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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 2023 HSC 13c

A particle of mass 1 kg is projected from the origin with speed 40 m s\( ^{-1}\) at an angle 30° to the horizontal plane.

  1. Use the information above to show that the initial velocity of the particle is
  2.     \(\mathbf{v}(0)={\displaystyle\left(\begin{array}{cc}20 \sqrt{3} \\ 20\end{array}\right)} \).   (1 mark)

The forces acting on the particle are gravity and air resistance. The air resistance is proportional to the velocity vector with a constant of proportionality 4 . Let the acceleration due to gravity be 10 m s \( ^{-2}\).

The position vector of the particle, at time \(t\) seconds after the particle is projected, is \(\mathbf{r}(t)\) and the velocity vector is \(\mathbf{v}(t)\).
 

  1. Show that  \(\mathbf{v}(t)={\displaystyle \left(\begin{array}{cc}20 \sqrt{3} e^{-4 t} \\ \dfrac{45}{2} e^{-4 t}-\dfrac{5}{2}\end{array}\right)}\)  (3 marks)

  2. Show that  \(\mathbf{r}(t)=\left(\begin{array}{c}5 \sqrt{3}\left(1-e^{-4 t}\right) \\ \dfrac{45}{8}\left(1-e^{-4 t}\right)-\dfrac{5}{2} t\end{array}\right)\)  (2 marks)

  3. The graphs  \(y=1-e^{-4 x}\)  and  \(y=\dfrac{4 x}{9}\) are given in the diagram below.
     
     
  4. Using the diagram, find the horizontal range of the particle, giving your answer rounded to one decimal place.  (2 marks)

Show Answers Only

  1. \(\text{Proof (See Worked Solutions)}\)
  2. \(\text{Proof (See Worked Solutions)}\)
  3. \(\text{Proof (See Worked Solutions)}\)
  4. \(8.7\ \text{metres}\)

Show Worked Solution

i. 

\(\underset{\sim}{v}(0)={\displaystyle\left(\begin{array}{cc} 40 \cos\ 30° \\ 40 \sin\ 30°\end{array}\right)} = {\displaystyle\left(\begin{array}{cc} 40 \times \frac{\sqrt3}{2} \\ 40 \times \frac{1}{2}\end{array}\right)}  = {\displaystyle\left(\begin{array}{cc}20 \sqrt{3} \\ 20\end{array}\right)} \)
 

ii.   \(\text{Air resistance:} \)

\(\underset{\sim}{F} = -4\underset{\sim}{v} = {\displaystyle\left(\begin{array}{cc} -4\dot{x} \\ -4\dot{y} \end{array}\right)} \)

\(\text{Horizontally:}\)

\(1 \times \ddot{x} \) \(=-4 \dot{x} \)  
\(\dfrac{d\dot{x}}{dt}\) \(=-4\dot{x}\)  
\(\dfrac{dt}{d\dot{x}}\) \(= -\dfrac{1}{4\dot{x}} \)  
\(t\) \(=-\dfrac{1}{4} \displaystyle \int \dfrac{1}{\dot{x}} \ d\dot{x} \)  
\(-4t\) \(=\ln |\dot{x}|+c \)  

 
\(\text{When}\ \ t=0, \ \dot{x}=20\sqrt3 \ \ \Rightarrow\ \ c=-\ln{20\sqrt3} \)

\(-4t\) \(=\ln|\dot{x}|-\ln 20\sqrt3 \)  
\(-4t\) \(=\ln\Bigg{|}\dfrac{\dot{x}}{20\sqrt{3}} \Bigg{|} \)  
\(\dfrac{\dot{x}}{20\sqrt{3}} \) \(=e^{-4t} \)  
\(\dot{x}\) \(=20\sqrt{3}e^{-4t}\)  

 
\(\text{Vertically:} \)

\(1 \times \ddot{y} \) \(=-1 \times 10-4 \dot{y} \)  
\(\dfrac{d\dot{y}}{dt}\) \(=-(10+4\dot{y})\)  
\(\dfrac{dt}{d\dot{y}}\) \(= -\dfrac{1}{10+4\dot{y}} \)  
\(t\) \(=- \displaystyle \int \dfrac{1}{10+4\dot{y}} \ d\dot{y} \)  
\(-4t\) \(=- \displaystyle \int \dfrac{4}{10+4\dot{y}} \ d\dot{y} \)  
\(-4t\) \(=\ln |10+4\dot{y}|+c \)  

 
\(\text{When}\ \ t=0, \ \dot{y}=20 \ \ \Rightarrow\ \ c=-\ln{90} \)

\(-4t\) \(=\ln|10+4\dot{y}|-\ln 90 \)  
\(-4t\) \(=\ln\Bigg{|}\dfrac{10+4\dot{y}}{\ln{90}} \Bigg{|} \)  
\(\dfrac{10+4\dot{y}}{90} \) \(=e^{-4t} \)  
\(4\dot{y}\) \(=90e^{-4t}-10\)  
\(\dot{y}\) \(=\dfrac{45}{2} e^{-4t}-\dfrac{5}{2} \)  

