 + u^2/(v_T^2))`
`(v_T^2)/(v_T^2-w^2)` `=(v_T^2 + u^2)/(v_T^2)`
`(v_T^2 + u^2)(v_T^2-w^2)` `=v_T^4`
`v_T^4-v_T^2 w^2+v_T^2 u^2-w^2 u^2` `=v_T^4`
`v_T^2 w^2+w^2 u^2` `=v_T^2 u^2`
`(v_T^2 w^2)/(v_T^2w^2u^2)+(w^2 u^2)/(v_T^2w^2u^2)` `=(v_T^2 u^2)/(v_T^2w^2u^2)`
`:.1/u^2 + 1/(v_T^2)` `=1/w^2`

Mechanics, EXT2 M1 2014 HSC 14c

A high speed train of mass `m` starts from rest and moves along a straight track. At time `t` hours, the distance travelled by the train from its starting point is `x` km, and its velocity is `v` km/h.

The train is driven by a constant force  `F`  in the forward direction. The resistive force in the opposite direction is  `Kv^2`, where  `K`  is a positive constant. The terminal velocity of the train is 300 km/h.

  1. Show that the equation of motion for the train is
     
         `m ddot x = F[1 − (v/300)^2]`.  (2 marks)

  2. Find, in terms of  `F`  and  `m`, the time it takes the train to reach a velocity of 200 km/h.  (4 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `(150 m ln5)/F`
Show Worked Solution

i.   `m ddot x = F – Kv^2,\ \ v_T = 300`

`text(At)\ \ v_T,\ \ ddotx=0`

`m(0)` `=F-K(300^2)`
`=>K` `=F/300^2`
`:. m ddot x` `= F – F/300^2 v^2`
  `=F[1 − (v/300)^2]\ \ \ text(… as required)`

 

ii. `m*(dv)/(dt)` `= F[1 − (v/(300))^2]`
  `(dv)/(1-(v/300)^2)` `=F/m\ dt`
  `(dv)/(300^2-v^2)` `=F/(300^2 m)\ dt`
  `dt` `=(300^2 m)/F xx (dv)/(300^2-v^2)`

 

`int_0^t dt` `=(300 m)/F  int_0^200  300/((300+v)(300-v))\ dv` 
 `:. t` `=(300 m)/F  int_0^200  (1/2)/(300+v) + (1/2)/(300-v)\ dv`
  `=(150 m)/F [ln(300+v) – ln(300-v)]_0^200`
  `=(150 m)/F [ln ((300+v)/(300-v))]_0^200`
  `=(150 m)/F  (ln5 – ln1)`
  `=(150 m ln5)/F`