 
\(\therefore \underset{\sim}v={\displaystyle \left(\begin{array}{cc}20 \sqrt{3} e^{-4 t} \\ \dfrac{45}{2} e^{-4 t}-\dfrac{5}{2}\end{array}\right)}\) 

 
iii.   \(\text{Horizontally:}\)

\(x\) \(= \displaystyle \int \dot{x}\ dx\)  
  \(= \displaystyle \int 20\sqrt3 e^{-4t}\ dt \)  
  \(=-5\sqrt3 e^{-4t}+c \)  

 
\(\text{When}\ \ t=0, \ x=0\ \ \Rightarrow\ \ c=5\sqrt3 \)

\(x\) \(=5\sqrt3-5\sqrt3 e^{-4t} \)  
  \(=5\sqrt3(1-e^{-4t}) \)  

 
\(\text{Vertically:}\)

\(y\) \(= \displaystyle \int \dot{y}\ dx\)  
  \(= \displaystyle \int \dfrac{45}{2} e^{-4t}-\dfrac{5}{2}\ dt \)  
  \(=-\dfrac{45}{8}e^{-4t}-\dfrac{5}{2}t+c \)  

 
\(\text{When}\ \ t=0, \ y=0\ \ \Rightarrow\ \ c= \dfrac{45}{8} \)

\(y\) \(=\dfrac{45}{8}-\dfrac{45}{8} e^{-4t}-\dfrac{5}{2}t \)  
  \(=\dfrac{45}{8}(1-e^{-4t})-\dfrac{5}{2} \)  

 
\(\therefore \underset{\sim}{r}=\left(\begin{array}{c}5 \sqrt{3}\left(1-e^{-4 t}\right) \\ \dfrac{45}{8}\left(1-e^{-4 t}\right)-\dfrac{5}{2} t\end{array}\right)\)
 

iv.   \(\text{Range}\ \Rightarrow\ \text{Find}\ \ t\ \ \text{when}\ \ y=0: \)

\(\dfrac{45}{8}(1-e^{-4t})-\dfrac{5}{2}t \) \(=0\)  
\(\dfrac{45}{8}(1-e^{-4t}) \) \(=\dfrac{5}{2}t \)  
\(1-e^{-4t}\) \(=\dfrac{4}{9}t \)  

 
\(\text{Graph shows intersection of these two graphs.}\)

\(\Rightarrow \text{Solution when}\ \ t\approx 2.25\)

\(\therefore\ \text{Range}\) \(=5\sqrt3(1-e^{(-4 \times 2.25)}) \)  
  \(=8.659…\)  
  \(=8.7\ \text{metres (to 1 d.p.)}\)  

♦ Mean mark (iv) 43%.
 

Mechanics, EXT2 M1 2022 HSC 16b

A projectile of mass `M` kg is launched vertically upwards from a horizontal plane with initial speed `v_0\ text{m s}^(-1)` which is less than 100`\ text{m s}^(-1)`. The projectile experiences a resistive force which has magnitude `0.1 M v` newtons, where `v\ text{m s}^(-1)` is the speed of the projectile. The acceleration due to gravity is 10`\ text{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.  (4 marks)

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`39.1\ text{m s}^(-1)`

Show Worked Solution

`text{Motion is only vertical.}`

`Mddoty=M(-g-0.1v)`

`(dv)/(dt)` `=-(g+0.1v)`  
`(dt)/(dv)` `=- 1/(g+0.1v)`  
`t` `=-int1/(g+0.1v) dv`  
  `=-10ln abs(g+0.1v)+c`  

 
`text{When}\ \ t=0, \ v=v_0`

`=>c=10ln abs(g+0.1v_0)`
 


♦ Mean mark 45%.
`t` `=10ln abs(g+0.1v_0)-10ln abs(g+0.1v)`  
  `=10ln abs((g+0.1v_0)/(g+0.1v))`  
`e^(t/10)` `=(g+0.1v_0)/(g+0.1v)`  
`(g+0.1v)` `=(g+0.1v_0)*e^(- t/10)\ \ \ (text{note}\ (g+0.1v)>0\ text{as}\ v<100)`  
`0.1v` `=-g+(g+0.1v_0)*e^(- t/10)`  
`v` `=10[-g+(g+0.1v_0)*e^(- t/10)]`  

 
`y=intv\ dt=10[-g t-10(g+0.1v_0)*e^(- t/10)]+c`

`text{When}\ \ t=0, \ y=0:`

`=> \ c=100(g+0.1v_0)`

`:.y=10[-g t-10(g+0.1v_0)*e^(- t/10)]+100(g+0.1v_0)`
 

`text{Find}\ v_0,\ text{given}\ y=0\ text{at}\ t=7:`

`0` `=10[-70-10(10+0.1v_0)*e^(-0.7)]+100(10+0.1v_0)`  
`0` `=-700-100(10+0.1v_0)*e^(-0.7)+1000+10v_0`  
`0` `=-700-1000e^(-0.7)-10v_0e^(-0.7)+1000+10v_0`  
`0` `=300-1000e^(-0.7)+v_0(10-10e^(-0.7))`  
`:.v_0` `=(1000e^(-0.7)-300)/(10-10e^(-0.7))`  
  `=39.0503…`  
  `=39.1\ text{m s}^(-1)\ \ text{(to 1 d.p.)}`  

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 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)

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`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 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 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`